ML Aggarwal Class 10 Maths Solutions Chapter 12 Equation of Straight Line

Question 1. Find the slope of a line whose inclination is
(i) 45°
(ii) 30°
Answer:
(i) Let m be the slope. Then \( m = \tan 45° = 1 \).
The slope is 1.

(ii) Let m be the slope. Then \( m = \tan 30° = \frac{1}{\sqrt{3}} \).
The slope is \( \frac{1}{\sqrt{3}} \).
In simple words: The slope of a line tells us how steep it is. When the line tips up at 45 degrees, the slope is 1. When it tips at 30 degrees, the slope is \( \frac{1}{\sqrt{3}} \).

Exam Tip: Remember that slope m always equals tan(θ), where θ is the inclination angle. Memorize the tan values for common angles like 30°, 45°, and 60°.

Question 2. Find the inclination of the line whose gradient is
(i) 1
(ii) \( \sqrt{3} \)
(iii) \( \frac{1}{\sqrt{3}} \)
Answer:
(i) Let the inclination be θ. We know that m = tan θ.
\( 1 = \tan θ \implies 1 = \tan 45° \)
\( \implies θ = 45° \)
The inclination is 45°.

(ii) Let the inclination be θ. We have m = tan θ.
\( \sqrt{3} = \tan θ \implies \sqrt{3} = \tan 60° \)
\( \implies θ = 60° \)
The inclination is 60°.

(iii) Let the inclination be θ. We have m = tan θ.
\( \frac{1}{\sqrt{3}} = \tan θ \implies \frac{1}{\sqrt{3}} = \tan 30° \)
\( \implies θ = 30° \)
The inclination is 30°.
In simple words: Inclination is the angle the line makes with the x-axis. If you know the slope (gradient), you can find the angle by taking the inverse tangent.

Exam Tip: Always use the relationship m = tan θ to move between slope and inclination. For part (iii), recognize that \( \frac{1}{\sqrt{3}} \) is the tangent of 30°, not 60°.

Question 3. Find the equation of a straight line parallel to x-axis which is at a distance
(i) 2 units above it
(ii) 3 units below it
Answer:
(i) Any line parallel to the x-axis has the form y = b, where b is a constant. Since this line is 2 units above the x-axis, b = 2.
The equation is \( y = 2 \) or \( y - 2 = 0 \).

(ii) For a line parallel to the x-axis that sits 3 units below it, we have b = -3.
The equation is \( y = -3 \) or \( y + 3 = 0 \).
In simple words: Lines parallel to the x-axis are always horizontal. If it is above the x-axis, y is positive. If it is below, y is negative.

Exam Tip: Lines parallel to the x-axis always have equations of the form y = constant. "Above" means positive y, "below" means negative y.

Question 4. Find the equation of a straight line parallel to y-axis which is at a distance
(i) 3 units to the right
(ii) 2 units to the left
Answer:
(i) Any line parallel to the y-axis has the form x = a, where a is a constant. Since this line is 3 units to the right of the y-axis, a = 3.
The equation is \( x = 3 \) or \( x - 3 = 0 \).

(ii) For a line parallel to the y-axis that sits 2 units to the left of it, we have a = -2.
The equation is \( x = -2 \) or \( x + 2 = 0 \).
In simple words: Lines parallel to the y-axis are always vertical. If it is to the right, x is positive. If it is to the left, x is negative.

Exam Tip: Lines parallel to the y-axis have equations x = constant. "Right" means positive x, "left" means negative x.

Question 5. Find the equation of a straight line parallel to y-axis and passing through the point (-3, 5).
Answer: For a line parallel to the y-axis, the equation is x = a. Since the line passes through (-3, 5), the x-coordinate must be -3.
Therefore, a = -3, and the equation is \( x = -3 \) or \( x + 3 = 0 \).
In simple words: Any point on a vertical line (parallel to the y-axis) always has the same x-value. So if the line passes through (-3, 5), all points on it have x = -3.

Exam Tip: When a vertical line passes through a given point, its equation is simply x = (the x-coordinate of that point).

Question 6. Find the equation of a line whose
(i) slope = 3, y-intercept = -5
(ii) slope = \( -\frac{2}{7} \), y-intercept = 3
(iii) gradient = \( \sqrt{3} \), y-intercept = \( -\frac{4}{3} \)
(iv) inclination = 30°, y-intercept = 2
Answer:
(i) Using the form \( y = mx + c \) where m = 3 and c = -5:
\( y = 3x - 5 \)

(ii) With m = \( -\frac{2}{7} \) and c = 3:
\( y = -\frac{2}{7}x + 3 \)

\[ \implies 7y = -2x + 21 \implies 2x + 7y - 21 = 0 \]

(iii) With m = \( \sqrt{3} \) and c = \( -\frac{4}{3} \):
\( y = \sqrt{3}x - \frac{4}{3} \)

\[ \implies 3y = 3\sqrt{3}x - 4 \implies 3\sqrt{3}x - 3y - 4 = 0 \]

(iv) Since inclination = 30°, the slope \( m = \tan 30° = \frac{1}{\sqrt{3}} \). With c = 2:
\( y = \frac{1}{\sqrt{3}}x + 2 \)

\[ \implies \sqrt{3}y = x + 2\sqrt{3} \implies x - \sqrt{3}y + 2\sqrt{3} = 0 \]
In simple words: To find a line's equation, use y = mx + c. Plug in the slope (m) and y-intercept (c), then rearrange if needed.

