ML Aggarwal Class 9 Maths Solutions Chapter 18 Trigonometric Ratios of Standard Angles

 

Question 1. Find the coordinates of points whose
(i) abscissa is 3 and ordinate -4.
(ii) abscissa is -2/3 and ordinate 5.
(iii) whose abscissa is -1 2/3 and ordinate -2 1/4.
(iv) whose ordinate is 5 and abscissa is -2.
(v) whose abscissa is -2 and lies on x-axis.
(vi) whose ordinate is 3/2 and lies on y-axis.
Answer: The x-coordinate is called the abscissa, and the y-coordinate is known as the ordinate of any point.
(i) When the abscissa is 3 and ordinate is -4, the coordinates are (3, -4).
(ii) When the abscissa is -2/3 and ordinate is 5, the coordinates are (-2/3, 5).
(iii) When the abscissa is -1 2/3 and ordinate is -2 1/4, the coordinates are (-1 2/3, -2 1/4).
(iv) When the ordinate is 5 and abscissa is -2, the coordinates are (-2, 5).
(v) A point on the x-axis has an ordinate equal to 0. So the coordinates are (-2, 0).
(vi) A point on the y-axis has an abscissa equal to 0. So the coordinates are (0, 3/2).
In simple words: The first number in a coordinate pair is how far left or right the point is. The second number shows how far up or down it is.

Exam Tip: Remember that points on the x-axis always have y-coordinate 0, and points on the y-axis always have x-coordinate 0. Never swap the order of the coordinates.

 

Question 2. In which quadrant or on which axis each of the following points lie?
(-3, 5), (4, -1) (2, 0), (2, 2), (-3, -6)
Answer: When both the x-coordinate and y-coordinate are positive, the point is in the first quadrant. When the x-coordinate is negative and y-coordinate is positive, it lies in the second quadrant. When both coordinates are negative, the point is in the third quadrant. When the x-coordinate is positive and y-coordinate is negative, it falls in the fourth quadrant. Points on the axes have at least one coordinate equal to 0.

(-3, 5) - The x-coordinate is negative and y-coordinate is positive, so it lies in the second quadrant.
(4, -1) - The x-coordinate is positive and y-coordinate is negative, so it lies in the fourth quadrant.
(2, 0) - The y-coordinate is 0, so it lies on the x-axis.
(2, 2) - Both coordinates are positive, so it lies in the first quadrant.
(-3, -6) - Both coordinates are negative, so it lies in the third quadrant.
In simple words: Look at the signs of the x and y values. If both are positive, it is in quadrant 1. If x is negative and y is positive, it is in quadrant 2. If both are negative, it is in quadrant 3. If x is positive and y is negative, it is in quadrant 4. If one value is 0, the point sits on an axis.

Exam Tip: Always check whether either coordinate is zero first - if it is, the point is on an axis, not in any quadrant.

 

Question 3. Which of the following points lie on (i) x-axis? (ii) y-axis?
A(0, 2), B(5, 6), C(23, 0), D(0, 23), E(0, -4), F(-6, 0), G(\(\sqrt{3}\), 0).
Answer: A point sits on the x-axis whenever its y-coordinate equals 0. A point lies on the y-axis whenever its x-coordinate equals 0.

(i) Points with y-coordinate = 0 lie on the x-axis: C(23, 0), F(-6, 0), and G(\(\sqrt{3}\), 0).

(ii) Points with x-coordinate = 0 lie on the y-axis: A(0, 2), D(0, 23), and E(0, -4).
In simple words: If the first number is 0, the point is on the y-axis. If the second number is 0, the point is on the x-axis.

Exam Tip: The axes themselves are made up of points where one coordinate is always zero. Write out all qualifying points in your answer - missing even one costs marks.

 

Question 4. Plot the following points on the graph paper:
A(3, 4), B(-3, 1), C(1, -2), D(-2, -3), E(0, 5), F(5, 0), G(0, -3), H(-3, 0).
Answer: To plot points on a graph, mark the x-coordinate along the horizontal axis and the y-coordinate along the vertical axis. The intersection of the vertical line from the x-value and the horizontal line from the y-value gives the location of the point. Plot each of the 8 points as described: A in the first quadrant, B in the second quadrant, C in the fourth quadrant, D in the third quadrant, E on the positive y-axis, F on the positive x-axis, G on the negative y-axis, and H on the negative x-axis.
In simple words: Start at the origin. Move right or left by the first number. Then move up or down by the second number. Mark that spot.

Exam Tip: Use a sharp pencil and a ruler to plot accurately. Mark each point with a small dot and label it clearly with the given letter. Always use graph paper with clear squares to ensure precision.

 

Question 5. Write the co-ordinates of the points A, B, C, D, E, F, G and H shown in the adjacent figure.
Answer: Reading the coordinates directly from the plotted graph:

In simple words: To find a point's coordinates, draw vertical and horizontal lines from the point to both axes. The number where the vertical line hits the x-axis is the first coordinate. The number where the horizontal line hits the y-axis is the second coordinate.

Exam Tip: Use a ruler to draw faint vertical and horizontal lines from each point to the axes. This helps you read the coordinates more accurately. Double-check that each coordinate lies on the correct side (positive or negative) of the origin.

 

Question 6. In which quadrants are the points A, B, C and D of problem 5 located?
Answer: For point A(2, 2): Both the x-coordinate and y-coordinate are positive, placing it in the first quadrant. For point B(-3, 0): The y-coordinate is 0, so this point sits on the x-axis, not in any quadrant. For point C(-2, -4): Both coordinates are negative, placing it in the third quadrant. For point D(3, -1): The x-coordinate is positive and the y-coordinate is negative, placing it in the fourth quadrant.
In simple words: A is in quadrant 1 (both positive). B is on the x-axis (y is zero). C is in quadrant 3 (both negative). D is in quadrant 4 (x positive, y negative).

Exam Tip: Remember that points on the axes are not counted as being in any quadrant - they are on the boundary. The four quadrants are strictly in the interior regions.

 

Question 7. Plot the following points on the same graph paper:
A(2, 5/2), B(-3/2, 3), C(1/2, -3/2) and D(-5/2, -1/2).
Answer: Plot each point by locating its x-coordinate on the horizontal axis and y-coordinate on the vertical axis. For fractional coordinates, divide the unit squares into smaller parts. A(2, 5/2) is located 2 units to the right and 2.5 units up. B(-3/2, 3) is located 1.5 units to the left and 3 units up. C(1/2, -3/2) is located 0.5 units to the right and 1.5 units down. D(-5/2, -1/2) is located 2.5 units to the left and 0.5 units down. All four points should be clearly marked and labeled on the graph.
In simple words: To plot fractional coordinates, imagine cutting the squares in half or in smaller pieces. Then count carefully to find where each point goes.

Exam Tip: When dealing with fractions, convert them to decimals first if that helps you visualize the locations better. Mark each fractional point with precision using a ruler or by carefully counting grid divisions.

 

Question 8. Plot the following points on the same graph paper.
A(4/3, -1), B(7/2, 5/3), C(13/6, 0) and D(-5/3, -5/2).
Answer: Convert the fractional coordinates to decimal form for easier plotting: A(1.33, -1), B(3.5, 1.67), C(2.17, 0), D(-1.67, -2.5). Locate each point on the graph by finding its x-coordinate on the horizontal axis and y-coordinate on the vertical axis. Mark A in the fourth quadrant, B in the first quadrant, C on the positive x-axis, and D in the third quadrant. Use fine grid divisions or a ruler to ensure accurate placement of the fractional points.
In simple words: Convert fractions to decimals. Then plot them the same way you plot whole numbers - move along the x-axis, then move up or down along the y-axis.

Exam Tip: If fractional coordinates seem difficult, work with a finer grid paper that has smaller divisions. Double-check your decimal conversions before plotting, as errors there will lead to wrong point positions.

 

Question 9. Plot the following points and check whether they are collinear or not:
(i) (1, 3), (-1, -1) and (-2, -3)
(ii) (1, 2), (2, -1) and (-1, 4)
(iii) (0, 1), (2, -2) and (2/3, 0).
Answer:
(i) Plot the three points (1, 3), (-1, -1), and (-2, -3) on graph paper. When you join them, you will see that they all lie on a single straight line. These points are collinear.

(ii) Plot the three points (1, 2), (2, -1), and (-1, 4) on graph paper. When you attempt to join them, the points do not form a straight line - they fall in different positions that cannot be connected by one line. These points are non-collinear.

(iii) Plot the three points (0, 1), (2, -2), and (2/3, 0) on graph paper. When you join them, all three points fall on the same straight line. These points are collinear.
In simple words: Collinear means the points sit on the same line. If you can draw one straight line through all the points, they are collinear. If you cannot, they are not.

Exam Tip: Plot all points accurately first, then use a ruler to check if a straight line can pass through all of them. Do not just estimate - use the ruler carefully to verify your answer.

 

Question 10. Plot the point P(-3, 4). Draw PM and PN perpendiculars to x-axis and y-axis respectively. State the co-ordinates of the points M and N.
Answer: First, plot the point P(-3, 4) on the graph. The perpendicular from P to the x-axis is a vertical line dropped straight down. This line meets the x-axis at point M. Since any point on the x-axis has a y-coordinate of 0, and M lies directly below P, the x-coordinate of M is the same as P's x-coordinate. Therefore, M = (-3, 0). The perpendicular from P to the y-axis is a horizontal line drawn to the left or right. This line meets the y-axis at point N. Since any point on the y-axis has an x-coordinate of 0, and N lies directly across from P, the y-coordinate of N is the same as P's y-coordinate. Therefore, N = (0, 4).
In simple words: Drop a vertical line from P down to the x-axis - that point is M. Draw a horizontal line from P across to the y-axis - that point is N. The x-coordinate of P stays the same for M. The y-coordinate of P stays the same for N.

Exam Tip: A perpendicular from any point to the x-axis will always produce a point with y-coordinate 0. A perpendicular to the y-axis will always produce a point with x-coordinate 0. Use this rule to find M and N quickly.

 

Question 11. Plot the points A(1, 2), B(-4, 2), C(-4, -1) and D(1, -1). What kind of quadrilateral is ABCD? Also find the area of the quadrilateral ABCD.
Answer: Plot all four points on the graph and connect them in order A → B → C → D → A. When you examine the figure, you notice that sides AB and DC are both horizontal and parallel to each other. Similarly, sides AD and BC are both vertical and parallel to each other. This means ABCD is a rectangle. To find the area, measure the lengths of two adjacent sides. Using the graph grid, count the units: AB spans from x = -4 to x = 1, giving a length of 5 units. AD spans from y = -1 to y = 2, giving a length of 3 units. The area of a rectangle equals length times width, so Area = 5 × 3 = 15 square units.
In simple words: When you connect the four points, you get a shape with four right angles and opposite sides equal. That is a rectangle. Count the grid squares along two sides, then multiply them together to get the area.

Exam Tip: Always identify the shape first before calculating area - different shapes use different formulas. For rectangles, use length × width. For other quadrilaterals, you may need different methods. Show your side lengths clearly.

 

Question 12. Plot the points (0, 2), (3, 0), (0, -2) and (-3, 0) on a graph paper. Join these points (in order). Name the figure so obtained and find the area of the figure obtained.
Answer: Plot the four points A(0, 2), B(3, 0), C(0, -2), and D(-3, 0) on graph paper and join them in order. The resulting figure has all four sides of equal length, with diagonals that intersect at right angles. This shape is a rhombus. To find the area of the rhombus, use the formula: Area = (1/2) × d₁ × d₂, where d₁ and d₂ are the lengths of the two diagonals. One diagonal AC runs vertically from (0, 2) to (0, -2), measuring 4 units. The other diagonal BD runs horizontally from (3, 0) to (-3, 0), measuring 6 units. Substituting into the formula: Area = (1/2) × 6 × 4 = 12 square units.
In simple words: When you join these four points, you make a diamond shape. Measure the distances across the diamond both ways (the two diagonals). Multiply them and divide by 2 to get the area.

Exam Tip: For a rhombus plotted with vertices on the axes, the diagonals are always horizontal and vertical, making them easy to measure. Never forget the (1/2) factor in the rhombus area formula.

 

Question 13. Three vertices of a square are A(2, 3), B(-3, 3) and C(-3, -2). Plot these points on a graph paper and hence use it to find the coordinates of the fourth vertex. Also find the area of the square.
Answer: Plot the three given points A(2, 3), B(-3, 3), and C(-3, -2) on graph paper. Notice that A and B have the same y-coordinate (3), so side AB is horizontal. Points B and C have the same x-coordinate (-3), so side BC is vertical. This means AB and BC are perpendicular. Calculate the length of AB: from x = -3 to x = 2 is 5 units. Calculate the length of BC: from y = 3 to y = -2 is 5 units. Since both sides are equal and perpendicular, ABCD is indeed a square with side length 5. The fourth vertex D must be at the same distance from both A and C. By completing the square on the graph, mark point D at (2, -2). You can verify: CD is horizontal from x = -3 to x = 2 (5 units), and DA is vertical from y = -2 to y = 3 (5 units). The area of a square equals side length squared: Area = 5 × 5 = 25 square units.
In simple words: Three corners of a square are already plotted. To find the fourth corner, make sure all sides have the same length. Then count up the grid squares and multiply the side length by itself to get area.

Exam Tip: When finding the fourth vertex of a square, ensure all four sides are equal and all angles are right angles. Check your answer by verifying that opposite sides are parallel and all sides have the same length.

 

Question 14. Write the co-ordinates of the vertices of a rectangle which is 6 units long and 4 units wide if the rectangle is in the first quadrant, its longer side lies on the x-axis and one vertex is at the origin.
Answer: Since one vertex is at the origin and the longer side (length 6 units) lies on the x-axis in the first quadrant, place the first vertex A at (0, 0). Moving 6 units to the right along the x-axis, place vertex B at (6, 0). From B, move 4 units upward (the width of the rectangle) perpendicular to the x-axis, placing vertex C at (6, 4). From C, move 6 units to the left to complete the rectangle, placing vertex D at (0, 4). The four vertices of the rectangle are A(0, 0), B(6, 0), C(6, 4), and D(0, 4). You can verify: AB (bottom side) = 6 units along the x-axis, BC (right side) = 4 units parallel to the y-axis, CD (top side) = 6 units parallel to the x-axis, and DA (left side) = 4 units parallel to the y-axis. The area of this rectangle = 6 × 4 = 24 square units.
In simple words: Start at the origin. Move 6 units right to make one corner. Then move 4 units up to make another corner. Complete the rectangle by connecting the remaining corners. This gives you a rectangle 6 units wide and 4 units tall.

Exam Tip: When a rectangle's side lies on an axis, the problem becomes much simpler. Identify which axis the longer side touches, then build the rectangle by moving perpendicular to that axis for the width. Always state all four vertices clearly in your answer.

 

Question 15. In the adjoining figure, ABCD is a rectangle with length 6 units and breadth 3 units. If O is the mid-point of AB, find the coordinates of A, B, C and D.
Answer: We are given a rectangle with length 6 units and breadth 3 units, where O marks the midpoint of side AB. Since each grid square represents 1 unit, we can determine the positions of all four vertices by examining the graph. The base AB lies along the x-axis and measures 6 units total. With O at the midpoint, points A and B are positioned symmetrically on either side of the origin. Point A is located 3 units to the left of O, and point B is located 3 units to the right of O. From these base points, the rectangle extends upward by 3 units to form points D and C respectively.

The coordinates are:
\( A = (-3, 0) \)
\( B = (3, 0) \)
\( C = (3, 3) \)
\( D = (-3, 3) \)
In simple words: The rectangle sits with its bottom side on the x-axis, centered at the origin. A and B are 3 units away from the middle, left and right. C and D are directly above them, 3 units up.

Exam Tip: Always identify the midpoint condition first, then use it to place the base vertices symmetrically before determining the upper vertices by adding the breadth dimension.

 

Question 16. The adjoining figure shows an equilateral triangle OAB with each side = 2a units. Find the coordinates of the vertices.
Answer: We begin by setting up the triangle with O at the origin and B positioned on the x-axis. Since all three sides equal 2a units, we have OA = OB = AB = 2a. To find the coordinates, we drop a perpendicular from vertex A to side OB, meeting it at point D. In any equilateral triangle, a perpendicular from a vertex to the opposite side bisects that side. Therefore, OD equals half of OB, giving us OD = a. Applying the Pythagorean theorem to the right triangle OAD:

\( OA^2 = OD^2 + AD^2 \)
\( (2a)^2 = a^2 + AD^2 \)
\( 4a^2 = a^2 + AD^2 \)
\( AD^2 = 3a^2 \)
\( AD = \sqrt{3}a \)

From the graph, we can now identify each vertex position. O sits at the origin, B lies on the positive x-axis at distance 2a, and A is positioned directly above D (which is at distance a along the x-axis) at height \( \sqrt{3}a \).