Exam Tip: For part (iv), convert the inclination angle to slope using m = tan(θ). Always clear fractions by multiplying through to get the final form.

Question 7. Find the slope and y-intercept of the following lines:
(i) x - 2y - 1 = 0
(ii) 4x - 5y - 9 = 0
(iii) 3x + 5y + 7 = 0
(iv) \( \frac{x}{3} + \frac{y}{4} = 1 \)
(v) y - 3 = 0
(vi) x - 3 = 0
Answer:
(i) Rearrange to slope-intercept form:
\( x - 2y - 1 = 0 \implies 2y = x - 1 \implies y = \frac{x}{2} - \frac{1}{2} \)
Comparing with y = mx + c: m = \( \frac{1}{2} \), c = \( -\frac{1}{2} \)

(ii) \( 4x - 5y - 9 = 0 \implies 5y = 4x - 9 \implies y = \frac{4x}{5} - \frac{9}{5} \)
m = \( \frac{4}{5} \), c = \( -\frac{9}{5} \)

(iii) \( 3x + 5y + 7 = 0 \implies 5y = -3x - 7 \implies y = -\frac{3x}{5} - \frac{7}{5} \)
m = \( -\frac{3}{5} \), c = \( -\frac{7}{5} \)

(iv) \( \frac{x}{3} + \frac{y}{4} = 1 \implies 4x + 3y = 12 \implies 3y = -4x + 12 \implies y = -\frac{4}{3}x + 4 \)
m = \( -\frac{4}{3} \), c = 4

(v) \( y - 3 = 0 \implies y = 3 = 0 \cdot x + 3 \)
m = 0, c = 3

(vi) \( x - 3 = 0 \implies x = 3 \)
This is a vertical line. The slope is undefined and there is no y-intercept because the line never crosses the y-axis.
In simple words: Convert every equation to the form y = mx + c. The coefficient of x is the slope. The constant term is the y-intercept. For vertical lines (x = constant), slope is undefined.

Exam Tip: Always isolate y to identify slope and y-intercept directly. For vertical lines, remember that slope is undefined and there is no y-intercept.

Question 8. The equation of the line PQ is 3y - 3x + 7 = 0.
(i) Write down the slope of the line PQ.
(ii) Calculate the angle that the line PQ makes with the positive direction of x-axis.
Answer:
(i) Rewrite in slope-intercept form:
\( 3y - 3x + 7 = 0 \implies 3y = 3x - 7 \implies y = x - \frac{7}{3} \)
Comparing with y = mx + c, we get m = 1.
The slope is 1.

(ii) Since m = tan θ:
\( \tan θ = 1 \implies θ = 45° \)
The angle is 45°.
In simple words: To find the slope, rewrite the equation as y = mx + c and read off m. To find the angle, set tan θ = slope and solve for θ.

Exam Tip: Recognize that tan 45° = 1, tan 60° = √3, and tan 30° = 1/√3. These are key angle-slope conversions.

Question 9. The given figure represents the lines y = x + 1 and y = √3x - 1. Write down the angles which the lines make with the positive direction of the x-axis. Hence, determine θ.
Answer: For the line y = x + 1: m₁ = 1, so \( \tan θ₁ = 1 \implies θ₁ = 45° \).

For the line y = √3x - 1: m₂ = √3, so \( \tan θ₂ = \sqrt{3} \implies θ₂ = 60° \).

From the figure, 60° is the exterior angle. By the exterior angle theorem:
\( 60° = θ + 45° \implies θ = 15° \)

The first line makes 45° with the x-axis, the second makes 60°, and θ = 15°.
In simple words: Compare each line's slope to tan of standard angles. The angle between two lines can be found using the exterior angle property.

Exam Tip: Identify slopes first, convert to inclination angles using inverse tan, then use geometric properties (like the exterior angle theorem) to find the angle between lines.

Question 10. Find the value of p, given that the line \( \frac{y}{2} = x - p \) passes through the point (-4, 4).
Answer: Since the point (-4, 4) lies on the line, it must satisfy the equation. Substitute x = -4 and y = 4:
\[ \frac{4}{2} = -4 - p \implies 2 = -4 - p \implies p = -4 - 2 = -6 \]
The value of p is -6.
In simple words: If a point is on a line, its coordinates must make the equation true. Plug the coordinates in and solve for the unknown.

Exam Tip: Always verify by substituting the point back into the equation after finding the unknown parameter.