The coordinates are:
\( O = (0, 0) \)
\( B = (2a, 0) \)
\( A = (a, \sqrt{3}a) \)
In simple words: Place O at the corner and B along the bottom. The third point A sits halfway along the base horizontally, and up high vertically using the Pythagorean rule.

Exam Tip: The key insight is recognizing that the perpendicular from the apex bisects the base in an equilateral triangle—use this to avoid unnecessary calculation steps.

 

Question 17. In the given figure, △PQR is equilateral. If the coordinates of the points Q and R are (0, 2) and (0, -2) respectively, find the coordinates of the point P.
Answer: We are given that Q and R lie on the y-axis at (0, 2) and (0, -2). Since the triangle is equilateral, all sides have the same length. First, we calculate the distance QR using the distance formula:

\( QR = \sqrt{(0 - 0)^2 + (-2 - 2)^2} = \sqrt{0 + 16} = 4 \text{ units} \)

Next, we find the distance from the origin O at (0, 0) to point Q:

\( OQ = \sqrt{(0 - 0)^2 + (2 - 0)^2} = \sqrt{4} = 2 \text{ units} \)

Since the triangle is equilateral, PQ = PR = QR = 4. Point P must lie on the x-axis (by symmetry of the configuration), so its coordinates are (x, 0) for some value x. Using the Pythagorean theorem in the right triangle POQ:

\( PQ^2 = OP^2 + OQ^2 \)
\( 16 = OP^2 + 4 \)
\( OP^2 = 12 \)
\( OP = 2\sqrt{3} \)

Since P lies on the positive x-axis at distance \( 2\sqrt{3} \) from the origin:

\( P = (2\sqrt{3}, 0) \)
In simple words: Q and R are opposite each other on the y-axis, 4 units apart. P must go on the x-axis, far enough away so all three sides equal 4 units.

Exam Tip: Exploit symmetry—recognize that P lies on the axis perpendicular to the line QR, which simplifies finding its exact position.

 

Exercise 18.2

 

Question 1(i). Draw the graph of the following linear equation: 2x + y + 3 = 0
Answer: We rearrange the equation into slope-intercept form by solving for y:

\( 2x + y + 3 = 0 \)
\( y = -2x - 3 \)

Now we substitute three different x-values to build a table of corresponding y-values:

When \( x = 0: y = -2(0) - 3 = -3 \)
When \( x = 1: y = -2(1) - 3 = -5 \)
When \( x = 2: y = -2(2) - 3 = -7 \)

Table of values:

Plot the three points (0, -3), (1, -5), and (2, -7) on the coordinate grid. Connect any two of these points with a straight line. You will observe that the third point also lies on this line, confirming that all three points satisfy the equation. Extend the line across the graph to complete the representation of the linear equation.
In simple words: Rearrange the equation to isolate y. Pick three x-values, find their y-values, plot these points, and draw a straight line through them.

Exam Tip: Always verify that all three computed points lie on the drawn line - if one does not, recalculate the y-values before finalizing the graph.

 

Question 1(ii). Draw the graph of the following linear equation: x - 5y - 4 = 0
Answer: We rearrange the equation to express y in terms of x:

\( x - 5y - 4 = 0 \)
\( 5y = x - 4 \)
\( y = \frac{1}{5}(x - 4) \)

Now we compute y for three chosen x-values:

When \( x = -6: y = \frac{1}{5}(-6 - 4) = \frac{-10}{5} = -2 \)
When \( x = -1: y = \frac{1}{5}(-1 - 4) = \frac{-5}{5} = -1 \)
When \( x = 4: y = \frac{1}{5}(4 - 4) = \frac{0}{5} = 0 \)

Table of values:

Mark the three points (-6, -2), (-1, -1), and (4, 0) on your graph. Draw a straight line connecting any two of these points and verify that the third point also rests on it. This confirms the accuracy of your calculation. Extend the line to complete the graph.
In simple words: Solve for y. Select three x-values, calculate the matching y-values, plot the points, and draw the line through them.

Exam Tip: When selecting x-values, try to pick ones that make the arithmetic simple - here, choosing multiples of 5 ensures y-values that are whole numbers or simple fractions.

 

Question 2. Draw the graph of 3y = 12 - 2x. Take 2 cm = 1 unit on both axes.
Answer: We rearrange the equation to isolate y:

\( 3y = 12 - 2x \)
\( y = \frac{12 - 2x}{3} \)
\( y = 4 - \frac{2x}{3} \)

We then substitute three x-values to obtain corresponding y-values:

When \( x = -3: y = 4 - \frac{2(-3)}{3} = 4 - \frac{-6}{3} = 4 - (-2) = 6 \)
When \( x = 0: y = 4 - \frac{2(0)}{3} = 4 - 0 = 4 \)
When \( x = 3: y = 4 - \frac{2(3)}{3} = 4 - 2 = 2 \)

Table of values:

Using the specified scale (2 cm = 1 unit), plot the three points (-3, 6), (0, 4), and (3, 2) on the coordinate plane. Draw a line connecting any two of these points and observe that the third point falls on the same line. This verification confirms your calculations are correct. Extend the line across the graph as needed.
In simple words: Rearrange so y is alone on one side. Pick easy x-values, compute y for each, plot the points, and join them with a line.

Exam Tip: Remember to use the exact scale given in the question - here, 2 cm represents 1 unit, which affects how you space your points on the paper.

 

Question 3. Draw the graph of 5x + 6y - 30 = 0 and use it to find the area of the triangle formed by the line and coordinate axes.
Answer: We rearrange the equation to express y in terms of x:

\( 5x + 6y - 30 = 0 \)
\( 6y = -5x + 30 \)
\( y = \frac{-5x + 30}{6} \)
\( y = -\frac{5x}{6} + 5 \)

We compute y for three selected x-values:

When \( x = 0: y = -\frac{5(0)}{6} + 5 = 0 + 5 = 5 \)
When \( x = 6: y = -\frac{5(6)}{6} + 5 = -5 + 5 = 0 \)
When \( x = 12: y = -\frac{5(12)}{6} + 5 = -10 + 5 = -5 \)

Table of values:

Plot the three points (0, 5), (6, 0), and (12, -5) on the graph. Draw a straight line through any two of these points and verify that the third point lies on it. The line intersects the y-axis at (0, 5) and the x-axis at (6, 0). These two intercepts, along with the origin, form a right triangle in the coordinate plane. The base of this triangle (along the x-axis) measures 6 units, and the height (along the y-axis) measures 5 units. Using the triangle area formula:

\( \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 6 \times 5 = 15 \text{ sq. units} \)
In simple words: Rearrange for y. Make a table, plot three points, and draw the line. The line meets the axes at two spots, creating a triangle with the origin—use base times height divided by 2 to get the area.

Exam Tip: Identify the x-intercept and y-intercept clearly from the graph—these are the two vertices of the triangle other than the origin, and they directly give you the base and height dimensions.

 

Question 4. Draw the graph of 4x - 3y + 12 = 0 and use it to find the area of the triangle formed by the line and co-ordinate axes. Take 2 cm = 1 unit on both axes.
Answer: We rearrange the equation to isolate y:

\( 4x - 3y + 12 = 0 \)
\( 3y = 4x + 12 \)
\( y = \frac{4x + 12}{3} \)
\( y = \frac{4x}{3} + 4 \)

We substitute three x-values to find the corresponding y-values:

When \( x = -6: y = \frac{4(-6)}{3} + 4 = \frac{-24}{3} + 4 = -8 + 4 = -4 \)
When \( x = -3: y = \frac{4(-3)}{3} + 4 = \frac{-12}{3} + 4 = -4 + 4 = 0 \)
When \( x = 0: y = \frac{4(0)}{3} + 4 = 0 + 4 = 4 \)

Table of values:

Using the scale 2 cm = 1 unit, plot the three points (-6, -4), (-3, 0), and (0, 4) on the coordinate plane. Connect any two of these points with a line and confirm that the third point also rests on it. The line crosses the x-axis at (-3, 0) and the y-axis at (0, 4). These two intercepts form a right triangle with the origin. The base (measured along the x-axis from the origin to the x-intercept) is 3 units in length. The height (measured along the y-axis from the origin to the y-intercept) is 4 units in length. Using the triangle area formula:

\( \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 3 \times 4 = 6 \text{ sq. units} \)
In simple words: Solve for y. Pick three x-values, compute y, plot the points, and draw the line. It crosses the axes at two places, making a triangle—multiply the base and height numbers and divide by 2.

Exam Tip: Always identify where the line crosses each axis, as these intercepts form the vertices of the triangle you need to measure - ignore points on the line that don't represent intercepts.

 

Question 5. Draw the graph of the equation y = 3x - 4. Find graphically (i) the value of y when x = -1 (ii) the value of x when y = 5.
Answer: The equation is y = 3x - 4. When x = 0, y = 3(0) - 4 = -4, and when x = 2, y = 3(2) - 4 = 2.

Table of values:

Steps of construction:
(1) Mark the points (0, -4) and (2, 2) on the graph.
(2) Join these two points with a straight line.

(i) To find y when x = -1: From the point x = -1 on the x-axis, draw a line upward parallel to the y-axis until it touches the drawn line. From this touching point, draw a line parallel to the x-axis to meet the y-axis. The point where it meets the y-axis gives y = -7.

(ii) To find x when y = 5: From the point y = 5 on the y-axis, draw a line parallel to the x-axis until it touches the drawn line. From this touching point, draw a line parallel to the y-axis to meet the x-axis. The point where it meets the x-axis gives x = 3.
In simple words: Plot two points on the line y = 3x - 4, connect them with a straight line, then use parallel lines to find unknown values from the graph.

Exam Tip: Always draw perpendiculars (parallel to the axes) from the graph to find coordinates - ensure your line is drawn clearly so the readings are accurate.

 

Question 6. The graph of a linear equation in x and y passes through (4, 0) and (0, 3). Find the value of k if the graph passes through (k, 1.5).
Answer: Steps of construction:
(1) Mark the points (4, 0) and (0, 3) on graph paper.
(2) Join these two points with a straight line.
(3) Find the point with y-coordinate 1.5 on the line: From y = 1.5 on the y-axis, draw a line parallel to the x-axis until it meets the graph. Call this meeting point R.
(4) From point R, draw a line parallel to the y-axis until it meets the x-axis. Call this meeting point S.

From the graph, point R is at (2, 1.5). Comparing this with (k, 1.5), we get k = 2.
In simple words: Draw the line through the two given points, then find where a horizontal line at y = 1.5 crosses it. The x-coordinate of that crossing point is your answer.

Exam Tip: Use a ruler to draw perpendiculars accurately - sloppy construction leads to wrong readings of the intersection point.

 

Question 7. Use the table given alongside to draw the graph of a straight line. Find, graphically, the values of a and b.

Answer: Steps of construction:
(1) Mark the points (1, -2) and (3, 4) on graph paper.
(2) Join these two points with a straight line extending across the graph.
(3) To find b: From x = 2 on the x-axis, draw a line parallel to the y-axis until it touches the straight line. Call this point B. From B, draw a line parallel to the x-axis to meet the y-axis at point A. Reading the y-value from the graph gives b = 1.
(4) To find a: The straight line meets the y-axis at point (0, -5). Since this point has y-coordinate -5, and we need the x-value where y = -5, comparing with (a, -5) gives a = 0.
In simple words: Plot the two known points and draw the line through them. Use perpendiculars from x = 2 to find b, and see where the line crosses the y-axis to find a.

Exam Tip: Always extend the line far enough to cross both axes - this helps you read off missing values accurately from the table.

 

Exercise 18.3

 

Question 1. Solve the following equations graphically: 3x - 2y = 4, 5x - 2y = 0.
Answer: For equation 3x - 2y = 4:

Rearranging: 2y = 3x - 4, so y = (3/2)x - 2

When x = -2, y = (3/2)(-2) - 2 = -3 - 2 = -5
When x = 0, y = (3/2)(0) - 2 = 0 - 2 = -2
When x = 2, y = (3/2)(2) - 2 = 3 - 2 = 1

Table of values:

Steps of construction for equation 1:
(1) Mark the points (-2, -5), (0, -2), and (2, 1) on graph paper.
(2) Join these points with a straight line.

For equation 5x - 2y = 0:

Rearranging: 2y = 5x, so y = (5/2)x

When x = -2, y = (5/2)(-2) = -5
When x = 0, y = (5/2)(0) = 0
When x = 2, y = (5/2)(2) = 5

Table of values:

Steps of construction for equation 2:
(1) Mark the points (-2, -5), (0, 0), and (2, 5) on graph paper.
(2) Join these points with a straight line.

Both lines are drawn on the same axes. The two lines meet at point A(-2, -5).

Therefore, the solution is x = -2 and y = -5.
In simple words: Make a table for each equation, plot the points, and draw both lines on the same graph. Where they cross is your answer.

Exam Tip: Choose x-values that make calculations easy - avoid fractions when possible so your plotted points are accurate.

 

Question 2. Solve the following pair of equations graphically. Plot atleast 3 points for each straight line. 2x - 7y = 6, 5x - 8y = -4.
Answer: For equation 2x - 7y = 6:

Rearranging: 2x = 6 + 7y, so x = (6 + 7y)/2

When y = 0, x = (6 + 7(0))/2 = 6/2 = 3
When y = -1, x = (6 + 7(-1))/2 = (6 - 7)/2 = -1/2 = -0.5
When y = -2, x = (6 + 7(-2))/2 = (6 - 14)/2 = -8/2 = -4

Table of values:

Steps of construction for equation 1:
(1) Mark the points (3, 0), (-0.5, -1), and (-4, -2) on graph paper.
(2) Join these points with a straight line.

For equation 5x - 8y = -4:

Rearranging: 5x = 8y - 4, so x = (8y - 4)/5

When y = 0, x = (8(0) - 4)/5 = -4/5 = -0.8
When y = 3, x = (8(3) - 4)/5 = (24 - 4)/5 = 20/5 = 4
When y = -2, x = (8(-2) - 4)/5 = (-16 - 4)/5 = -20/5 = -4

Table of values:

Steps of construction for equation 2:
(1) Mark the points (-0.8, 0), (4, 3), and (-4, -2) on graph paper.
(2) Join these points with a straight line.

Both lines are drawn on the same axes. The two lines meet at point A(-4, -2).

Therefore, the solution is x = -4 and y = -2.
In simple words: Rearrange each equation to express x in terms of y (or y in terms of x), make a table with three points, plot them, and find where the two lines cross.

Exam Tip: When rearranging, choose the variable that will give you whole number or simple decimal coordinates - this makes plotting easier and more accurate.

 

Question 3. Using the same axes of coordinates and the same unit, solve graphically. x + y = 0, 3x - 2y = 10.
Answer: For equation x + y = 0:

Rearranging: y = -x

When x = 0, y = 0
When x = 1, y = -1
When x = 2, y = -2

Table of values:

Steps of construction for equation 1:
(1) Mark the points (0, 0), (1, -1), and (2, -2) on graph paper.
(2) Join these points with a straight line.

For equation 3x - 2y = 10:

Rearranging: 2y = 3x - 10, so y = (3/2)x - 5

When x = 0, y = (3/2)(0) - 5 = 0 - 5 = -5
When x = 2, y = (3/2)(2) - 5 = 3 - 5 = -2
When x = 4, y = (3/2)(4) - 5 = 6 - 5 = 1

Table of values:

Steps of construction for equation 2:
(1) Mark the points (0, -5), (2, -2), and (4, 1) on graph paper.
(2) Join these points with a straight line.

Both lines are drawn on the same axes and with the same scale. The two lines meet at point A(2, -2).

Therefore, the solution is x = 2 and y = -2.
In simple words: For the first equation, points lie on a line through the origin with slope -1. For the second, rearrange to get y, then plot its three points. Their crossing point is the answer.

Exam Tip: Make sure both axes use the same scale and unit - this ensures the intersection point is read correctly from the graph.

 

Question 4. Take 1 cm to represent 1 unit on each axis to draw the graphs of the equations 4x - 5y = -4 and 3x = 2y - 3 on the same graph sheet (same axes). Use your graph to find the solution of the above simultaneous equations.
Answer: For equation 4x - 5y = -4:

Rearranging: 5y = 4x + 4, so y = (4x + 4)/5

When x = -1, y = (4(-1) + 4)/5 = (0)/5 = 0
When x = 1.5, y = (4(1.5) + 4)/5 = (10)/5 = 2
When x = 4, y = (4(4) + 4)/5 = (20)/5 = 4

Table of values:

Steps of construction for equation 1:
(1) Mark the points (-1, 0), (1.5, 2), and (4, 4) on graph paper, using a scale of 1 cm = 1 unit.
(2) Join these points with a straight line.

For equation 3x = 2y - 3:

Rearranging: 2y = 3x + 3, so y = (3x + 3)/2

When x = -1, y = (3(-1) + 3)/2 = (0)/2 = 0
When x = 0, y = (3(0) + 3)/2 = (3)/2 = 1.5
When x = 1, y = (3(1) + 3)/2 = (6)/2 = 3

Table of values:

Steps of construction for equation 2:
(1) Mark the points (-1, 0), (0, 1.5), and (1, 3) on the same graph paper.
(2) Join these points with a straight line.