Question 11. Given that (a, 2a) lies on the line \( \frac{y}{2} = 3x - 6 \), find the value of a.
Answer: Since (a, 2a) lies on the line, substitute x = a and y = 2a:
\[ \frac{2a}{2} = 3a - 6 \implies a = 3a - 6 \implies 6 = 3a - a = 2a \implies a = 3 \]
The value of a is 3.
In simple words: Plug the point's coordinates into the equation. The x-value is a and the y-value is 2a. Then solve the resulting equation.

Exam Tip: After solving, substitute back to confirm: when a = 3, the point is (3, 6). Check: \( \frac{6}{2} = 3(3) - 6 = 3 \). ✓

Question 12. The graph of the equation y = mx + c passes through the points (1, 4) and (-2, -5). Determine the values of m and c.
Answer: Since both points lie on the line, they satisfy the equation.

For (1, 4): \( 4 = m(1) + c \implies 4 = m + c \) ... (1)

For (-2, -5): \( -5 = m(-2) + c \implies -5 = -2m + c \) ... (2)

Subtract equation (2) from equation (1):
\[ 4 - (-5) = (m + c) - (-2m + c) \implies 9 = 3m \implies m = 3 \]

Substitute m = 3 into equation (1):
\[ 4 = 3 + c \implies c = 1 \]

Therefore, m = 3 and c = 1. The equation is \( y = 3x + 1 \).
In simple words: When a line passes through two points, use both points to write two equations. Solve the system to find m and c.

Exam Tip: Always verify your answer by checking both points: (1, 4): y = 3(1) + 1 = 4 ✓; (-2, -5): y = 3(-2) + 1 = -5 ✓.

Question 12. If points (1, 4) and (-2, -5) lie on the line y = mx + c, find m and c.
Answer: Because both points lie on the line, they must satisfy the equation y = mx + c. Substituting (1, 4): 4 = m + c, which gives m = 4 - c (Equation 1). Substituting (-2, -5): -5 = -2m + c. Replacing m with (4 - c) from Equation 1: -5 = -2(4 - c) + c simplifies to -5 = -8 + 3c, so c = 1. Putting c = 1 back into Equation 1: m = 3. Therefore, m = 3 and c = 1.
In simple words: Put each point into the line equation one at a time. You get two equations with m and c. Solve them together to find m = 3 and c = 1.

Exam Tip: Always substitute both given points carefully — a sign error in substitution is the most common mistake here.

Question 13. Find the equation of the line passing through the point (2, -5) and making an intercept of -3 on the y-axis.
Answer: Since the y-intercept is -3, we have c = -3 in the equation y = mx + c. The line passes through (2, -5), so: -5 = 2m - 3, which gives 2m = -2, so m = -1. Substituting m = -1 and c = -3 into y = mx + c: y = -x - 3, or x + y + 3 = 0.
In simple words: The y-intercept tells us c = -3. Use the given point to find m = -1. Then write the full line equation.

Exam Tip: Remember that the y-intercept is the value of c directly — do not substitute to find it.

Question 14. Find the equation of the straight line passing through (-1, 2) and whose slope is \( \frac{2}{5} \).
Answer: Using the point-slope form y - y₁ = m(x - x₁) with m = \( \frac{2}{5} \), point (-1, 2): y - 2 = \( \frac{2}{5} \)(x - (-1)) becomes y - 2 = \( \frac{2}{5} \)(x + 1). Multiplying both sides by 5: 5(y - 2) = 2(x + 1), which expands to 5y - 10 = 2x + 2. Rearranging: 2x - 5y + 12 = 0.
In simple words: Plug the point and slope into the point-slope formula. Multiply to clear fractions. Rearrange to standard form.

Exam Tip: Always clear fractions by multiplying — this avoids errors when rearranging the final equation.

Question 15. Find the equation of a straight line whose inclination is 60° and which passes through the point (0, -3).
Answer: The inclination is 60°, so the slope m = tan 60° = \( \sqrt{3} \). The line passes through (0, -3), which is on the y-axis, so the y-intercept c = -3. Using y = mx + c: y = \( \sqrt{3} \)x - 3, or \( \sqrt{3} \)x - y - 3 = 0.
In simple words: Find the slope from the inclination angle using tangent. Since the point is (0, -3), the intercept c = -3 right away. Write the equation.

Exam Tip: When a point lies on the y-axis, its x-coordinate is 0 — use this to spot the y-intercept instantly.

Question 16. Find the gradient of a line passing through the following pairs of points:
(i) (0, -2), (3, 4)
(ii) (3, -7), (-1, 8)
Answer:
(i) Using the gradient formula \( m = \frac{y_2 - y_1}{x_2 - x_1} \), we have m = \( \frac{4 - (-2)}{3 - 0} = \frac{6}{3} = 2 \). The gradient of the line through (0, -2) and (3, 4) is 2.
(ii) Using the same formula: m = \( \frac{8 - (-7)}{-1 - 3} = \frac{15}{-4} = -\frac{15}{4} \). The gradient of the line through (3, -7) and (-1, 8) is \( -\frac{15}{4} \).
In simple words: Subtract y-values to get the "rise". Subtract x-values to get the "run". Divide rise by run to find gradient.