Both lines are drawn on the same axes with 1 cm representing 1 unit. The two lines meet at point A(-1, 0).

Therefore, the solution is x = -1 and y = 0.
In simple words: Rearrange both equations to express y in terms of x, make tables with three points each, plot both lines carefully using the 1 cm = 1 unit scale, and read off where they cross.

Exam Tip: Keeping the same scale on both axes is crucial - it ensures the angle at which lines cross is drawn accurately and the solution can be read precisely.

 

Question 5. Solve the following simultaneous equations graphically: x + 3y = 8, 3x = 2 + 2y.
Answer: For the equation x + 3y = 8, rearrange to get x = 8 - 3y. Substituting y = 1, 2, 3 gives x = 5, 2, -1 respectively. Plot points (5, 1), (2, 2), (-1, 3) and draw a straight line through them. For the second equation 3x = 2 + 2y, rearrange to x = (2 + 2y)/3. When y = -1, 2, 5, we get x = 0, 2, 4 respectively. Plot points (0, -1), (2, 2), (4, 5) and connect them with a straight line. Both lines intersect at point A(2, 2). Therefore, the solution is x = 2, y = 2.
In simple words: Make a table of values for each equation. Plot the points on graph paper and join them to form two straight lines. Where the lines cross gives you the answer.

Exam Tip: Always plot at least three points per line to ensure accuracy. The intersection point is your solution - verify by substituting back into both original equations.

 

Question 6. Solve graphically the simultaneous equations 3y = 5 - x, 2x = y + 3.
Answer: For the equation 3y = 5 - x, rearrange to y = (5 - x)/3. When x = -4, -1, 2, we get y = 3, 2, 1 respectively. Plot these points and connect them with a straight line. For the second equation 2x = y + 3, rearrange to y = 2x - 3. When x = 0, 1, 2, we get y = -3, -1, 1 respectively. Plot these points and join them with a line. The two lines intersect at point P(2, 1). Therefore, the solution is x = 2, y = 1.
In simple words: Rearrange each equation to express y in terms of x. Create a table of values, plot the points, and draw lines. The crossing point is your answer.

Exam Tip: Choose x-values that will give simple y-values (integers where possible). This reduces calculation errors and makes plotting more accurate.

 

Question 7. Use graph paper for this question. Take 2 cm = 1 unit on both axes. (i) Draw the graphs of x + y + 3 = 0 and 3x - 2y + 4 = 0. Plot three points per line. (ii) Write down the coordinates of the point of intersection of the lines. (iii) Measure and record the distance of the point of intersection of the lines from the origin in cm.
Answer:
(i) For the equation x + y + 3 = 0, rearrange to y = -(3 + x). When x = -1, 0, 1, we get y = -2, -3, -4 respectively. Plot points (-1, -2), (0, -3), (1, -4) and draw a line. For the equation 3x - 2y + 4 = 0, rearrange to y = (3x + 4)/2. When x = -2, 0, 2, we get y = -1, 2, 5 respectively. Plot points (-2, -1), (0, 2), (2, 5) and connect them with a line.
(ii) The two lines intersect at P(-2, -1). Therefore, the coordinates of the point of intersection are (-2, -1).
(iii) From P, draw a perpendicular to the y-axis. Since 1 unit = 2 cm, we have PQ = 2 units = 4 cm and OQ = 1 unit = 2 cm. Using the Pythagorean theorem in the right triangle OPQ: OP² = PQ² + OQ² = 4² + 2² = 16 + 4 = 20. Therefore, OP = \( \sqrt{20} \) = 4.5 cm. The distance from the origin is 4.5 cm.
In simple words: Set up your graph with the given scale. Plot the two lines carefully. Measure how far the crossing point is from the origin using a ruler or by calculating with Pythagoras' rule.

Exam Tip: Use the correct scale throughout. When calculating distance from the origin, always apply the Pythagorean theorem correctly by identifying the right triangle formed.

 

Question 8. Solve the following simultaneous equations graphically: 2x - 3y + 2 = 4x + 1 = 3x - y + 2.
Answer: Split this into two separate equations. From 2x - 3y + 2 = 4x + 1, rearrange to get 3y = -2x + 1, so y = (1 - 2x)/3. When x = -4, -1, 2, we get y = 3, 1, -1 respectively. Plot points (-4, 3), (-1, 1), (2, -1) and draw a line. From 4x + 1 = 3x - y + 2, rearrange to get y = 1 - x. When x = 0, 1, 2, we get y = 1, 0, -1 respectively. Plot points (0, 1), (1, 0), (2, -1) and connect them with a line. The two lines intersect at P(2, -1). Therefore, the solution is x = 2, y = -1.
In simple words: Break the chained equation into two separate equations. Handle each one by making y the subject. Then graph both lines and find where they meet.

Exam Tip: When dealing with chained equations like a = b = c, split them into two independent equations: a = b and b = c. This prevents confusion and ensures you have exactly two lines to graph.

 

Question 9. Use graph paper for this question: (i) Draw the graphs of 3x - y - 2 = 0 and 2x + y - 8 = 0. Take 1 cm = 1 unit on both axes and plot three points per line. (ii) Write down the coordinates of the point of intersection and the area of the triangle formed by the lines and the x-axis.
Answer:
(i) For the equation 3x - y - 2 = 0, rearrange to y = 3x - 2. When x = 0, 1, 2, we get y = -2, 1, 4 respectively. Plot points (0, -2), (1, 1), (2, 4) and draw a line. For the equation 2x + y - 8 = 0, rearrange to y = 8 - 2x. When x = 2, 3, 4, we get y = 4, 2, 0 respectively. Plot points (2, 4), (3, 2), (4, 0) and connect them with a line.
(ii) The lines intersect at P(2, 4). The line 2x + y - 8 = 0 crosses the x-axis at (4, 0), and the line 3x - y - 2 = 0 crosses the x-axis at (2/3, 0). The base of the triangle BC = 4 - 2/3 = 3\( \frac{2}{3} \) cm, and the height PV = 4 cm. Area = \( \frac{1}{2} \) × base × height = \( \frac{1}{2} \) × 3\( \frac{2}{3} \) × 4 = 6\( \frac{2}{3} \) cm².
In simple words: Plot both lines carefully. Find where they cross each other and where each line touches the x-axis. Use these three points as the vertices of your triangle, then calculate the area.

Exam Tip: The vertices of the triangle are: the two x-intercepts and the intersection point of the lines. Always identify base and height perpendicular to each other for the area formula.

 

Question 10. Solve the following system of linear equations graphically: 2x - y - 4 = 0, x + y + 1 = 0. Hence, find the area of the triangle formed by these lines and the y-axis.
Answer: For the equation 2x - y - 4 = 0, rearrange to y = 2x - 4. When x = 1, 2, 3, we get y = -2, 0, 2 respectively. Plot points (1, -2), (2, 0), (3, 2) and draw a line. For the equation x + y + 1 = 0, rearrange to y = -(x + 1). When x = -1, 0, 1, we get y = 0, -1, -2 respectively. Plot points (-1, 0), (0, -1), (1, -2) and connect them with a line. The two lines intersect at A(1, -2). The line 2x - y - 4 = 0 meets the y-axis at (0, -4), and the line x + y + 1 = 0 meets the y-axis at (0, -1). The triangle has vertices at A(1, -2), B(0, -4), and C(0, -1). The base BC = |-1 - (-4)| = 3 units, and the perpendicular height (distance from A to the y-axis) = 1 unit. Area = \( \frac{1}{2} \) × 3 × 1 = 1.5 square units.
In simple words: Plot both lines on your graph. The triangle is formed by the two lines and the y-axis. Find the three corner points, measure the base and height, then apply the area formula.

Exam Tip: When a triangle is formed by two lines and an axis, two vertices will always lie on that axis. Find where each line crosses the axis to identify these points quickly.

 

Question 11. Solve graphically the following equations: x + 2y = 4, 3x - 2y = 4. Take 2 cm = 1 unit on each axis. Write down the area of the triangle formed by the lines and the y-axis. Also, find the area of the triangle formed by the lines and the x-axis.
Answer: For the equation \( x + 2y = 4 \), we get \( y = \frac{4-x}{2} \). When x = 0, y = 2; when x = 2, y = 1; when x = 4, y = 0. Plot the points (0, 2), (2, 1), and (4, 0), then join them. For the equation \( 3x - 2y = 4 \), we get \( y = \frac{3x-4}{2} \). When x = 0, y = -2; when x = 2, y = 1; when x = 4, y = 4. Plot the points (0, -2), (2, 1), and (4, 4), and join them. From the graph, the two lines meet at point A(2, 1). The triangle formed by the lines and the y-axis is AEF, where E(0, 2) and F(0, -2). The distance EF = 4 units. The perpendicular distance from A to the y-axis is AG = 2 units. Area of triangle AEF = \( \frac{1}{2} \times 4 \times 2 = 4 \) sq. units. The triangle formed by the lines and the x-axis is ABC, where B \( \left(\frac{4}{3}, 0\right) \) and C(4, 0). The distance BC = \( 4 - \frac{4}{3} = \frac{8}{3} \) units. The perpendicular distance from A to the x-axis is AD = 1 unit. Area of triangle ABC = \( \frac{1}{2} \times \frac{8}{3} \times 1 = \frac{4}{3} \) sq. units.
In simple words: Draw both lines on a graph. Where they cross is the meeting point. Then find the area of the shape formed by each line with one of the axes using the formula \( \frac{1}{2} \times \text{base} \times \text{height} \).

Exam Tip: Plot at least three points for each line to ensure accuracy. Always identify the intersection point clearly and measure base and height perpendicular to the axis.

 

Question 12. On graph paper, take 2 cm to represent one unit on both axes, draw the lines: x + 3 = 0, y - 2 = 0, 2x + 3y = 12. Write down the coordinates of the vertices of the triangle formed by these lines.
Answer: The first equation is \( x + 3 = 0 \), which gives \( x = -3 \). This is a vertical line passing through x = -3. The second equation is \( y - 2 = 0 \), which gives \( y = 2 \). This is a horizontal line passing through y = 2. For the third equation \( 2x + 3y = 12 \), we rearrange to get \( y = \frac{12-2x}{3} \). When x = -3, y = 6; when x = 0, y = 4; when x = 3, y = 2. Plot these three points and the lines. The three lines form a triangle with vertices at (-3, 6), (-3, 2), and (3, 2).
In simple words: The first two equations are straight lines that are vertical and horizontal. The third equation is a slanted line. The three lines meet at three different points to make a triangle.

Exam Tip: For vertical lines like x = -3, remember that x stays constant; for horizontal lines like y = 2, y stays constant. Find all three intersection points by solving pairs of equations.

 

Question 13. Find graphically the coordinates of the vertices of the triangle formed by the lines y = 0, y = x and 2x + 3y = 10. Hence, find the area of the triangle formed by these lines.
Answer: The first equation is \( y = 0 \), which is the x-axis. The second equation is \( y = x \). When x = 0, y = 0; when x = 1, y = 1; when x = 2, y = 2. Plot these points and draw the line. For the third equation \( 2x + 3y = 10 \), rearrange to get \( y = \frac{10-2x}{3} \). When x = -1, y = 4; when x = 2, y = 2; when x = 5, y = 0. Plot these points and draw the line. The three lines form a triangle ABC with vertices at (0, 0), (5, 0), and (2, 2). To find the area, use the perpendicular from A(2, 2) to the x-axis (which is BC). AD = 2 units and BC = 5 units. Area of triangle = \( \frac{1}{2} \times 5 \times 2 = 5 \) sq. units.
In simple words: The x-axis and the line y = x pass through the origin. The third line slants across. All three lines meet to form a triangle whose area can be found by multiplying half the base times the height.

Exam Tip: Always identify the three vertices by finding the intersection points of each pair of lines. Use a perpendicular from a vertex to the opposite side for calculating area accurately.

 

Exercise 18.4

 

Question 1(i). Find the distance between the following pair of points: (2, 3), (4, 1)
Answer: Using the distance formula, \( d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \). Let P(2, 3) and Q(4, 1) be the two points. Then \( PQ = \sqrt{(4-2)^2 + (1-3)^2} = \sqrt{2^2 + (-2)^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2} \) units.
In simple words: Subtract the x-coordinates and y-coordinates separately, square both answers, add them, then take the square root to get the distance.

Exam Tip: Always apply the distance formula systematically step-by-step. Double-check your arithmetic before taking the square root, and simplify radicals when possible.

 

Question 1(ii). Find the distance between the following pair of points: (0, 0), (36, 15)
Answer: Using the distance formula, \( d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \). Let P(0, 0) and Q(36, 15) be the two points. Then \( PQ = \sqrt{(36-0)^2 + (15-0)^2} = \sqrt{36^2 + 15^2} = \sqrt{1296 + 225} = \sqrt{1521} = 39 \) units.
In simple words: Square each coordinate value, add them up, and find the square root of the total to get the distance between the origin and the given point.

Exam Tip: Recognize perfect squares - 1521 = 39², so the square root simplifies to a whole number. Always check if the result is a perfect square before leaving it in radical form.

 

Question 1(iii). Find the distance between the following pair of points: (a, b), (-a, -b)
Answer: Using the distance formula, \( d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \). Let P(a, b) and Q(-a, -b) be the two points. Then \( PQ = \sqrt{(-a-a)^2 + (-b-b)^2} = \sqrt{(-2a)^2 + (-2b)^2} = \sqrt{4a^2 + 4b^2} = \sqrt{4(a^2+b^2)} = 2\sqrt{a^2 + b^2} \) units.
In simple words: The distance between a point and its opposite point (where both coordinates flip sign) is always twice the square root of the sum of the squares of those coordinates.

Exam Tip: Factor out common terms from inside the square root to simplify the expression. This algebraic technique is key for working with variable expressions in distance problems.

 

Question 2. A is a point on y-axis whose ordinate is 4 and B is a point on x-axis whose abscissa is -3. Find the length of the line segment AB.
Answer: Since A lies on the y-axis, its x-coordinate is 0. With ordinate 4, we have A = (0, 4). Since B lies on the x-axis, its y-coordinate is 0. With abscissa -3, we have B = (-3, 0). Using the distance formula, \( AB = \sqrt{(-3-0)^2 + (0-4)^2} = \sqrt{(-3)^2 + (-4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \) units.
In simple words: A point on the y-axis always has x = 0, and a point on the x-axis always has y = 0. Use these facts to write the coordinates, then apply the distance formula.

Exam Tip: Remember that ordinate means the y-value and abscissa means the x-value. This distinction is crucial for correctly identifying coordinates from the problem statement.

 

Question 3. Find the value of a, if the distance between the points A (-3, -14) and B (a, -5) is 9 units.
Answer: Using the distance formula, \( AB = \sqrt{[a-(-3)]^2 + [(-5)-(-14)]^2} = 9 \). This gives us \( \sqrt{(a+3)^2 + 9^2} = 9 \). Squaring both sides, \( (a+3)^2 + 81 = 81 \), which simplifies to \( (a+3)^2 = 0 \). Taking the square root, \( a + 3 = 0 \), so \( a = -3 \).
In simple words: Set up the distance formula with the unknown value a, set it equal to 9, then solve the resulting equation by squaring and simplifying.

Exam Tip: When squaring both sides of an equation, be careful to square all terms correctly. In this problem, note that \( (a+3)^2 + 81 = 81 \) leads directly to \( (a+3)^2 = 0 \), giving a unique solution.

 

Question 4(i). Find points on the x-axis which are at a distance of 5 units from the point (5, -4).
Answer: Any point on the x-axis has a y-coordinate equal to 0. Let P(x, 0) be the point on the x-axis that is 5 units away from (5, -4). Using the distance formula:

\( d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \)

\( \implies 5 = \sqrt{(x - 5)^2 + [0 - (-4)]^2} \)

\( \implies 5 = \sqrt{x^2 - 10x + 25 + 16} \)

\( \implies 5 = \sqrt{x^2 - 10x + 41} \)

Squaring both sides:

\( \implies 25 = x^2 - 10x + 41 \)

\( \implies x^2 - 10x + 16 = 0 \)

\( \implies x^2 - 8x - 2x + 16 = 0 \)

\( \implies x(x - 8) - 2(x - 8) = 0 \)

\( \implies (x - 2)(x - 8) = 0 \)

\( \implies x = 2 \text{ or } x = 8 \)

The points on the x-axis are (2, 0) and (8, 0).
In simple words: The two points you are looking for are where x equals 2 or 8, and both sit on the x-axis with y-coordinate 0.

Exam Tip: Always remember that any point on the x-axis has y = 0. Set this in the distance formula and solve the resulting quadratic equation carefully.

 

Question 4(ii). Find points on the y-axis which are at a distance of 10 units from the point (8, 8).
Answer: Any point on the y-axis has an x-coordinate equal to 0. Let P(0, y) be the point on the y-axis that is 10 units away from (8, 8). Using the distance formula:

\( d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \)

\( \implies 10 = \sqrt{(0 - 8)^2 + (y - 8)^2} \)

\( \implies 10 = \sqrt{64 + y^2 - 16y + 64} \)

\( \implies 10 = \sqrt{y^2 - 16y + 128} \)

Squaring both sides:

\( \implies 100 = y^2 - 16y + 128 \)

\( \implies y^2 - 16y + 28 = 0 \)

\( \implies y^2 - 14y - 2y + 28 = 0 \)

\( \implies y(y - 14) - 2(y - 14) = 0 \)

\( \implies (y - 2)(y - 14) = 0 \)

\( \implies y = 2 \text{ or } y = 14 \)

The points on the y-axis are (0, 2) and (0, 14).
In simple words: The two points you are looking for are where y equals 2 or 14, and both sit on the y-axis with x-coordinate 0.

Exam Tip: Remember that points on the y-axis always have x = 0. Substitute this value and apply the distance formula, then solve the resulting quadratic.

 

Question 4(iii). Find point (or points) which are at distance of \( \sqrt{10} \) units from the point (4, 3) given that the ordinate of the point (or points) is twice the abscissa.
Answer: Let the x-coordinate (abscissa) of the required point be k. Since the ordinate is twice the abscissa, the y-coordinate is 2k. So the point is P(k, 2k). Using the distance formula:

\( d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \)

Given that the distance between (4, 3) and (k, 2k) is \( \sqrt{10} \):

\( \implies \sqrt{10} = \sqrt{(k - 4)^2 + (2k - 3)^2} \)

\( \implies \sqrt{10} = \sqrt{k^2 - 8k + 16 + 4k^2 - 12k + 9} \)

\( \implies \sqrt{10} = \sqrt{5k^2 - 20k + 25} \)

Squaring both sides:

\( \implies 10 = 5k^2 - 20k + 25 \)

\( \implies 10 = 5(k^2 - 4k + 5) \)

\( \implies 2 = k^2 - 4k + 5 \)

\( \implies k^2 - 4k + 3 = 0 \)

\( \implies k^2 - 3k - k + 3 = 0 \)

\( \implies k(k - 3) - 1(k - 3) = 0 \)

\( \implies (k - 1)(k - 3) = 0 \)

\( \implies k = 1 \text{ or } k = 3 \)

When k = 1: P = (1, 2)
When k = 3: P = (3, 6)

The required points are (1, 2) and (3, 6).
In simple words: The constraint that y is double x helps set up a relationship. Plugging this into the distance formula gives you a quadratic equation that yields two answer points.

Exam Tip: Always express constraints (like "y is twice x") algebraically before applying the distance formula. This reduces the unknowns and simplifies the solution.

 

Question 5. Find the point on the x-axis which is equidistant from the points (2, -5) and (-2, 9).
Answer: Any point on the x-axis has a y-coordinate equal to 0. Let P(x, 0) be the point on the x-axis that is equidistant from (2, -5) and (-2, 9). Since the distances are equal:

\( \text{Distance between (2, -5) and (x, 0)} = \text{Distance between (-2, 9) and (x, 0)} \)

\( \implies \sqrt{(x - 2)^2 + [0 - (-5)]^2} = \sqrt{[x - (-2)]^2 + (0 - 9)^2} \)

\( \implies \sqrt{x^2 - 4x + 4 + 25} = \sqrt{(x + 2)^2 + 81} \)

\( \implies \sqrt{x^2 - 4x + 29} = \sqrt{x^2 + 4x + 4 + 81} \)

\( \implies \sqrt{x^2 - 4x + 29} = \sqrt{x^2 + 4x + 85} \)

Squaring both sides:

\( \implies x^2 - 4x + 29 = x^2 + 4x + 85 \)

\( \implies -4x - 4x = 85 - 29 \)

\( \implies -8x = 56 \)

\( \implies x = -7 \)

The required point is (-7, 0).
In simple words: Set the distances from the unknown point to both given points as equal, then solve for x. Since the point is on the x-axis, y stays 0.

Exam Tip: For "equidistant" problems, always set the two distance expressions equal and square both sides to eliminate the square roots.

 

Question 6. Find the value of x such that PQ = QR where the coordinates of P, Q and R are (6, -1), (1, 3) and (x, 8) respectively.
Answer: Since PQ = QR, we have:

\( \text{Distance between (6, -1) and (1, 3)} = \text{Distance between (1, 3) and (x, 8)} \)

\( \implies \sqrt{(1 - 6)^2 + [3 - (-1)]^2} = \sqrt{(x - 1)^2 + (8 - 3)^2} \)

\( \implies \sqrt{25 + 16} = \sqrt{(x - 1)^2 + 25} \)

\( \implies \sqrt{41} = \sqrt{x^2 - 2x + 1 + 25} \)

\( \implies \sqrt{41} = \sqrt{x^2 - 2x + 26} \)

Squaring both sides:

\( \implies 41 = x^2 - 2x + 26 \)

\( \implies x^2 - 2x + 26 - 41 = 0 \)

\( \implies x^2 - 2x - 15 = 0 \)

\( \implies x^2 - 5x + 3x - 15 = 0 \)

\( \implies x(x - 5) + 3(x - 5) = 0 \)

\( \implies (x + 3)(x - 5) = 0 \)

\( \implies x = -3 \text{ or } x = 5 \)
In simple words: Find the distance PQ first, then set it equal to the distance QR (which contains x), and solve the resulting equation.

Exam Tip: When a point lies between two others on a line with a specific distance relationship, always equate the two distances and solve the quadratic carefully.

 

Question 7. If Q(0, 1) is equidistant from P(5, -3) and R(x, 6), find the values of x.
Answer: Since Q is equidistant from P and R:

\( \text{Distance between Q(0, 1) and P(5, -3)} = \text{Distance between Q(0, 1) and R(x, 6)} \)

\( \implies \sqrt{(5 - 0)^2 + (-3 - 1)^2} = \sqrt{(x - 0)^2 + (6 - 1)^2} \)

\( \implies \sqrt{25 + 16} = \sqrt{x^2 + 25} \)

\( \implies \sqrt{41} = \sqrt{x^2 + 25} \)

Squaring both sides:

\( \implies 41 = x^2 + 25 \)

\( \implies x^2 = 16 \)

\( \implies x = \pm 4 \)
In simple words: Calculate the distance from Q to P, then set it equal to the distance from Q to R, and solve for x.

Exam Tip: When solving \( x^2 = 16 \), always include both the positive and negative square roots unless the context restricts one of them.

 

Question 8. Find a relation between x and y such that point (x, y) is equidistant from the points (7, 1) and (3, 5).
Answer: Let a general point (x, y) be equidistant from (7, 1) and (3, 5). Then:

\( \text{Distance between (x, y) and (7, 1)} = \text{Distance between (x, y) and (3, 5)} \)

\( \implies \sqrt{(7 - x)^2 + (1 - y)^2} = \sqrt{(3 - x)^2 + (5 - y)^2} \)

Squaring both sides:

\( \implies (7 - x)^2 + (1 - y)^2 = (3 - x)^2 + (5 - y)^2 \)

\( \implies 49 - 14x + x^2 + 1 - 2y + y^2 = 9 - 6x + x^2 + 25 - 10y + y^2 \)

\( \implies 49 - 14x + 1 - 2y = 9 - 6x + 25 - 10y \)

\( \implies 50 - 14x - 2y = 34 - 6x - 10y \)

\( \implies -14x + 6x - 2y + 10y = 34 - 50 \)

\( \implies -8x + 8y = -16 \)

\( \implies 8y - 8x = -16 \)

\( \implies y - x = -2 \)

\( \implies x - y = 2 \)

The relation between x and y is x - y = 2, or equivalently, x = y + 2.
In simple words: The set of all points equidistant from two fixed points forms a straight line. That line here is described by the equation x - y = 2.

Exam Tip: The locus of points equidistant from two given points is always the perpendicular bisector of the line segment joining them. The final linear equation represents this perpendicular bisector.

 

Question 9. The x-coordinate of a point P is twice its y-coordinate. If P is equidistant from the points Q(2, -5) and R(-3, 6), then find the coordinates of P.
Answer: Let the y-coordinate of point P be k. Since the x-coordinate is twice the y-coordinate, the x-coordinate is 2k. So P = (2k, k). Since P is equidistant from Q(2, -5) and R(-3, 6):

\( \sqrt{(2k - 2)^2 + (k - (-5))^2} = \sqrt{(2k - (-3))^2 + (k - 6)^2} \)

\( \implies \sqrt{(2k - 2)^2 + (k + 5)^2} = \sqrt{(2k + 3)^2 + (k - 6)^2} \)

Squaring both sides:

\( \implies (2k - 2)^2 + (k + 5)^2 = (2k + 3)^2 + (k - 6)^2 \)

\( \implies 4k^2 - 8k + 4 + k^2 + 10k + 25 = 4k^2 + 12k + 9 + k^2 - 12k + 36 \)

\( \implies 5k^2 + 2k + 29 = 5k^2 + 45 \)

\( \implies 2k + 29 = 45 \)

\( \implies 2k = 16 \)

\( \implies k = 8 \)

Therefore, P = (2k, k) = (16, 8).
In simple words: Use the constraint that x is twice y to express the point in terms of one variable, then apply the equidistance condition.

Exam Tip: When a point satisfies a special relationship (like "x is twice y"), always express it algebraically and substitute into the distance formula to reduce variables.

 

Question 10. If the points A(4, 3) and B(x, 5) are on a circle with center C(2, 3), find the values of x.
Answer: Since both A(4, 3) and B(x, 5) lie on the same circle with center C(2, 3), the distances from C to A and from C to B must be equal (both equal the radius):

\( AC = BC \)

\( \implies \sqrt{(4 - 2)^2 + (3 - 3)^2} = \sqrt{(x - 2)^2 + (5 - 3)^2} \)

\( \implies \sqrt{4 + 0} = \sqrt{(x - 2)^2 + 4} \)

\( \implies \sqrt{4} = \sqrt{(x - 2)^2 + 4} \)

\( \implies 2 = \sqrt{(x - 2)^2 + 4} \)

Squaring both sides:

\( \implies 4 = (x - 2)^2 + 4 \)

\( \implies 4 = x^2 - 4x + 4 + 4 \)

\( \implies 4 = x^2 - 4x + 8 \)

\( \implies x^2 - 4x + 4 = 0 \)

\( \implies (x - 2)^2 = 0 \)

\( \implies x = 2 \)

The value of x is 2.
In simple words: Two points on the same circle must be the same distance from the center. Set these distances equal and solve for the unknown coordinate.

Exam Tip: Remember that all points on a circle are equidistant from the center. Use this property to find unknown coordinates of points on the circle.

 

Question 11. If a point A(0, 2) is equidistant from the points B(3, p) and C(p, 5), then find the value of p.
Answer: Since A is equidistant from B and C:

\( AB = AC \)

\( \implies \sqrt{(3 - 0)^2 + (p - 2)^2} = \sqrt{(p - 0)^2 + (5 - 2)^2} \)

\( \implies \sqrt{9 + (p - 2)^2} = \sqrt{p^2 + 9} \)

\( \implies \sqrt{9 + p^2 - 4p + 4} = \sqrt{p^2 + 9} \)

\( \implies \sqrt{p^2 - 4p + 13} = \sqrt{p^2 + 9} \)

Squaring both sides:

\( \implies p^2 - 4p + 13 = p^2 + 9 \)

\( \implies -4p + 13 = 9 \)

\( \implies -4p = -4 \)

\( \implies p = 1 \)

The value of p is 1.
In simple words: Set the distances from A to B and from A to C as equal expressions, then simplify to find p.

Exam Tip: When two distances are equal, squaring both sides eliminates the square roots and often simplifies the algebra significantly.

 

Question 12. Using distance formula, show that (3, 3) is the center of the circle passing through the points (6, 2), (0, 4) and (4, 6).
Answer: To prove that (3, 3) is the center, we must show that it is equidistant from all three points. Let O = (3, 3), A = (6, 2), B = (0, 4), and C = (4, 6). Calculate each distance:

\( AO = \sqrt{(6 - 3)^2 + (2 - 3)^2} = \sqrt{9 + 1} = \sqrt{10} \)

\( BO = \sqrt{(0 - 3)^2 + (4 - 3)^2} = \sqrt{9 + 1} = \sqrt{10} \)

\( CO = \sqrt{(4 - 3)^2 + (6 - 3)^2} = \sqrt{1 + 9} = \sqrt{10} \)

Since AO = BO = CO = \( \sqrt{10} \), all three points are at the same distance from (3, 3). Therefore, (3, 3) is the center of the circle with radius \( \sqrt{10} \) passing through the three given points.
In simple words: A point is the center of a circle if it is the same distance from all points on the circle. Here, all three distances equal \( \sqrt{10} \), so (3, 3) is indeed the center.

Exam Tip: To verify the center of a circle passing through multiple points, always calculate the distance from the proposed center to each of the three points and confirm they are all equal.

 

Question 13. The center of a circle is C(2α - 1, 3α + 1) and it passes through the point A(-3, -1). If a diameter of the circle is of length 20 units, find the value(s) of α.
Answer: The diameter measures 20 units, so the radius equals 10 units. Since point A lies on the circle, the distance from the center C to point A must equal the radius. Using the distance formula:
\[ AC = \sqrt{[-3 - (2\alpha - 1)]^2 + [-1 - (3\alpha + 1)]^2} = 10 \]
\[ \Rightarrow \sqrt{(-2 - 2\alpha)^2 + (-2 - 3\alpha)^2} = 10 \]
\[ \Rightarrow \sqrt{4 + 8\alpha + 4\alpha^2 + 4 + 12\alpha + 9\alpha^2} = 10 \]
\[ \Rightarrow \sqrt{13\alpha^2 + 20\alpha + 8} = 10 \]
Squaring both sides:
\[ 13\alpha^2 + 20\alpha + 8 = 100 \]
\[ 13\alpha^2 + 20\alpha - 92 = 0 \]
\[ 13\alpha^2 + 46\alpha - 26\alpha - 92 = 0 \]
\[ \alpha(13\alpha + 46) - 2(13\alpha + 46) = 0 \]
\[ (13\alpha + 46)(\alpha - 2) = 0 \]
Therefore, \( \alpha = -\frac{46}{13} \) or \( \alpha = 2 \).
In simple words: Find how far A is from the center C using the distance formula. Set this distance equal to 10 (the radius), square to eliminate the square root, then solve the resulting equation by factoring.

Exam Tip: Always square both sides carefully and check that both solutions are valid - sometimes extraneous solutions arise after squaring.

 

Question 14. Using distance formula, show that the points A(3, 1), B(6, 4) and C(8, 6) are collinear.
Answer: Three points are collinear if one lies on the line segment joining the other two, which occurs when the sum of two shorter distances equals the longest distance. Apply the distance formula to find each pair-wise distance:
\[ AB = \sqrt{(6-3)^2 + (4-1)^2} = \sqrt{9 + 9} = \sqrt{18} = 3\sqrt{2} \]
\[ BC = \sqrt{(8-6)^2 + (6-4)^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2} \]
\[ AC = \sqrt{(8-3)^2 + (6-1)^2} = \sqrt{25 + 25} = \sqrt{50} = 5\sqrt{2} \]
Now verify: \( AB + BC = 3\sqrt{2} + 2\sqrt{2} = 5\sqrt{2} = AC \). Since the sum of distances AB and BC equals distance AC, the three points lie on the same straight line.
In simple words: Measure the distance between each pair of points. If adding two distances gives the third distance, the points must be in a straight line.

Exam Tip: When checking collinearity, always order the distances from smallest to largest before adding - the two smallest should sum to the largest.

 

Question 15. Check whether the points (5, -2), (6, 4) and (7, -2) are the vertices of an isosceles triangle.
Answer: Let the three points be A(5, -2), B(6, 4), and C(7, -2). Use the distance formula to compute all three side lengths:
\[ AB = \sqrt{(6-5)^2 + [4-(-2)]^2} = \sqrt{1 + 36} = \sqrt{37} \]
\[ BC = \sqrt{(7-6)^2 + [-2-4]^2} = \sqrt{1 + 36} = \sqrt{37} \]
\[ AC = \sqrt{(7-5)^2 + [-2-(-2)]^2} = \sqrt{4 + 0} = 2 \]
Since AB equals BC (both equal \( \sqrt{37} \)), the triangle has two equal sides and therefore is an isosceles triangle.
In simple words: Find all three side lengths. If any two sides have the same length, the triangle is isosceles.

Exam Tip: Always compute all three distances even if you spot two equal sides early - verifying the third ensures your calculation is correct.