Exam Tip: Keep the order of subtraction consistent for both coordinates — y₂ - y₁ in the numerator, x₂ - x₁ in the denominator.

Question 17. The coordinates of two points E and F are (0, 4) and (3, 7) respectively. Find:
(i) the gradient of EF.
(ii) the equation of EF.
(iii) the coordinates of the point where the line EF intersects the x-axis.
Answer:
(i) The gradient m = \( \frac{7 - 4}{3 - 0} = \frac{3}{3} = 1 \). The gradient of EF is 1.
(ii) Using y - y₁ = m(x - x₁) with point E(0, 4) and m = 1: y - 4 = 1(x - 0), so y - 4 = x, giving x - y + 4 = 0. The equation of EF is x - y + 4 = 0.
(iii) At the x-axis, y = 0. Substituting into x - y + 4 = 0: x - 0 + 4 = 0, so x = -4. The line EF crosses the x-axis at (-4, 0).
In simple words: Find gradient from the two points. Use one point and the gradient to write the line equation. Set y = 0 to find where it hits the x-axis.

Exam Tip: Always verify your equation by checking that both given points satisfy it.

Question 18. Find the intercepts made by the line 2x - 3y + 12 = 0 on the coordinate axes.
Answer: To find the x-intercept, set y = 0: 2x - 3(0) + 12 = 0 gives 2x = -12, so x = -6. To find the y-intercept, set x = 0: 2(0) - 3y + 12 = 0 gives -3y = -12, so y = 4. The x-intercept is -6 and the y-intercept is 4.
In simple words: To find where a line hits the x-axis, set y to 0 and solve for x. To find where it hits the y-axis, set x to 0 and solve for y.

Exam Tip: Write the intercepts as coordinates for full clarity: x-intercept is (-6, 0) and y-intercept is (0, 4).

Question 19. Find the equation of the line passing through the points P(5, 1) and Q(1, -1). Hence, show that the points P, Q and R(11, 4) are collinear.
Answer: First, find the gradient: m = \( \frac{-1 - 1}{1 - 5} = \frac{-2}{-4} = \frac{1}{2} \). Using y - y₁ = m(x - x₁) with P(5, 1): y - 1 = \( \frac{1}{2} \)(x - 5). Multiplying by 2: 2(y - 1) = x - 5, so 2y - 2 = x - 5, giving x - 2y - 3 = 0. This is the equation of line PQ. To check if R(11, 4) is collinear, substitute into the equation: 11 - 2(4) - 3 = 11 - 8 - 3 = 0. Since R satisfies the equation, all three points are collinear.
In simple words: Find the line equation for P and Q. Then plug in R's coordinates. If the result equals 0, then R lies on the same line.

Exam Tip: Three points are collinear if and only if the third point satisfies the line equation formed by the first two points.

Question 20. Find the value of 'a' for which the following points A(a, 3), B(2, 1) and C(5, a) are collinear. Hence, find the equation of the line.
Answer: For three points to be collinear, the gradient AB must equal the gradient BC. So \( \frac{1 - 3}{2 - a} = \frac{a - 1}{5 - 2} \), which gives \( \frac{-2}{2 - a} = \frac{a - 1}{3} \). Cross-multiplying: -2 × 3 = (a - 1)(2 - a), so -6 = 2a - a² - 2 + a, which simplifies to a² - 3a - 4 = 0. Factoring: (a + 1)(a - 4) = 0, so a = -1 or a = 4. For a = -1: using points A(-1, 3) and B(2, 1), the equation is 2x + 3y - 7 = 0. For a = 4: using points A(4, 3) and B(2, 1), the equation is x - y - 1 = 0.
In simple words: Set the two gradients equal. Solve the algebra to find a. Then write the line equation for each value of a.

Exam Tip: When a problem has two solutions, always verify both by checking collinearity or substitution in the equation.

Question 21. Use a graph paper for this question. The graph of a linear equation in x and y, passes through A(-1, -1) and B(2, 5). From your graph, find the values of h and k, if the line passes through (h, 4) and \( (\frac{1}{2}, k) \).
Answer: Plot points A(-1, -1) and B(2, 5) on graph paper and draw the straight line through them. The line passes through (h, 4), so locate where y = 4 intersects the line and read the x-coordinate: h = \( \frac{3}{2} \). The line passes through \( (\frac{1}{2}, k) \), so locate x = \( \frac{1}{2} \) on the line and read the y-coordinate: k = 2. Therefore, h = \( \frac{3}{2} \) and k = 2.
In simple words: Draw the line through the two given points. Find where y = 4 to get h. Find where x = 1/2 to get k.