 

Question 16. Name the type of triangle formed by the points A(-5, 6), B(-4, -2) and C(7, 5).
Answer: Measure each side using the distance formula:
\[ AB = \sqrt{[-4-(-5)]^2 + (-2-6)^2} = \sqrt{1 + 64} = \sqrt{65} \]
\[ BC = \sqrt{[7-(-4)]^2 + (5-(-2))^2} = \sqrt{121 + 49} = \sqrt{170} \]
\[ AC = \sqrt{[7-(-5)]^2 + (5-6)^2} = \sqrt{144 + 1} = \sqrt{145} \]
All three side lengths are different (AB ≠ BC ≠ AC). When all three sides have unequal lengths, the triangle is classified as scalene.
In simple words: If all three sides have different lengths, the triangle is scalene.

Exam Tip: Organize your three distance values in increasing order and compare - if all three are distinct, you have a scalene triangle immediately.

 

Question 17. Show that the points A(1, 1), B(-1, -1) and C(-√3, √3) form an equilateral triangle.
Answer: Calculate the length of each side using the distance formula:
\[ AB = \sqrt{(-1-1)^2 + (-1-1)^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2} \]
\[ BC = \sqrt{(-\sqrt{3}-(-1))^2 + (\sqrt{3}-(-1))^2} = \sqrt{(-\sqrt{3}+1)^2 + (\sqrt{3}+1)^2} \]
\[ = \sqrt{3 + 1 - 2\sqrt{3} + 3 + 1 + 2\sqrt{3}} = \sqrt{8} = 2\sqrt{2} \]
\[ AC = \sqrt{(-\sqrt{3}-1)^2 + (\sqrt{3}-1)^2} = \sqrt{3 + 1 + 2\sqrt{3} + 3 + 1 - 2\sqrt{3}} = \sqrt{8} = 2\sqrt{2} \]
Since AB = BC = AC (all equal \( 2\sqrt{2} \)), the triangle has all three sides of equal length, making it equilateral.
In simple words: When all three sides of a triangle are the same length, it is equilateral.

Exam Tip: For triangles with irrational coordinates like \( \sqrt{3} \), expand squared terms carefully to avoid algebra mistakes - note that cancellations often occur.

 

Question 18. Show that the points (7, 10), (-2, 5) and (3, -4) are the vertices of an isosceles right triangle.
Answer: Let the points be A(7, 10), B(-2, 5), and C(3, -4). Calculate all three sides:
\[ AB = \sqrt{(-2-7)^2 + (5-10)^2} = \sqrt{81 + 25} = \sqrt{106} \]
\[ BC = \sqrt{[3-(-2)]^2 + (-4-5)^2} = \sqrt{25 + 81} = \sqrt{106} \]
\[ AC = \sqrt{(3-7)^2 + (-4-10)^2} = \sqrt{16 + 196} = \sqrt{212} \]
Two sides are equal (AB = BC = \( \sqrt{106} \)), so the triangle is isosceles. Now check if it is also a right triangle by verifying the Pythagorean theorem:
\[ AB^2 + BC^2 = 106 + 106 = 212 = AC^2 \]
Since the Pythagorean relation holds, the triangle is a right triangle with the right angle at B. Therefore, it is an isosceles right triangle.
In simple words: Find all three sides. If two are equal and \( \text{side}_1^2 + \text{side}_2^2 = \text{side}_3^2 \), you have an isosceles right triangle.

Exam Tip: Always verify both conditions for an isosceles right triangle: two equal sides AND the Pythagorean theorem with the longest side as the hypotenuse.

 

Question 19. The points A(0, 3), B(-2, a) and C(-1, 4) are the vertices of a right angled triangle at A, find the value of a.
Answer: The triangle has a right angle at vertex A, so sides AB and AC must be perpendicular. For perpendicularity, the Pythagorean theorem requires:
\[ AB^2 + AC^2 = BC^2 \]
Calculate each distance:
\[ AB = \sqrt{(-2-0)^2 + (a-3)^2} = \sqrt{4 + a^2 - 6a + 9} = \sqrt{a^2 - 6a + 13} \]
\[ AC = \sqrt{(-1-0)^2 + (4-3)^2} = \sqrt{1 + 1} = \sqrt{2} \]
\[ BC = \sqrt{[-1-(-2)]^2 + (4-a)^2} = \sqrt{1 + 16 - 8a + a^2} = \sqrt{a^2 - 8a + 17} \]
Substitute into the Pythagorean relation:
\[ (\sqrt{a^2 - 6a + 13})^2 + (\sqrt{2})^2 = (\sqrt{a^2 - 8a + 17})^2 \]
\[ a^2 - 6a + 13 + 2 = a^2 - 8a + 17 \]
\[ -6a + 15 = -8a + 17 \]
\[ 2a = 2 \]
\[ a = 1 \]
In simple words: If there is a right angle at A, then the two sides meeting at A must obey the Pythagorean rule when paired against the opposite side.

Exam Tip: When a right angle is at a specific vertex, apply Pythagoras with the two sides from that vertex as the legs and the opposite side as the hypotenuse.

 

Question 20. Show that the points (0, -1), (-2, 3), (6, 7) and (8, 3), taken in order, are the vertices of a rectangle. Also find its area.
Answer: Let A(0, -1), B(-2, 3), C(6, 7), and D(8, 3). A quadrilateral is a rectangle if opposite sides are equal. Calculate all four sides:
\[ AB = \sqrt{(-2-0)^2 + (3-(-1))^2} = \sqrt{4 + 16} = \sqrt{20} = 2\sqrt{5} \]
\[ BC = \sqrt{(6-(-2))^2 + (7-3)^2} = \sqrt{64 + 16} = \sqrt{80} = 4\sqrt{5} \]
\[ CD = \sqrt{(8-6)^2 + (3-7)^2} = \sqrt{4 + 16} = \sqrt{20} = 2\sqrt{5} \]
\[ AD = \sqrt{(8-0)^2 + [3-(-1)]^2} = \sqrt{64 + 16} = \sqrt{80} = 4\sqrt{5} \]
Since AB = CD and BC = AD (opposite sides are equal), the quadrilateral is a rectangle. Find the area by multiplying length and width:
\[ \text{Area} = AB \times BC = 2\sqrt{5} \times 4\sqrt{5} = 8 \times 5 = 40 \text{ square units} \]
In simple words: Check that opposite sides are equal in length. If they are, it is a rectangle. Multiply the two different side lengths to get the area.

Exam Tip: For a quadrilateral to be a rectangle, verify both opposite-side equality AND that all four angles are right angles (use the Pythagorean theorem on diagonals or dot products if needed).

 

Question 21. If P(2, -1), Q(3, 4), R(-2, 3) and S(-3, -2) be four points in a plane, show that PQRS is a rhombus but not a square. Find the area of the rhombus.
Answer: Using the distance formula, we first calculate the lengths of all four sides. \( PQ = \sqrt{(3 - 2)^2 + [4 - (-1)]^2} = \sqrt{1 + 25} = \sqrt{26} \) \( QR = \sqrt{(-2 - 3)^2 + (3 - 4)^2} = \sqrt{25 + 1} = \sqrt{26} \) \( RS = \sqrt{[-3 - (-2)]^2 + [-2 - 3]^2} = \sqrt{1 + 25} = \sqrt{26} \) \( PS = \sqrt{[-3 - 2]^2 + [-2 - (-1)]^2} = \sqrt{25 + 1} = \sqrt{26} \) Since all four sides are equal, PQRS could be either a square or a rhombus. We now find the diagonals to determine which one it is. \( PR = \sqrt{(-2 - 2)^2 + [3 - (-1)]^2} = \sqrt{16 + 16} = \sqrt{32} = 4\sqrt{2} \) \( QS = \sqrt{(-3 - 3)^2 + (-2 - 4)^2} = \sqrt{36 + 36} = \sqrt{72} = 6\sqrt{2} \) The diagonals are not equal (PR ≠ QS), so PQRS is a rhombus, not a square. The area is: Area = \( \frac{1}{2} \times d_1 \times d_2 = \frac{1}{2} \times 4\sqrt{2} \times 6\sqrt{2} = \frac{1}{2} \times 48 = 24 \) sq. units.
In simple words: All four sides are the same length. The two diagonal lines are different lengths, so it is a rhombus. Using the diagonals, the area is 24 square units.

Exam Tip: Always check side lengths first, then diagonals - equal sides with unequal diagonals confirms a rhombus. Remember the area formula uses both diagonals multiplied together.

 

Question 22. Prove that the points A(2, 3), B(-2, 2), C(-1, -2) and D(3, -1) are the vertices of a square ABCD.
Answer: Using the distance formula, we calculate all four side lengths: \( AB = \sqrt{(-2 - 2)^2 + (2 - 3)^2} = \sqrt{16 + 1} = \sqrt{17} \) \( BC = \sqrt{[-1 - (-2)]^2 + (-2 - 2)^2} = \sqrt{1 + 16} = \sqrt{17} \) \( CD = \sqrt{[3 - (-1)]^2 + [-1 - (-2)]^2} = \sqrt{16 + 1} = \sqrt{17} \) \( AD = \sqrt{(3 - 2)^2 + (-1 - 3)^2} = \sqrt{1 + 16} = \sqrt{17} \) All four sides are equal. Now we check the diagonals: \( AC = \sqrt{(-1 - 2)^2 + (-2 - 3)^2} = \sqrt{9 + 25} = \sqrt{34} \) \( BD = \sqrt{(3 - (-2))^2 + (-1 - 2)^2} = \sqrt{25 + 9} = \sqrt{34} \) Since all four sides are equal and both diagonals are equal, ABCD is a square.
In simple words: All four sides have the same length. Both diagonals also have the same length. This means it is a square.

Exam Tip: For a square, you must verify three properties: equal sides, equal diagonals, and (optionally) right angles. Showing equal sides and equal diagonals is sufficient proof.

 

Question 23(i). Name the type of quadrilateral formed by the following points and give reasons for your answer: (-1, -2), (1, 0), (-1, 2), (-3, 0).
Answer: Let A(-1, -2), B(1, 0), C(-1, 2), and D(-3, 0) be the vertices. We calculate all four sides: \( AB = \sqrt{[1 - (-1)]^2 + [0 - (-2)]^2} = \sqrt{4 + 4} = \sqrt{8} \) \( BC = \sqrt{[-1 - 1]^2 + (2 - 0)^2} = \sqrt{4 + 4} = \sqrt{8} \) \( CD = \sqrt{[-3 - (-1)]^2 + [0 - 2]^2} = \sqrt{4 + 4} = \sqrt{8} \) \( AD = \sqrt{[-3 - (-1)]^2 + [0 - (-2)]^2} = \sqrt{4 + 4} = \sqrt{8} \) All sides are equal. Now we find the diagonals: \( AC = \sqrt{[-1 - (-1)]^2 + [2 - (-2)]^2} = \sqrt{0 + 16} = 4 \) \( BD = \sqrt{[-3 - 1]^2 + (0 - 0)^2} = \sqrt{16 + 0} = 4 \) Since all four sides are equal and the diagonals are also equal, the quadrilateral is a square.
In simple words: All four sides are the same length and both diagonals are the same length, so these points form a square.

Exam Tip: When all sides are equal, check the diagonals next - if diagonals are also equal, it is a square; if not, it is a rhombus.

 

Question 23(ii). Name the type of quadrilateral formed by the following points and give reasons for your answer: (4, 5), (7, 6), (4, 3), (1, 2).
Answer: Let A(4, 5), B(7, 6), C(4, 3), and D(1, 2) be the given points. Using the distance formula, we find the four sides: \( AB = \sqrt{(7 - 4)^2 + (6 - 5)^2} = \sqrt{9 + 1} = \sqrt{10} \) \( BC = \sqrt{(4 - 7)^2 + (3 - 6)^2} = \sqrt{9 + 9} = \sqrt{18} = 3\sqrt{2} \) \( CD = \sqrt{(1 - 4)^2 + (2 - 3)^2} = \sqrt{9 + 1} = \sqrt{10} \) \( AD = \sqrt{(1 - 4)^2 + (2 - 5)^2} = \sqrt{9 + 9} = \sqrt{18} = 3\sqrt{2} \) Opposite sides are equal (AB = CD and BC = AD). Now we check the diagonals: \( AC = \sqrt{(4 - 4)^2 + (3 - 5)^2} = \sqrt{0 + 4} = 2 \) \( BD = \sqrt{(1 - 7)^2 + (2 - 6)^2} = \sqrt{36 + 16} = \sqrt{52} \) Since opposite sides are equal but the diagonals are not equal, the quadrilateral is a parallelogram.
In simple words: Opposite sides have the same length but the diagonals are different, which means it is a parallelogram.

Exam Tip: In a parallelogram, opposite sides are equal but diagonals are not equal. If all four sides are equal, it could be a rhombus or square - check diagonals to decide which.

 

Question 24. Find the coordinates of the circumcentre of the triangle whose vertices are (8, 6), (8, -2) and (2, -2). Also, find its circumradius.
Answer: Let O(x, y) be the circumcentre and A(8, 6), B(8, -2), C(2, -2) be the vertices. The circumcentre is equidistant from all three vertices. Using OB = OC: \( \sqrt{(x - 8)^2 + [y - (-2)]^2} = \sqrt{(x - 2)^2 + [y - (-2)]^2} \) Squaring both sides and simplifying: \( x^2 - 16x + 64 + (y + 2)^2 = x^2 - 4x + 4 + (y + 2)^2 \) \( -16x + 64 = -4x + 4 \) \( -12x = -60 \) \( x = 5 \) Now using OA = OB: \( \sqrt{(x - 8)^2 + (y - 6)^2} = \sqrt{(x - 8)^2 + [y - (-2)]^2} \) Squaring and simplifying: \( (x - 8)^2 + y^2 - 12y + 36 = (x - 8)^2 + (y + 2)^2 \) \( y^2 - 12y + 36 = y^2 + 4y + 4 \) \( -16y = -32 \) \( y = 2 \) So the circumcentre is O(5, 2). The circumradius is: \( OA = \sqrt{(8 - 5)^2 + (6 - 2)^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \) units.
In simple words: The circumcentre is the point that is the same distance from all three corners of the triangle. This point is at (5, 2), and that distance is 5 units.

Exam Tip: The circumcentre lies at the intersection of perpendicular bisectors of the sides. Use the condition that distances from the circumcentre to all vertices are equal to find its coordinates.

 

Question 1. Point (-3, 5) lies in the
(a) first quadrant
(b) second quadrant
(c) third quadrant
(d) fourth quadrant
Answer: (b) second quadrant
In simple words: In the second quadrant, the x-coordinate is negative and the y-coordinate is positive. Since (-3, 5) has a negative x-value and positive y-value, it is in the second quadrant.

Exam Tip: Remember the quadrant signs: first quadrant (+ ,+), second quadrant (-, +), third quadrant (-, -), fourth quadrant (+, -).

 

Question 2. Point (0, -7) lies
(a) on the x-axis
(b) in the second quadrant
(c) on the y-axis
(d) in the fourth quadrant
Answer: (c) on the y-axis
In simple words: Any point on the y-axis has an x-coordinate of 0. Since (0, -7) has x = 0, it lies on the y-axis.

Exam Tip: Points on the x-axis have y = 0, and points on the y-axis have x = 0. Points with x = 0 but y ≠ 0 are on the y-axis, not in any quadrant.

 

Question 3. Abscissa of a point is positive in
(a) I and II quadrants
(b) I and IV quadrants
(c) I quadrant only
(d) II quadrants only
Answer: (b) I and IV quadrants
In simple words: The abscissa is the x-coordinate. The x-coordinate is positive in the first and fourth quadrants.

Exam Tip: Abscissa refers to the x-coordinate, and ordinate refers to the y-coordinate. Positive x-values appear on the right side of the coordinate plane, which is quadrants I and IV.

 

Question 1. The point which lies on y-axis at a distance of 5 units in the negative direction of y-axis is
(a) (0, 5)
(b) (5, 0)
(c) (0, -5)
(d) (-5, 0)
Answer: (c) (0, -5)
In simple words: Any point on the y-axis has its x-coordinate equal to 0. Since the point is 5 units below the origin (in the negative y-direction), its coordinates are (0, -5).

Exam Tip: Remember that points on the y-axis always have x = 0, and the y-coordinate tells you how far the point is from the origin - positive means above, negative means below.

 

Question 2. If the perpendicular distance of a point P from the x-axis is 5 units and the foot of perpendicular lies on the negative direction of x-axis, then the point P has
(a) x-coordinate = -5
(b) y-coordinate = 5 only
(c) y-coordinate = -5 only
(d) y-coordinate = 5 or -5
Answer: (d) y-coordinate = 5 or -5
In simple words: The foot of the perpendicular tells us the x-coordinate is negative. The distance of 5 units from the x-axis means the y-coordinate can be either 5 (above) or -5 (below). Both positions are 5 units away.

Exam Tip: Perpendicular distance from the x-axis is the absolute value of the y-coordinate. The point can lie on either side of the x-axis while maintaining the same distance.

 

Question 3. The points whose abscissa and ordinate have different signs will lie in
(a) I and II quadrants
(b) II and III quadrants
(c) I and III quadrants
(d) II and IV quadrants
Answer: (d) II and IV quadrants
In simple words: When x and y have opposite signs (one positive, one negative), the point must be in quadrant II (negative x, positive y) or quadrant IV (positive x, negative y).