Exam Tip: Use a sharp pencil and a ruler for accuracy. Read coordinates from the grid lines, not just by eye.

Question 22. ABCD is a parallelogram where A(x, y), B(5, 8), C(4, 7) and D(2, -4). Find
(i) the coordinates of A.
(ii) the equation of the diagonal BD.
Answer:
(i) In a parallelogram, the diagonals bisect each other. Let O be the intersection point of diagonals AC and BD. The midpoint of BD is \( O = \left(\frac{5 + 2}{2}, \frac{8 - 4}{2}\right) = \left(\frac{7}{2}, 2\right) \). Since O is also the midpoint of AC: \( \left(\frac{7}{2}, 2\right) = \left(\frac{x + 4}{2}, \frac{y + 7}{2}\right) \). This gives \( \frac{7}{2} = \frac{x + 4}{2} \) and \( 2 = \frac{y + 7}{2} \), so 7 = x + 4 and 4 = y + 7. Therefore, x = 3 and y = -3. The coordinates of A are (3, -3).
(ii) Using the two-point form with B(5, 8) and D(2, -4): y - 8 = \( \frac{-4 - 8}{2 - 5} \)(x - 5) = \( \frac{-12}{-3} \)(x - 5) = 4(x - 5). So y - 8 = 4x - 20, giving 4x - y - 12 = 0. The equation of diagonal BD is 4x - y - 12 = 0.
In simple words: The diagonals of a parallelogram cross at their midpoints. Use this to find A. Then use two points to find the diagonal equation.

Exam Tip: Always use the midpoint property of parallelogram diagonals — this is the key insight for finding missing vertices.

Question 23. In △ABC, A(3, 5), B(7, 8) and C(1, -10). Find the equation of the median through A.
Answer: A median connects a vertex to the midpoint of the opposite side. The midpoint D of BC is \( D = \left(\frac{7 + 1}{2}, \frac{8 - 10}{2}\right) = (4, -1) \). The median AD passes through A(3, 5) and D(4, -1). Using the two-point form: y - 5 = \( \frac{-1 - 5}{4 - 3} \)(x - 3) = \( \frac{-6}{1} \)(x - 3) = -6(x - 3). So y - 5 = -6x + 18, giving 6x + y - 23 = 0. The equation of the median through A is 6x + y - 23 = 0.
In simple words: Find the midpoint of the opposite side. Then write the line equation through the vertex and that midpoint.

Exam Tip: Do not confuse the median with the altitude — a median goes to the opposite side's midpoint, an altitude is perpendicular.

Question 24. Find the equation of a line passing through the point (-2, 3) and having x-intercept 4 units.
Answer: If the x-intercept is 4, the line crosses the x-axis at (4, 0). The line passes through both (-2, 3) and (4, 0). Using the two-point form: y - 3 = \( \frac{0 - 3}{4 - (-2)} \)(x - (-2)) = \( \frac{-3}{6} \)(x + 2) = \( \frac{-1}{2} \)(x + 2). Multiplying by 2: 2(y - 3) = -(x + 2), so 2y - 6 = -x - 2, giving x + 2y - 4 = 0. The equation of the line is x + 2y - 4 = 0.
In simple words: The x-intercept gives you a second point on the line: (4, 0). Then use both points to find the equation.

Exam Tip: An intercept of k means the line crosses that axis at a distance k from the origin — remember to form the point correctly.

Question 25. Find the equation of the line whose x-intercept is 6 and y-intercept is -4.
Answer: The line crosses the x-axis at (6, 0) and the y-axis at (0, -4). Using the two-point formula for a line passing through these two points:
\( y - 0 = \frac{-4 - 0}{0 - 6}(x - 6) \)
\( y = \frac{-4}{-6}(x - 6) \)
\( y = \frac{2}{3}(x - 6) \)
\( 3y = 2(x - 6) \)
\( 3y = 2x - 12 \)
\( 2x - 3y = 12 \)
In simple words: To find the line's equation when you know where it hits both axes, use those two points and apply the two-point formula to get the answer.

Exam Tip: Always identify the intercept points first - they give you two points to work with immediately, making the two-point formula the quickest approach.

Question 26. A(2, 5), B(-1, 2) and C(5, 8) are the vertices of a triangle ABC, 'M' is a point on AB such that AM : MB = 1 : 2. Find the coordinates of 'M'. Hence, find the equation of the line passing through C and M.
Answer: Since M divides AB in the ratio 1 : 2, applying the section formula:
\( M = \left(\frac{1 \times (-1) + 2 \times 2}{1 + 2}, \frac{1 \times 2 + 2 \times 5}{1 + 2}\right) \)
\( M = \left(\frac{-1 + 4}{3}, \frac{2 + 10}{3}\right) \)
\( M = \left(\frac{3}{3}, \frac{12}{3}\right) \)
\( M = (1, 4) \)
Now, finding the equation of line CM using points C(5, 8) and M(1, 4):
\( y - 8 = \frac{4 - 8}{1 - 5}(x - 5) \)
\( y - 8 = \frac{-4}{-4}(x - 5) \)
\( y - 8 = 1(x - 5) \)
\( y - 8 = x - 5 \)
\( x - y + 3 = 0 \)
In simple words: Use the section formula to find where M sits on the line segment, then use that point with C to write the equation of line CM.

Exam Tip: Always verify your section formula work by checking that the coordinates satisfy the given ratio before proceeding to find the line equation.

Question 27. Find the equation of the line passing through the point (1, 4) and intersecting the line x - 2y - 11 = 0 on the y-axis.
Answer: First, locate where the given line meets the y-axis. When x = 0:
\( 0 - 2y - 11 = 0 \)
\( -2y = 11 \)
\( y = -\frac{11}{2} \)
So the intersection point is \( \left(0, -\frac{11}{2}\right) \). The required line passes through (1, 4) and \( \left(0, -\frac{11}{2}\right) \). Using the two-point formula:
\( y - 4 = \frac{-\frac{11}{2} - 4}{0 - 1}(x - 1) \)
\( y - 4 = \frac{-\frac{11}{2} - \frac{8}{2}}{-1}(x - 1) \)
\( y - 4 = \frac{-\frac{19}{2}}{-1}(x - 1) \)
\( y - 4 = \frac{19}{2}(x - 1) \)
\( 2(y - 4) = 19(x - 1) \)
\( 2y - 8 = 19x - 19 \)
\( 19x - 2y - 11 = 0 \)
In simple words: Find where the given line crosses the y-axis, then treat that point and the given point as two points to build your new line.

Exam Tip: When a line "intersects on the y-axis," remember that the x-coordinate there is always 0 - this saves calculation time.

Question 28. Find the equation of the straight line containing the point (3, 2) and making positive equal intercepts on axes.
Answer: Let the line pass through (x, 0) on the x-axis and (0, y) on the y-axis. Since the intercepts are equal and positive, x = y. The slope of the line connecting (x, 0) and (0, x) is:
\( m = \frac{0 - x}{x - 0} = \frac{-x}{x} = -1 \)
Using the point-slope form with point (3, 2) and slope -1:
\( y - 2 = -1(x - 3) \)
\( y - 2 = -x + 3 \)
\( x + y - 5 = 0 \)
In simple words: When a line makes equal positive intercepts, the slope will always be -1. Use this fact with the given point to find the equation quickly.

Exam Tip: Lines with equal intercepts always have slope -1; lines with equal but opposite intercepts have slope 1. These are key shortcuts for intercept-based questions.

Question 29. Three vertices of a parallelogram ABCD taken in order are A(3, 6), B(5, 10) and C(3, 2) find:
(i) the coordinates of the fourth vertex D.
(ii) length of diagonal BD.
(iii) equation of side AB of the parallelogram ABCD.
Answer:
(i) In a parallelogram, the diagonals bisect each other. The midpoint of diagonal AC is:
\( \left(\frac{3 + 3}{2}, \frac{6 + 2}{2}\right) = (3, 4) \)
The midpoint of diagonal BD should also be (3, 4). If D = (x, y):
\( \frac{5 + x}{2} = 3 \text{ and } \frac{10 + y}{2} = 4 \)
\( 5 + x = 6 \text{ and } 10 + y = 8 \)
\( x = 1 \text{ and } y = -2 \)
So D = (1, -2).
(ii) Using the distance formula for B(5, 10) and D(1, -2):
\( BD = \sqrt{(1 - 5)^2 + (-2 - 10)^2} \)
\( BD = \sqrt{(-4)^2 + (-12)^2} \)
\( BD = \sqrt{16 + 144} \)
\( BD = \sqrt{160} = 4\sqrt{10} \) units
(iii) For line AB through A(3, 6) and B(5, 10):
\( y - 6 = \frac{10 - 6}{5 - 3}(x - 3) \)
\( y - 6 = \frac{4}{2}(x - 3) \)
\( y - 6 = 2(x - 3) \)
\( y - 6 = 2x - 6 \)
\( 2x - y = 0 \)
In simple words: The diagonals of a parallelogram always meet at their midpoints - use this to find the missing vertex. Then apply distance and line formulas as usual.

Exam Tip: When finding a fourth vertex of a parallelogram, the diagonal bisection property is faster than side-slope methods - always use it.