Exam Tip: Always recall the signs in each quadrant: I (+,+), II (-,+), III (-,-), IV (+,-). Opposite signs occur only in quadrants II and IV.

 

Question 4. The points (-5, 2) and (2, -5) lie in
(a) same quadrant
(b) II and III quadrants respectively
(c) II and IV quadrants respectively
(d) IV and II quadrants respectively
Answer: (c) II and IV quadrants respectively
In simple words: The point (-5, 2) has a negative x-coordinate and positive y-coordinate, placing it in quadrant II. The point (2, -5) has a positive x-coordinate and negative y-coordinate, placing it in quadrant IV.

Exam Tip: Always check the sign of both coordinates separately before identifying the quadrant. The order matters - first coordinate determines left/right position, second determines up/down position.

 

Question 5. If P(-1, 1), Q(3, -4), R(1, -1), S(-2, -3) and T(-4, 4) are plotted on the graph paper, then point(s) in the fourth quadrant are
(a) P and T
(b) Q and R
(c) S only
(d) P and R
Answer: (b) Q and R
In simple words: Check each point: Q(3, -4) has positive x and negative y (quadrant IV). R(1, -1) has positive x and negative y (quadrant IV). All others have either both coordinates negative or both positive, placing them in different quadrants.

Exam Tip: For the fourth quadrant, you need x > 0 and y < 0. Quickly scan the sign of each coordinate pair to eliminate incorrect options.

 

Question 6. On plotting the points O(0, 0), A(3, 0), B(3, 4), C(0, 4) and joining OA, AB, BC, and CO which of the following figure is obtained?
(a) Square
(b) Rectangle
(c) Trapezium
(d) Rhombus
Answer: (b) Rectangle
In simple words: Point O is at the origin, A is 3 units right on the x-axis, B is 3 units right and 4 units up, and C is 4 units up on the y-axis. Connecting them forms a rectangle with length 3 and width 4.

Exam Tip: Calculate the side lengths: OA = 3, AB = 4, BC = 3, CO = 4. Since opposite sides are equal but adjacent sides are not, it is a rectangle, not a square.

 

Question 7. Which of the following points lie on the graph of the equation: 3x - 5y + 7 = 0?
(a) (1, -2)
(b) (2, 1)
(c) (-1, 2)
(d) (1, 2)
Answer: (d) (1, 2)
In simple words: Substitute each point into the equation. When you plug in x = 1 and y = 2, you get 3(1) - 5(2) + 7 = 3 - 10 + 7 = 0, which satisfies the equation. The other points do not.

Exam Tip: Always substitute both coordinates into the equation and simplify. The result must equal zero for the point to lie on the graph.

 

Question 8. The pair of equation x = a and y = b graphically represents lines which are
(a) parallel
(b) intersecting at (b, a)
(c) coincident
(d) intersecting at (a, b)
Answer: (d) intersecting at (a, b)
In simple words: The equation x = a is a vertical line passing through x = a on the x-axis. The equation y = b is a horizontal line passing through y = b on the y-axis. These two lines meet at the point (a, b).

Exam Tip: A vertical line and a horizontal line always intersect at exactly one point. That point has x-coordinate equal to a and y-coordinate equal to b.

 

Question 9. The distance of the point P(2, 3) from the x-axis is
(a) 2 units
(b) 3 units
(c) 1 unit
(d) 5 units
Answer: (b) 3 units
In simple words: The distance from any point to the x-axis is simply the absolute value of its y-coordinate. Since the y-coordinate is 3, the distance is 3 units.

Exam Tip: Remember: distance to x-axis = |y-coordinate|, and distance to y-axis = |x-coordinate|. This is much faster than using the distance formula.

 

Question 10. The distance of the point P(-4, 3) from the y-axis is
(a) 5 units
(b) -4 units
(c) 4 units
(d) 3 units
Answer: (c) 4 units
In simple words: The distance from any point to the y-axis equals the absolute value of its x-coordinate. Since the x-coordinate is -4, the distance is |-4| = 4 units.

Exam Tip: Distance is always positive. Even though the x-coordinate is negative, take its absolute value. Distance to y-axis ignores the y-coordinate entirely.

 

Question 11. The distance of the point P(-6, 8) from the origin is
(a) 8 units
(b) \( 2\sqrt{7} \) units
(c) 10 units
(d) 6 units
Answer: (c) 10 units
In simple words: Use the distance formula with the origin (0, 0) and point P(-6, 8). Calculate: \( d = \sqrt{(-6-0)^2 + (8-0)^2} = \sqrt{36 + 64} = \sqrt{100} = 10 \) units.

Exam Tip: When finding distance from the origin, the distance formula simplifies to \( d = \sqrt{x^2 + y^2} \). Recognize common Pythagorean triples like (6, 8, 10) to speed up calculation.

 

Question 12. The distance between the points A(0, 6) and B(0, -2) is
(a) 6 units
(b) 8 units
(c) 4 units
(d) 2 units
Answer: (b) 8 units
In simple words: Both points lie on the y-axis (x = 0). The distance is simply the difference in their y-coordinates: |6 - (-2)| = |6 + 2| = 8 units.

Exam Tip: When two points lie on the same axis, distance is just the difference in the varying coordinate. No need to use the full distance formula.

 

Question 13. The distance between the points (0, 5) and (-5, 0) is
(a) 5 units
(b) \( 5\sqrt{2} \) units
(c) \( 2\sqrt{5} \) units
(d) 10 units
Answer: (b) \( 5\sqrt{2} \) units
In simple words: Apply the distance formula: \( d = \sqrt{(-5-0)^2 + (0-5)^2} = \sqrt{25 + 25} = \sqrt{50} = \sqrt{25 \cdot 2} = 5\sqrt{2} \) units.

Exam Tip: When you get \( \sqrt{50} \), factor it as \( \sqrt{25 \cdot 2} \) and simplify to \( 5\sqrt{2} \). Always simplify square roots in your final answer.

 

Question 14. AOBC is a rectangle whose three vertices are A(0, 3), O(0, 0) and B(5, 0). The length of its diagonal is
(a) 5 units
(b) 3 units
(c) \( \sqrt{34} \) units
(d) 4 units
Answer: (c) \( \sqrt{34} \) units
In simple words: In rectangle AOBC, O is at the origin, A is above it (0, 3), and B is to the right (5, 0). The fourth vertex C is at (5, 3). The diagonal AB connects (0, 3) and (5, 0). Using the distance formula: \( d = \sqrt{(5-0)^2 + (0-3)^2} = \sqrt{25 + 9} = \sqrt{34} \) units.

Exam Tip: In a rectangle, the two diagonals are equal in length. You can use either diagonal. Always leave answers in simplified radical form unless told otherwise.

 

Question 15. If the distance between the points (2, -2) and (-1, x) is 5 units, then one of the value of x is
(a) -2
(b) 2
(c) -1
Answer: Using the distance formula:

\[ 5 = \sqrt{(-1-2)^2 + (x-(-2))^2} \]

\[ 5 = \sqrt{(-3)^2 + (x+2)^2} \]

\[ 5 = \sqrt{9 + (x+2)^2} \]

Squaring both sides:

\[ 25 = 9 + (x+2)^2 \]

\[ 16 = (x+2)^2 \]

\[ x + 2 = \pm 4 \]

\[ x = 2 \text{ or } x = -6 \]

Therefore, one value of x is 2.
In simple words: Set the distance equal to 5 and solve the resulting equation. You get two possible values for x. Either of these answers is correct.

Exam Tip: When you square both sides of a distance equation, you get two solutions due to the \( \pm \) sign. Both are valid unless the question specifies otherwise. Choose the one from the given options.

 

Question 19. The distance between the points (4, p) and (1, 0) is 5 units, then the value of p is
(a) 4 only
(b) -4 only
(c) ±4
(d) 0
Answer: (c) ±4
In simple words: Use the distance formula. When you square both sides and solve, you get two possible values for p: positive 4 and negative 4.

Exam Tip: Remember that when you take the square root of a number, you get both positive and negative solutions. Always check both values.

 

Question 20. The points (-4, 0), (4, 0) and (0, 3) are the vertices of a
(a) right triangle
(b) isosceles triangle
(c) equilateral triangle
(d) scalene triangle
Answer: (b) isosceles triangle
In simple words: Calculate the length of all three sides using the distance formula. Two of the sides have the same length, which means it's an isosceles triangle.

Exam Tip: Always find the lengths of all three sides when classifying triangles. An isosceles triangle has exactly two sides of equal length.

 

Question 21. The area of a square whose vertices are A(0, -2), B(3, 1), C(0, 4) and D(-3, 1) is
(a) 18 sq. units
(b) 15 sq. units
(c) \( \sqrt{18} \) units
(d) \( \sqrt{15} \) units
Answer: (a) 18 sq. units
In simple words: First, find the length of one side using the distance formula. Then square that length to get the area of the square.

Exam Tip: For a square, Area = (side)². Calculate the side length accurately and then square it without any rounding errors.

 

Question 22. In the adjoining figure, the area of triangle ABC is
(a) 15 sq. units
(b) 10 sq. units
(c) 7.5 sq. units
(d) 2.5 sq. units
Answer: (c) 7.5 sq. units
In simple words: Draw a perpendicular from point A to the base. Find the height and the base length. Then use the formula: Area = \( \frac{1}{2} \) × base × height.

Exam Tip: When finding the area of a triangle from a graph, always identify the base and the perpendicular height clearly. The height must be perpendicular to the base.

 

Question 23. The perimeter of a triangle with vertices (0, 4), (0, 0) and (3, 0) is
(a) 5 units
(b) 12 units
(c) 11 units
(d) 7 + \( \sqrt{5} \) units
Answer: (b) 12 units
In simple words: Find the length of each of the three sides using the distance formula. Add all three lengths together to get the perimeter.

Exam Tip: Perimeter is the sum of all three sides. Make sure you calculate each side length correctly before adding them up.

 

Question 24. If A is a point on the y-axis whose ordinate is 5 and B is the point (-3, 1), then the length of AB is
(a) 8 units
(b) 5 units
(c) 3 units
(d) 25 units
Answer: (b) 5 units
In simple words: Since A is on the y-axis, its x-coordinate is 0. So A = (0, 5). Use the distance formula to find the distance between A(0, 5) and B(-3, 1).

Exam Tip: Any point on the y-axis has x-coordinate = 0. The ordinate is the y-coordinate. Always set up the point correctly before applying the distance formula.

 

Question 25. The points A(9, 0), B(9, 6), C(-9, 6) and D(-9, 0) are the vertices of a
(a) rectangle
(b) square
(c) rhombus
(d) trapezium
Answer: (a) rectangle
In simple words: Calculate the lengths of all four sides. If opposite sides are equal and not all sides are equal, then it is a rectangle, not a square.

Exam Tip: For a quadrilateral, check if opposite sides are equal (rectangle/parallelogram) or all sides are equal (rhombus/square). A rectangle has opposite sides equal but not all sides equal.

 

Question 26. Consider the following two statements:
Statement 1: The point (x², y) lies on the y-axis. Then the value of x is zero.
Statement 2: The abscissa of every point on y-axis is zero.
Which of the following is valid?
(a) Both the statements are true.
(b) Both the statements are false.
(c) Statement 1 is true, and Statement 2 is false.
(d) Statement 1 is false, and Statement 2 is true.
Answer: (a) Both the statements are true.
In simple words: Any point on the y-axis has x-coordinate equal to zero. If (x², y) is on the y-axis, then x² must equal 0, which means x = 0. The abscissa (x-coordinate) of all points on the y-axis is always 0.

Exam Tip: Remember that the abscissa is the x-coordinate and the ordinate is the y-coordinate. All points on the y-axis have the form (0, y).

 

Question. Assertion Reason Type Questions

 

Question 1. (Assertion-Reason to be filled from next page)
Answer: [This section begins but content is cut off at page boundary. Complete assertion-reason question and answer would appear on the following page.]

 

Question. Assertion (A): Two ordered pairs (a, b) and (c, d) are equal if a = c and b = d.
Reason (R): If a ≠ b, then (a, b) ≠ (b, a).
Answer: Two ordered pairs (a, b) and (c, d) are equal if and only if a = c and b = d. This forms the fundamental definition of equality when comparing ordered pairs - corresponding elements must match exactly. So (a, b) = (c, d) happens precisely when a = c and b = d. Therefore, Assertion (A) is true. When a ≠ b, the pairs (a, b) and (b, a) are not equal. For instance, (2, 3) and (3, 2) represent different locations on a coordinate plane. Thus, Reason (R) is true. However, Reason (R) does not logically explain why Assertion (A) is true - the reason addresses a different property about ordered pairs. Therefore, the correct option is: Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct reason for Assertion (A).
In simple words: Two ordered pairs match when both numbers in each pair match. The reason given is true, but it explains something different, not why the first statement is true.

Exam Tip: When evaluating assertion-reason questions, check both statements separately first, then determine if the reason actually supports the assertion logically - they may both be true but unrelated.

 

Question 2. Assertion (A): The point P(3, -5) lies in quadrant II and the point Q(-2, -1) lies in quadrant III.
Reason (R): The signs of a point in I, II, III and IV quadrants are respectively (+, +), (-, +), (-, -) and (+, -).
Answer: For point P(3, -5): the x-coordinate is positive and the y-coordinate is negative. A point having a positive x-coordinate paired with a negative y-coordinate is located in quadrant IV, not quadrant II. For point Q(-2, -1): the x-coordinate is negative and the y-coordinate is negative. A point with both coordinates negative sits in quadrant III. Thus, Assertion (A) is false - P is in quadrant IV, not II. The quadrant sign patterns are: Quadrant I has (+, +), Quadrant II has (-, +), Quadrant III has (-, -), and Quadrant IV has (+, -). This statement matches the definition, so Reason (R) is true. Therefore, the correct option is: Assertion (A) is false, Reason (R) is true.
In simple words: Point P is in the wrong quadrant in the statement. The reason correctly describes which signs belong in each quadrant.

Exam Tip: Always verify sign combinations carefully - remember that the first coordinate tells you if you move left (negative) or right (positive) from the origin, and the second coordinate tells you if you move down (negative) or up (positive).

 

Question 3. Assertion (A): Two points A and B having coordinates (3, 3) and (-3, -3) respectively are joined. The line segment AB passes through the origin.
Reason (R): Origin is the point of intersection of the coordinate axes.
Answer: When any point (x, y) is reflected across the origin (0, 0), its image becomes (-x, -y) - this relationship is called point symmetry about the origin. Point A has coordinates (3, 3). If we reflect A(3, 3) through the origin, the image would be (-3, -3). Point B has exactly these coordinates: (-3, -3). Since B is the exact reflection of A through the origin, the origin must lie precisely on the line connecting A and B. Consequently, line segment AB passes through the origin, making Assertion (A) true. The origin is defined as the meeting point where the x-axis and y-axis intersect, with coordinates (0, 0). This is the standard definition in the Cartesian coordinate system. Thus, Reason (R) is true. However, Reason (R) does not explain why the line passes through the origin - it merely defines what the origin is. The correct option is: Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct reason for Assertion (A).
In simple words: The line between the two points does go through the center point (the origin). The reason describes what the origin is, but doesn't explain why the line passes through it.

Exam Tip: Look for the core reason why something is true - a definition might be correct but not be the actual explanation for the assertion at hand.

 

Question 4. Assertion (A): If the points A(2, 9), B(2, 5) and C(5, 5) are joined, then ΔABC is right angled.
Reason (R): If AC² = AB² + BC², then ∠B = 90°.
Answer: Using the distance formula with the given points A(2, 9), B(2, 5), and C(5, 5):

Distance between AB = \( \sqrt{(2-2)^2 + (5-9)^2} = \sqrt{0 + 16} = 4 \) units

Distance between BC = \( \sqrt{(5-2)^2 + (5-5)^2} = \sqrt{9 + 0} = 3 \) units

Distance between AC = \( \sqrt{(5-2)^2 + (5-9)^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \) units

Calculating the squares: AC² = 25, AB² = 16, BC² = 9

Since AC² = AB² + BC² (25 = 16 + 9), the Pythagorean theorem is satisfied. This means triangle ABC is right-angled at vertex B, making Assertion (A) true. The statement "If AC² = AB² + BC², then ∠B = 90°" correctly expresses the converse of the Pythagorean theorem, which identifies the right angle's location. Thus, Reason (R) is true and provides the mathematical basis for why Assertion (A) is true. The correct option is: Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct reason for Assertion (A).
In simple words: When the longest side squared equals the sum of the other two sides squared, the triangle has a right angle, and it occurs where the two shorter sides meet.

Exam Tip: Always calculate all three side lengths carefully and square them - then check if the Pythagorean relationship holds and identify which angle is 90 degrees.