Question 30. A and B are the two points on the x-axis and y-axis respectively. P(2, -3) is the mid-point of AB. Find:
(i) the coordinates of A and B.
(ii) the slope of the line AB.
(iii) the equation of the line AB.
Answer:
(i) Let A = (x, 0) on the x-axis and B = (0, y) on the y-axis. Since P(2, -3) is the midpoint:
\( 2 = \frac{x + 0}{2} \text{ and } -3 = \frac{0 + y}{2} \)
\( 2 = \frac{x}{2} \text{ and } -3 = \frac{y}{2} \)
\( x = 4 \text{ and } y = -6 \)
So A = (4, 0) and B = (0, -6).
(ii) The slope of AB is:
\( m = \frac{-6 - 0}{0 - 4} = \frac{-6}{-4} = \frac{3}{2} \)
(iii) Using the point-slope form with A(4, 0) and slope \( \frac{3}{2} \):
\( y - 0 = \frac{3}{2}(x - 4) \)
\( 2y = 3(x - 4) \)
\( 2y = 3x - 12 \)
\( 3x - 2y = 12 \)
In simple words: When a point is the midpoint of a segment with endpoints on the axes, you can find both endpoints by doubling the midpoint's coordinates.

Exam Tip: For axis-intercept problems, always remember: points on the x-axis have y = 0, and points on the y-axis have x = 0 - this constraint directly helps you find coordinates.

Question 31. M and N are two points on the x-axis and y-axis respectively. P(3, 2) divides the line segment MN in the ratio 2 : 3. Find:
(i) the coordinates of M and N.
(ii) slope of the line MN.
Answer:
(i) Let M = (x, 0) on the x-axis and N = (0, y) on the y-axis. By the section formula, P divides MN in ratio 2 : 3:
\( P = \left(\frac{2 \times 0 + 3 \times x}{2 + 3}, \frac{2 \times y + 3 \times 0}{2 + 3}\right) \)
\( P = \left(\frac{3x}{5}, \frac{2y}{5}\right) \)
Since P = (3, 2):
\( 3 = \frac{3x}{5} \text{ and } 2 = \frac{2y}{5} \)
\( 3x = 15 \text{ and } 2y = 10 \)
\( x = 5 \text{ and } y = 5 \)
So M = (5, 0) and N = (0, 5).
(ii) The slope of line MN is:
\( m = \frac{5 - 0}{0 - 5} = \frac{5}{-5} = -1 \)
In simple words: Apply the section formula directly with the division ratio and the given point to find where M and N sit on the axes.

Exam Tip: When using the section formula with axis points, substitute 0 for the axis-coordinate from the start to simplify your calculations.

Question 32. The line through P(5, 3) intersects y-axis at Q.
(i) Write the slope of the line.
(ii) Write the equation of the line.
(iii) Find the co-ordinates of Q.
Answer:
(i) From the diagram, the line makes a 45° angle with the x-axis, so the slope is \( m = \tan 45° = 1 \).
(ii) Using the point-slope form with P(5, 3) and slope 1:
\( y - 3 = 1(x - 5) \)
\( y - 3 = x - 5 \)
\( x - y - 2 = 0 \)
(iii) At the y-axis, x = 0. Substituting into the equation:
\( 0 - y - 2 = 0 \)
\( y = -2 \)
So Q = (0, -2).
In simple words: A 45° angle always gives a slope of 1 or -1. Once you have the slope and a point, the point-slope form quickly gives the equation.

Exam Tip: For lines at special angles (30°, 45°, 60°), learn the slope values - tan 45° = 1 and tan 60° = √3 are exam favourites.

Question 33. (i) Write down the coordinates of point P that divides the line joining A(-4, 1) and B(17, 10) in the ratio 1 : 2.
(ii) Calculate the distance OP, where O is the origin.
(iii) In what ratio does the y-axis divide the line AB?
Answer:
(i) Using the section formula for P dividing AB in ratio 1 : 2:
\( P = \left(\frac{1 \times 17 + 2 \times (-4)}{1 + 2}, \frac{1 \times 10 + 2 \times 1}{1 + 2}\right) \)
\( P = \left(\frac{17 - 8}{3}, \frac{10 + 2}{3}\right) \)
\( P = \left(\frac{9}{3}, \frac{12}{3}\right) \)
\( P = (3, 4) \)
(ii) Using the distance formula:
\( OP = \sqrt{(0 - 3)^2 + (0 - 4)^2} \)
\( OP = \sqrt{9 + 16} \)
\( OP = \sqrt{25} = 5 \) units
(iii) The y-axis is the line x = 0. Let the y-axis divide AB in ratio m : n. Using the section formula with x-coordinate = 0:
\( 0 = \frac{m \times 17 + n \times (-4)}{m + n} \)
\( 17m - 4n = 0 \)
\( 17m = 4n \)
\( \frac{m}{n} = \frac{4}{17} \)
So the ratio is 4 : 17.
In simple words: For finding where a line crosses the y-axis and its division ratio, set the x-coordinate of the section formula equal to 0 and solve.

Exam Tip: When asked "in what ratio does the y-axis divide," always set x = 0 in the section formula - this directly gives you the ratio without needing to find the actual intersection point first.