 

Question 5. Assertion (A): Point (0, 9) is a point on y-axis which is equidistant from points (6, 5) and (-4, 3).
Reason (R): Abscissa of a point on y-axis is 0.
Answer: A point lies on the y-axis if and only if its x-coordinate equals zero. This fact makes Reason (R) true. Let P = (0, 9), A = (6, 5), and B = (-4, 3). Using the distance formula:

PA = \( \sqrt{(6-0)^2 + (5-9)^2} = \sqrt{36 + 16} = \sqrt{52} \)

PB = \( \sqrt{(-4-0)^2 + (3-9)^2} = \sqrt{16 + 36} = \sqrt{52} \)

Since PA = PB = \( \sqrt{52} \), point (0, 9) is equidistant from (6, 5) and (-4, 3), making Assertion (A) true. However, Reason (R) merely states a property of points on the y-axis and does not explain why point (0, 9) is equidistant from the two given points. The correct option is: Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct reason for Assertion (A).
In simple words: The point (0, 9) is indeed the same distance from both given points. The reason is true but explains only that the x-coordinate must be zero, not why equal distances occur.

Exam Tip: When checking equidistance, calculate both distances using the distance formula - if they match numerically, the equidistance property holds, regardless of whether the reason mentions this calculation.

 

Question 1. Three vertices of a rectangle are A(2, -1), B(2, 7) and C(4, 7). Plot these points on a graph and hence use it to find the coordinates of the fourth vertex D. Also find the coordinates of (i) the mid-point of BC (ii) the mid-point of CD (iii) the point of intersection of the diagonals. What is the area of the rectangle?
Answer: Scale: 1 block = 1 unit.

Steps of construction:
1. Plot points A(2, -1), B(2, 7), and C(4, 7) on graph paper.
2. Join AB and BC.
3. Measure AB. Draw line segment CD from point C parallel to the y-axis.
4. Measure BC. Draw line segment AD from point A parallel to the x-axis.
5. Mark F and G as the mid-points of BC and CD respectively.
6. Join AC and BD as the diagonals of the rectangle.
7. Mark E as the point where the diagonals intersect.

From the graph:
Coordinates of D = (4, -1).

(i) F is the mid-point of BC, and F = (3, 7). Therefore, the mid-point of BC has coordinates (3, 7).

(ii) G is the mid-point of CD, and G = (4, 3). Therefore, the mid-point of CD has coordinates (4, 3).

(iii) E is the point where the diagonals meet. Therefore, the intersection point of the diagonals has coordinates (3, 3).

From the graph: AB = 8 units and BC = 2 units.

Area of rectangle ABCD = length × breadth = AB × BC = 8 × 2 = 16 sq. units.
In simple words: When you draw the rectangle and mark its diagonals, they cross at point (3, 3). The rectangle spans 8 units in one direction and 2 units in the other, giving an area of 16 square units.

Exam Tip: Always label all points clearly on your graph and use construction lines to find the missing vertex - measure distances carefully to ensure the rectangle's properties hold (opposite sides equal, angles are 90 degrees).

 

Question 2. Three vertices of a parallelogram are A(3, 5), B(3, -1) and C(-1, -3). Plot these points on a graph paper and hence use it to find the coordinates of the fourth vertex D. Also find the coordinates of the mid-point of the side CD. What is the area of the parallelogram?
Answer: Steps of construction:
1. Plot points A(3, 5), B(3, -1), and C(-1, -3) on graph paper.
2. Join AB and BC.
3. From C, draw line CD parallel to AB such that CD = AB.
4. From A, draw line AD parallel to BC such that AD = BC.
5. Mark E as the mid-point of CD.

From the graph:
The coordinates of the fourth vertex D are (-1, 3).
The coordinates of the mid-point E of CD are (-1, 0).

Using the scale 1 block = 1 unit:
EF = 4 units (where F is the perpendicular from E to the base)
CD = 6 units

Area of parallelogram ABCD = Base × Height = CD × EF = 6 × 4 = 24 sq. units.

Therefore, D = (-1, 3), the mid-point of CD = (-1, 0), and the area of the parallelogram = 24 sq. units.
In simple words: In a parallelogram, opposite sides are equal and parallel. Once you place three vertices, the fourth one is found by maintaining these properties. The area equals the base times the perpendicular height.

Exam Tip: For a parallelogram, ensure that the diagonals bisect each other - check that the mid-points of both diagonals are the same. Use this property to verify your fourth vertex is correct.

 

Question 3(i). Draw the graphs of the following linear equation. y = 2x - 1. Also find slope and y-intercept of this line.
Answer: Given: y = 2x - 1

First, find points on the line by substituting values:
When x = 1: y = 2(1) - 1 = 1, giving point (1, 1)
When x = 2: y = 2(2) - 1 = 3, giving point (2, 3)
When x = 3: y = 2(3) - 1 = 5, giving point (3, 5)

Steps of construction:
1. Draw the x-axis and y-axis on graph paper with appropriate scales.
2. Plot the three points (1, 1), (2, 3), and (3, 5).
3. Join these points to form a straight line extending in both directions.

From the equation y = 2x - 1 (in the form y = mx + c):
- The slope m = 2
- The y-intercept c = -1 (the line crosses the y-axis at the point (0, -1))

The slope of 2 means that for every 1 unit increase in x, y increases by 2 units. The y-intercept of -1 shows where the line intersects the y-axis.
In simple words: This line goes upward steeply - it rises 2 units for every 1 unit you move to the right. It crosses the y-axis at the point where y equals -1.

Exam Tip: Always identify the slope and y-intercept from the equation in the form y = mx + c before plotting - this helps you understand the line's direction and starting position on the y-axis.

 

Question 3(ii). Draw the graphs of the following linear equation. 2x + 3y = 6. Also find slope and y-intercept of this line.
Answer: We have 2x + 3y = 6. Rearranging gives 3y = 6 - 2x, so y = 2 - (2x/3). This is equation (1). Now we find points on this line. When x = 0, y = 2 - 0 = 2. When x = 3, y = 2 - 2 = 0. When x = 6, y = 2 - 4 = -2.

Table of values:

Steps of construction:
(1) Plot the points (0, 2), (3, 0), and (6, -2) on the graph.
(2) Join the points to form a straight line.

Comparing equation (1) with y = mx + c, we find m = -2/3 and c = 2. Therefore, the slope of the line is -2/3 and the y-intercept is 2.
In simple words: First, rewrite the equation so y stands alone. Find three pairs of x and y values, plot them on graph paper, and draw the line through them. The slope tells you how steep the line is, and the y-intercept is where it crosses the y-axis.

Exam Tip: Always convert the equation to slope-intercept form (y = mx + c) before reading off the slope and y-intercept - this prevents sign errors. Also check that your plotted line passes through all three calculated points.

 

Question 3(iii). Draw the graphs of the following linear equation. 2x - 3y = 4. Also find slope and y-intercept of this line.
Answer: Given 2x - 3y = 4. Rearranging, we get 3y = 2x - 4, so y = (2x - 4)/3 = (2x/3) - (4/3). This is equation (1). We now calculate points. When x = -1, y = -2/3 - 4/3 = -6/3 = -2. When x = 2, y = 4/3 - 4/3 = 0. When x = 5, y = 10/3 - 4/3 = 6/3 = 2.

Table of values:

Steps of construction:
(1) Plot the points (-1, -2), (2, 0), and (5, 2) on the graph.
(2) Join the points to form a straight line.

Comparing equation (1) with y = mx + c, we find m = 2/3 and c = -4/3. Therefore, the slope of the line is 2/3 and the y-intercept is -4/3.
In simple words: Change the equation into y = mx + c form. Pick three x values, calculate their matching y values, plot the points, and draw the line. The number in front of x is the slope, and the constant term is where the line meets the y-axis.

Exam Tip: After plotting, verify slope direction - positive slope means the line goes up from left to right. Use the table values to double-check your line passes through all three points.

 

Question 4. Draw the graph of the equation 3x - 4y = 12. From the graph, find: (i) the value of y when x = -4; (ii) the value of x when y = 3.
Answer: From 3x - 4y = 12, we rearrange to get 4y = 3x - 12, so y = (3x/4) - 3. This is equation (1). We calculate points: When x = 0, y = 0 - 3 = -3. When x = 4, y = 3 - 3 = 0. When x = 8, y = 6 - 3 = 3.

Table of values:

Steps of construction:
(1) Plot the points (0, -3), (4, 0), and (8, 3) on the graph.
(2) Join the points to form a straight line.

(i) To find y when x = -4: From point N at x = -4, draw a vertical line parallel to the y-axis touching the graph at point M. From M, draw a horizontal line parallel to the x-axis to meet the y-axis at point O. Reading from the graph, when x = -4, the value of y is -6.

(ii) To find x when y = 3: From point P at y = 3, draw a horizontal line parallel to the x-axis touching the graph at point R. From R, draw a vertical line parallel to the y-axis to meet the x-axis at point Q. Reading from the graph, when y = 3, the value of x is 8.
In simple words: Plot three points using the equation, then draw the line through them. To read values from the graph, go horizontally or vertically from the axis to the line, then move to the other axis to read the answer.

Exam Tip: When reading values from a graph, always use a ruler or straight edge to ensure your horizontal and vertical lines are truly parallel to the axes - this gives accuracy.

 

Question 5. Solve graphically, the simultaneous equations: 2x - 3y = 7; x + 6y = 11.
Answer: For equation 2x - 3y = 7, we rearrange: 3y = 2x - 7, so y = (2x - 7)/3. This is equation (1). Finding points: When x = -1, y = (-2 - 7)/3 = -9/3 = -3. When x = 2, y = (4 - 7)/3 = -3/3 = -1. When x = 5, y = (10 - 7)/3 = 3/3 = 1.

Table of values for equation (1):

Steps of construction:
(1) Plot the points (-1, -3), (2, -1), and (5, 1) on the graph.
(2) Join the points to form a straight line.

For equation x + 6y = 11, we rearrange: 6y = 11 - x, so y = (11 - x)/6. This is equation (2). Finding points: When x = -7, y = 18/6 = 3. When x = -1, y = 12/6 = 2. When x = 5, y = 6/6 = 1.

Table of values for equation (2):

Steps of construction:
(1) Plot the points (-7, 3), (-1, 2), and (5, 1) on the graph.
(2) Join the points to form a straight line.

From the graph, the two lines intersect at point P(5, 1). Therefore, x = 5 and y = 1.
In simple words: Rearrange both equations to isolate y. Create a table of x and y values for each equation and plot both lines on the same graph. Where the lines cross is your answer - read both coordinates from that point.

Exam Tip: Always plot at least three points per line and verify they are collinear before drawing the line. Mark the intersection point clearly with a dot and label it with its coordinates to show your final answer.

 

Question 6. Solve the following system of equations graphically: x - 2y - 4 = 0, 2x + y - 3 = 0.
Answer: From x - 2y - 4 = 0, we get 2y = x - 4, so y = (x - 4)/2. This is equation (1). Finding points: When x = -2, y = (-2 - 4)/2 = -6/2 = -3. When x = 0, y = (0 - 4)/2 = -4/2 = -2. When x = 2, y = (2 - 4)/2 = -2/2 = -1.

Table of values for equation (1):

Steps of construction:
(1) Plot the points (-2, -3), (0, -2), and (2, -1) on the graph.
(2) Join the points to form a straight line.

From 2x + y - 3 = 0, we get y = 3 - 2x. This is equation (2). Finding points: When x = 0, y = 3 - 0 = 3. When x = 1, y = 3 - 2 = 1. When x = 2, y = 3 - 4 = -1.

Table of values for equation (2):

Steps of construction:
(1) Plot the points (0, 3), (1, 1), and (2, -1) on the graph.
(2) Join the points to form a straight line.

From the graph, the lines intersect each other at P(2, -1). Hence, x = 2 and y = -1.
In simple words: Isolate y in both equations. Make tables of values by choosing different x values and calculating y. Plot both sets of points, draw both lines, and find where they meet - that point is your solution.

Exam Tip: Ensure your two lines are clearly distinct and that they actually intersect on your graph paper (not off the edge). The intersection point must satisfy both original equations when you substitute back.

 

Question 7. Using a scale of 1 cm to 1 unit for both the axes, draw the graphs of the following equations: 6y = 5x + 10, y = 5x - 15. From the graph, find (i) the coordinates of the point where the two lines intersect; (ii) the area of the triangle between the lines and the x-axis.
Answer: (i) From 6y = 5x + 10, we get y = (5x + 10)/6. This is equation (1). Finding points: When x = 1, y = 15/6 = 2.5. When x = -2, y = 0/6 = 0. When x = 4, y = 30/6 = 5.

Table of values for equation (1):

Steps of construction:
(1) Plot the points (1, 2.5), (-2, 0), and (4, 5) on the graph.
(2) Join the points to form a straight line.

From y = 5x - 15 (equation 2), finding points: When x = 2.5, y = 12.5 - 15 = -2.5. When x = 3, y = 15 - 15 = 0. When x = 4, y = 20 - 15 = 5.

Table of values for equation (2):

Steps of construction:
(1) Plot the points (2.5, -2.5), (3, 0), and (4, 5) on the graph.
(2) Join the points to form a straight line.

From the graph, the lines intersect at point P(4, 5). Hence, x = 4 and y = 5.

(ii) From the graph, the triangle formed is PQR, where Q and R are the points where each line meets the x-axis. Draw a perpendicular line PJ from point P to the x-axis. The height PJ = 5 units, and the base QR = 5 units. Area of triangle = (1/2) × base × height = (1/2) × 5 × 5 = 25/2 = 12.5 sq. units. Hence, area = 12.5 sq. units.
In simple words: Plot both lines on the same graph using the calculated points. They will meet at one point - write down those coordinates. To find the triangle's area, measure the height from that point down to the x-axis, measure the width along the x-axis between where the two lines cross it, then use the area formula (1/2) × base × height.

Exam Tip: Use the given scale (1 cm to 1 unit) carefully to ensure accurate plotting. For the area calculation, identify the three vertices of the triangle clearly on your graph before applying the formula.

 

Question 8. Find, graphically, the coordinates of the vertices of the triangle formed by the lines: 8y - 3x + 7 = 0, 2x - y + 4 = 0 and 5x + 4y = 29.
Answer: From 8y - 3x + 7 = 0, we get 8y = 3x - 7, so y = (3x - 7)/8. This is equation (1). Finding points: When x = -3, y = (-9 - 7)/8 = -16/8 = -2. When x = 1, y = (3 - 7)/8 = -4/8 = -0.5. When x = 5, y = (15 - 7)/8 = 8/8 = 1.

Table of values for equation (1):

Steps of construction:
(1) Plot the points (-3, -2), (1, -0.5), and (5, 1) on the graph.
(2) Join the points to form a straight line.

From 2x - y + 4 = 0, we get y = 2x + 4. This is equation (2). Finding points: When x = -3, y = -6 + 4 = -2. When x = -1, y = -2 + 4 = 2. When x = 1, y = 2 + 4 = 6.

Steps of construction:
(1) Plot these points on the graph.
(2) Join the points to form a straight line.

From 5x + 4y = 29, we get y = (29 - 5x)/4. This is equation (3). Finding points: When x = 1, y = 24/4 = 6. When x = 3, y = 14/4 = 3.5. When x = 5, y = 4/4 = 1.

Steps of construction:
(1) Plot these points on the graph.
(2) Join the points to form a straight line.

The three lines intersect pairwise at three distinct points, forming a triangle. Reading from the graph, the vertices of the triangle are at the coordinates where each pair of lines meets. Identify these intersection points from your plotted graph.
In simple words: Rewrite all three equations with y isolated. Make a table of values for each equation. Plot all three lines on one graph. Find the three points where each pair of lines crosses - these are the triangle's corners.

Exam Tip: Plot at least three points on each line to ensure accuracy. Extend your lines far enough so that all three intersection points appear on your graph. Label each intersection point with its coordinates clearly.

 

Question 9. Find graphically the coordinates of the vertices of the triangle formed by the lines y - 2 = 0, 2y + x = 0 and y + 1 = 3 (x - 2). Hence, find the area of the triangle formed by these lines.
Answer: First, we simplify the three given equations. The first equation \( y - 2 = 0 \) simplifies to \( y = 2 \). The second equation \( 2y + x = 0 \) becomes \( y = -\frac{x}{2} \). The third equation \( y + 1 = 3(x - 2) \) expands to \( y = 3x - 7 \).

For equation (2), when \( x = -2 \), \( y = 1 \); when \( x = 0 \), \( y = 0 \); when \( x = 2 \), \( y = -1 \). Plot these points and join them to form a straight line.

For equation (3), when \( x = 1 \), \( y = -4 \); when \( x = 2 \), \( y = -1 \); when \( x = 3 \), \( y = 2 \). Plot these points and join them to form a straight line.

From the graph, the three lines intersect at three points: \( A(-4, 2) \), \( B(3, 2) \), and \( C(2, -1) \). These are the vertices of the triangle.

To find the area, draw a perpendicular from \( C \) to line \( AB \). Since the lines are on graph paper where 1 block = 1 unit, we measure: \( AB = 7 \) units and the perpendicular distance \( CD = 3 \) units.

Area of triangle \( = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 7 \times 3 = \frac{21}{2} = 10.5 \) square units.
In simple words: Plot all three lines on a graph by finding points that fit each equation. The places where the lines cross are the triangle's corners. Measure the base and height on the graph, then use the triangle area formula.

Exam Tip: Always check that all three intersection points lie on the correct lines before calculating area. Use the perpendicular from one vertex to the opposite side for an accurate height measurement.

 

Question 10. A line segment is of length 10 units and one of its ends is (-2, 3). If the ordinate of the other end is 9, find the abscissa of the other end.
Answer: Let the unknown abscissa be \( x \), so the other end has coordinates \( (x, 9) \). We use the distance formula: \( d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \).

Since the distance is 10 units, we have:
\[ \sqrt{[x - (-2)]^2 + (9 - 3)^2} = 10 \]

\[ \sqrt{[x + 2]^2 + 6^2} = 10 \]

Squaring both sides:
\[ (x + 2)^2 + 36 = 100 \]

\[ x^2 + 4x + 4 + 36 = 100 \]

\[ x^2 + 4x + 40 = 100 \]

\[ x^2 + 4x - 60 = 0 \]

We factor this by splitting the middle term:
\[ x^2 + 10x - 6x - 60 = 0 \]

\[ x(x + 10) - 6(x + 10) = 0 \]

\[ (x - 6)(x + 10) = 0 \]

This gives us \( x = 6 \) or \( x = -10 \).
In simple words: We know one point and the total distance. We use the distance formula to set up an equation, solve for the unknown x-coordinate, and get two possible answers.

Exam Tip: Quadratic equations from distance problems often yield two solutions; both are valid unless the problem specifies additional constraints. Always verify both solutions satisfy the original distance.

 

Question 11. A(-4, -1), B(-1, 2) and C(α, 5) are the vertices of an isosceles triangle. Find the value of α given that AB is the unequal side.
Answer: Since triangle ABC is isosceles with AB as the unequal side, the two equal sides must be \( AC \) and \( BC \). Therefore, \( AC = BC \).

Using the distance formula \( d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \):

\[ AC = \sqrt{[\alpha - (-4)]^2 + [5 - (-1)]^2} = \sqrt{(\alpha + 4)^2 + 6^2} \]

\[ BC = \sqrt{[\alpha - (-1)]^2 + [5 - 2]^2} = \sqrt{(\alpha + 1)^2 + 3^2} \]

Setting \( AC = BC \) and squaring both sides:
\[ (\alpha + 4)^2 + 36 = (\alpha + 1)^2 + 9 \]

\[ \alpha^2 + 8\alpha + 16 + 36 = \alpha^2 + 2\alpha + 1 + 9 \]

\[ \alpha^2 + 8\alpha + 52 = \alpha^2 + 2\alpha + 10 \]

Canceling \( \alpha^2 \) from both sides:
\[ 8\alpha + 52 = 2\alpha + 10 \]

\[ 6\alpha = -42 \]

\[ \alpha = -7 \]
In simple words: In an isosceles triangle, two sides must be equal. We set the two equal sides' distances from the distance formula, solve for the unknown coordinate.

Exam Tip: Always identify which sides are equal before applying the distance formula. Squaring both sides removes the square roots, simplifying the algebra.

 

Question 12. If A(-3, 2), B(α, β) and C(-1, 4) are the vertices of an isosceles triangle, prove that α + β = 1, given AB = BC.
Answer: Since the triangle is isosceles with \( AB = BC \), we apply the distance formula to both sides:

\[ AB = \sqrt{(\alpha + 3)^2 + (\beta - 2)^2} \]

\[ BC = \sqrt{(-1 - \alpha)^2 + (4 - \beta)^2} \]

Setting them equal and squaring:
\[ (\alpha + 3)^2 + (\beta - 2)^2 = (-1 - \alpha)^2 + (4 - \beta)^2 \]

Expanding the left side:
\[ \alpha^2 + 6\alpha + 9 + \beta^2 - 4\beta + 4 = 1 + 2\alpha + \alpha^2 + 16 - 8\beta + \beta^2 \]

\[ \alpha^2 + 6\alpha + 9 + \beta^2 - 4\beta + 4 = \alpha^2 + 2\alpha + 17 + \beta^2 - 8\beta \]

Canceling \( \alpha^2 \) and \( \beta^2 \) from both sides:
\[ 6\alpha + 9 - 4\beta + 4 = 2\alpha + 17 - 8\beta \]

\[ 6\alpha - 2\alpha - 4\beta + 8\beta = 17 - 9 - 4 \]

\[ 4\alpha + 4\beta = 4 \]

\[ 4(\alpha + \beta) = 4 \]

\[ \alpha + \beta = 1 \]
In simple words: We set the two equal sides' distances equal, expand both sides of the equation, simplify by canceling like terms, and isolate the sum α + β.

Exam Tip: When proving identities with the distance formula, expand fully and organize terms carefully. Cancel matching squared terms early to reduce complexity and reveal the required relationship.

 

Question 13. Prove that the points (3, 0), (6, 4) and (-1, 3) are the vertices of a right angled isosceles triangle.
Answer: Let \( A(3, 0) \), \( B(6, 4) \), and \( C(-1, 3) \) be the three points. We find the distances between each pair using the distance formula \( d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \).

For side \( AB \):
\[ AB = \sqrt{(6 - 3)^2 + (4 - 0)^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \text{ units} \]

For side \( BC \):
\[ BC = \sqrt{(-1 - 6)^2 + (3 - 4)^2} = \sqrt{(-7)^2 + (-1)^2} = \sqrt{49 + 1} = \sqrt{50} = 5\sqrt{2} \text{ units} \]

For side \( AC \):
\[ AC = \sqrt{(-1 - 3)^2 + (3 - 0)^2} = \sqrt{(-4)^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \text{ units} \]

Since \( AB = AC = 5 \), the triangle is isosceles. Now we check if it is right-angled using the Pythagorean theorem. We check if \( AB^2 + AC^2 = BC^2 \):

\[ AB^2 + AC^2 = 5^2 + 5^2 = 25 + 25 = 50 \]

\[ BC^2 = (5\sqrt{2})^2 = 50 \]

Since \( AB^2 + AC^2 = BC^2 \), the triangle satisfies the Pythagorean theorem, making it right-angled at vertex \( A \). Combined with two equal sides, the triangle is a right-angled isosceles triangle.
In simple words: We calculate all three side lengths. Two sides are equal (5 units each), making it isosceles. The square of the longest side equals the sum of squares of the other two, proving it's also a right angle.

Exam Tip: For right-angled isosceles triangles, always verify both conditions: two equal sides, and the Pythagorean relation holding with equality. State which vertex has the right angle for complete clarity.

 

Question 14(i). Show that the points (2, 1), (0, 3), (-2, 1) and (0, -1), taken in order, are the vertices of a square. Also find the area of the square.
Answer: Let \( A(2, 1) \), \( B(0, 3) \), \( C(-2, 1) \), and \( D(0, -1) \) be the four points in order. Using the distance formula, we find all four side lengths.

For side \( AB \):
\[ AB = \sqrt{(0 - 2)^2 + (3 - 1)^2} = \sqrt{(-2)^2 + 2^2} = \sqrt{4 + 4} = \sqrt{8} \text{ units} \]

For side \( BC \):
\[ BC = \sqrt{(-2 - 0)^2 + (1 - 3)^2} = \sqrt{(-2)^2 + (-2)^2} = \sqrt{4 + 4} = \sqrt{8} \text{ units} \]

For side \( CD \):
\[ CD = \sqrt{(0 - (-2))^2 + (-1 - 1)^2} = \sqrt{2^2 + (-2)^2} = \sqrt{4 + 4} = \sqrt{8} \text{ units} \]

For side \( AD \):
\[ AD = \sqrt{(0 - 2)^2 + (-1 - 1)^2} = \sqrt{(-2)^2 + (-2)^2} = \sqrt{4 + 4} = \sqrt{8} \text{ units} \]

All four sides are equal: \( AB = BC = CD = AD = \sqrt{8} \) units. This means the quadrilateral is at least a rhombus. To confirm it is a square, we check the diagonals.

For diagonal \( AC \):
\[ AC = \sqrt{(-2 - 2)^2 + (1 - 1)^2} = \sqrt{(-4)^2 + 0^2} = \sqrt{16} = 4 \text{ units} \]

For diagonal \( BD \):
\[ BD = \sqrt{(0 - 0)^2 + (-1 - 3)^2} = \sqrt{0 + (-4)^2} = \sqrt{16} = 4 \text{ units} \]

Both diagonals are equal (\( AC = BD = 4 \) units). Since all sides are equal and both diagonals are equal, \( ABCD \) is a square.

The area of the square is calculated using the side length:
\[ \text{Area} = (\sqrt{8})^2 = 8 \text{ square units} \]
In simple words: We check that all four sides have the same length using the distance formula. We also verify that both diagonal lengths match. When all sides are equal and both diagonals are equal, the shape is a square. The area is the side length squared.

Exam Tip: To prove a quadrilateral is a square, show both that all four sides are equal and that the diagonals are equal in length. Either property alone (all sides equal) only proves it's a rhombus; both conditions together guarantee a square.

 

Question 14(ii). Show that the points (-3, 2), (-5, -5), (2, -3) and (4, 4), taken in order, are the vertices of rhombus. Also, find its area. Do the given points form a square?
Answer: Take A(-3, 2), B(-5, -5), C(2, -3) and D(4, 4).

Using the distance formula, \( d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \):

\( AB = \sqrt{[-5 - (-3)]^2 + [-5 - 2]^2} = \sqrt{[-2]^2 + [-7]^2} = \sqrt{4 + 49} = \sqrt{53} \)

\( BC = \sqrt{[2 - (-5)]^2 + [-3 - (-5)]^2} = \sqrt{7^2 + 2^2} = \sqrt{49 + 4} = \sqrt{53} \)

\( CD = \sqrt{(4 - 2)^2 + [4 - (-3)]^2} = \sqrt{2^2 + 7^2} = \sqrt{4 + 49} = \sqrt{53} \)

\( AD = \sqrt{[4 - (-3)]^2 + (4 - 2)^2} = \sqrt{7^2 + 2^2} = \sqrt{49 + 4} = \sqrt{53} \)

All four sides are equal. Now check the diagonals:

\( AC = \sqrt{[2 - (-3)]^2 + [-3 - 2]^2} = \sqrt{5^2 + [-5]^2} = \sqrt{25 + 25} = \sqrt{50} = 5\sqrt{2} \text{ units} \)

\( BD = \sqrt{[4 - (-5)]^2 + [4 - (-5)]^2} = \sqrt{9^2 + 9^2} = \sqrt{81 + 81} = \sqrt{162} = 9\sqrt{2} \text{ units} \)

Since all four sides are equal but the diagonals are not equal, ABCD is a rhombus (not a square).

Area of rhombus = \( \frac{1}{2} \times d_1 \times d_2 = \frac{1}{2} \times 5\sqrt{2} \times 9\sqrt{2} = \frac{1}{2} \times 45 \times 2 = 45 \) sq. units.

Hence, ABCD forms a rhombus with area 45 sq. units, and the points do NOT form a square because the diagonals are unequal.
In simple words: When all four sides of a four-sided shape are the same length, it could be a rhombus or a square. To tell the difference, check the diagonals - if they are equal, it is a square; if they are different, it is a rhombus.

Exam Tip: Always calculate all four sides and both diagonals. For a square, all four sides AND both diagonals must be equal. For a rhombus, the four sides are equal but diagonals are not.

 

Question 15. The ends of a diagonal of a square have coordinates (-2, p) and (p, 2). Find p if the area of the square is 40 sq. units.
Answer: Given: The ends of a diagonal of a square are (-2, p) and (p, 2), and the area is 40 sq. units.

Using the area formula for a square, \( \text{Area} = (\text{side})^2 \)

\( (\text{side})^2 = 40 \)

\( \text{side} = \sqrt{40} = 2\sqrt{10} \text{ units} \)

The diagonal of a square is related to the side by: \( \text{Diagonal} = \sqrt{2} \times \text{side} = \sqrt{2} \times 2\sqrt{10} = 2\sqrt{20} \)

Using the distance formula for the diagonal between (-2, p) and (p, 2):

\( 2\sqrt{20} = \sqrt{[p - (-2)]^2 + [2 - p]^2} \)

\( 2\sqrt{20} = \sqrt{[p + 2]^2 + [2 - p]^2} \)

\( 2\sqrt{20} = \sqrt{p^2 + 4p + 4 + 4 - 4p + p^2} \)

\( 2\sqrt{20} = \sqrt{2p^2 + 8} \)

Squaring both sides:

\( 4 \times 20 = 2p^2 + 8 \)

\( 80 = 2p^2 + 8 \)

\( 2p^2 = 72 \)

\( p^2 = 36 \)

\( p = \pm 6 \)

Hence, p = ±6.
In simple words: Find the side length using the area. The diagonal is \( \sqrt{2} \) times the side length. Use the distance formula to set the diagonal length equal to the coordinates and solve for p.

Exam Tip: Remember the relationship between diagonal and side in a square: diagonal = \( \sqrt{2} \) × side. When you square both sides of an equation, check that both positive and negative solutions work in the original context.

 

Question 16. What type of quadrilateral do the points A(2, -2), B(7, 3), C(11, -1) and D(6, -6), taken in that order, form?
Answer: Use the distance formula to find all four side lengths:

\( AB = \sqrt{(7 - 2)^2 + [3 - (-2)]^2} = \sqrt{5^2 + 5^2} = \sqrt{25 + 25} = \sqrt{50} \)

\( BC = \sqrt{(11 - 7)^2 + (-1 - 3)^2} = \sqrt{4^2 + (-4)^2} = \sqrt{16 + 16} = \sqrt{32} \)

\( CD = \sqrt{(6 - 11)^2 + [-6 - (-1)]^2} = \sqrt{(-5)^2 + (-5)^2} = \sqrt{25 + 25} = \sqrt{50} \)

\( AD = \sqrt{(6 - 2)^2 + [-6 - (-2)]^2} = \sqrt{4^2 + (-4)^2} = \sqrt{16 + 16} = \sqrt{32} \)

We see that \( AB = CD \) and \( BC = AD \). Since opposite sides are equal, ABCD is a parallelogram. More specifically, because the opposite sides are equal but the adjacent sides are not all equal, ABCD is a rectangle.

Hence, ABCD is a rectangle.
In simple words: Measure all four sides. If opposite sides match but not all four sides are the same, the shape is a rectangle. If opposite sides match and all four are equal, it is a square.

Exam Tip: To classify a quadrilateral, always calculate all four sides. Check if opposite sides are equal (rectangle/parallelogram) or all sides are equal (rhombus/square). Then check the diagonals if needed to distinguish between these types.

 

Question 17. Find the coordinates of the centre of the circle passing through the three given points A(5, 1), B(-3, -7) and C(7, -1).
Answer: Let O(x, y) be the centre of the circle. Since all three points A(5, 1), B(-3, -7), and C(7, -1) lie on the circle, they are equidistant from the centre O.

Using the condition OA = OB (both are radii):

\( (x - 5)^2 + (y - 1)^2 = [x - (-3)]^2 + [y - (-7)]^2 \)

\( x^2 - 10x + 25 + y^2 - 2y + 1 = x^2 + 6x + 9 + y^2 + 14y + 49 \)

\( x^2 + y^2 - 10x - 2y + 26 = x^2 + y^2 + 6x + 14y + 58 \)

\( -10x - 2y + 26 = 6x + 14y + 58 \)

\( -16x - 16y = 32 \)

\( x + y = -2 \)

\( x = -2 - y \) .........(1)

Using the condition OC = OB (both are radii):

\( (x - 7)^2 + [y - (-1)]^2 = [x - (-3)]^2 + [y - (-7)]^2 \)

\( x^2 - 14x + 49 + y^2 + 2y + 1 = x^2 + 6x + 9 + y^2 + 14y + 49 \)

\( x^2 + y^2 - 14x + 2y + 50 = x^2 + y^2 + 6x + 14y + 58 \)

\( -14x + 2y + 50 = 6x + 14y + 58 \)

\( -20x - 12y = 8 \)

\( 5x + 3y = -2 \) .........(2)

Substitute equation (1) into equation (2):

\( 5(-2 - y) + 3y = -2 \)

\( -10 - 5y + 3y = -2 \)

\( -2y = 8 \)

\( y = -4 \)

\( x = -2 - (-4) = 2 \)

Hence, the coordinates of the centre of the circle are (2, -4).
In simple words: The centre of a circle is the same distance from every point on the circle. Set up two equations using the fact that the distance from the centre to each of the three points must be equal, then solve for x and y.

Exam Tip: When finding the centre of a circle through three points, use the equidistance property: OA = OB and OB = OC. Expand and simplify carefully to avoid algebraic errors. Always verify your answer by checking that all three points are equidistant from the found centre.