Question 34. Find the equations of the diagonals of a rectangle whose sides are x = -1, x = 2, y = -2 and y = 6.
Answer: The four lines form a rectangle at their intersections. The four vertices are:
A = (-1, -2), B = (2, -2), C = (2, 6), D = (-1, 6)
The diagonals of the rectangle are AC and BD.
For diagonal AC from A(-1, -2) to C(2, 6):
\( y - (-2) = \frac{6 - (-2)}{2 - (-1)}(x - (-1)) \)
\( y + 2 = \frac{8}{3}(x + 1) \)
\( 3(y + 2) = 8(x + 1) \)
\( 3y + 6 = 8x + 8 \)
\( 8x - 3y + 2 = 0 \)
For diagonal BD from B(2, -2) to D(-1, 6):
\( y - (-2) = \frac{6 - (-2)}{-1 - 2}(x - 2) \)
\( y + 2 = \frac{8}{-3}(x - 2) \)
\( 3(y + 2) = -8(x - 2) \)
\( 3y + 6 = -8x + 16 \)
\( 8x + 3y - 10 = 0 \)
In simple words: First identify the vertices where the four given lines meet, then use the two-point formula to find equations of both diagonals.

Exam Tip: For rectangles formed by two pairs of parallel lines (x = a, x = b and y = c, y = d), the vertices are always at the four intersections - visualizing or sketching helps avoid errors.

Question 35. Find the equation of the straight line passing through the origin and through the point of intersection of the lines 5x + 7y = 3 and 2x - 3y = 7.
Answer: Start by finding where the two lines meet. Use the given equations:
5x + 7y = 3 ....(i)
2x - 3y = 7 ....(ii)
Multiply (i) by 3 and (ii) by 7:
15x + 21y = 9 ....(iii)
14x - 21y = 49 ....(iv)
Add (iii) and (iv):
\( \implies 29x = 58 \)
\( \implies x = 2 \)
Substitute x = 2 into (i):
\( \implies 5(2) + 7y = 3 \)
\( \implies 10 + 7y = 3 \)
\( \implies 7y = -7 \)
\( \implies y = -1 \)
The intersection point is (2, -1). Now find the line through (2, -1) and (0, 0) using the two-point form:
\( y - y_1 = \frac{y_2 - y_1}{x_2 - x_1}(x - x_1) \)
\( \implies y - (-1) = \frac{0 - (-1)}{0 - 2}(x - 2) \)
\( \implies y + 1 = \frac{1}{-2}(x - 2) \)
\( \implies -2(y + 1) = x - 2 \)
\( \implies -2y - 2 = x - 2 \)
\( \implies x + 2y = 0 \)
In simple words: First find where the two given lines cross. That point is (2, -1). Then write the equation of a line through this point and the origin (0, 0). The answer is x + 2y = 0.

Exam Tip: Remember to find the intersection point first by solving the system of equations, then apply the two-point form carefully. Always verify your final equation by checking both points satisfy it.

Exercise 12.2

Question 1. State which one of the following is true: The straight lines y = 3x - 5 and 2y = 4x + 7 are
(i) parallel
(ii) perpendicular
(iii) neither parallel nor perpendicular
Answer: Rewrite both lines in slope-intercept form. The first line is y = 3x - 5. Rewrite the second line:
\( 2y = 4x + 7 \)
\( \implies y = 2x + \frac{7}{2} \)
Compare each with y = mx + c. The slopes are 3 and 2. Since the slopes are not equal, the lines are not parallel. Check if they are perpendicular by multiplying the slopes: \( 3 \times 2 = 6 \neq -1 \), so they are not perpendicular either.
In simple words: Two lines are parallel if their slopes match. They are perpendicular if the slope product equals -1. Here the slopes are 3 and 2, so neither condition holds.

Exam Tip: Always convert both line equations to y = mx + c form before comparing slopes. Check parallelism (equal slopes) and perpendicularity (product of slopes equals -1) systematically.

Question 2. If 6x + 5y - 7 = 0 and 2px + 5y + 1 = 0 are parallel lines, find the value of p.
Answer: Convert the first equation to slope form:
\( 6x + 5y - 7 = 0 \)
\( \implies 5y = -6x + 7 \)
\( \implies y = -\frac{6}{5}x + \frac{7}{5} \)
The slope is \( m_1 = -\frac{6}{5} \). For the second equation:
\( 2px + 5y + 1 = 0 \)
\( \implies 5y = -2px - 1 \)
\( \implies y = -\frac{2p}{5}x - \frac{1}{5} \)
The slope is \( m_2 = -\frac{2p}{5} \). Since the lines are parallel, their slopes must be equal:
\( m_1 = m_2 \)
\( \implies -\frac{6}{5} = -\frac{2p}{5} \)
\( \implies 2p = 6 \)
\( \implies p = 3 \)
In simple words: Parallel lines have identical slopes. Set the slopes equal to each other and solve for the unknown value p.

Exam Tip: For parallel lines, equate the slope coefficients directly without calculating numerical values if the equations have a similar structure. This saves time and reduces arithmetic errors.

Question 3. If the straight lines 3x - 5y + 7 = 0 and 4x + ay + 9 = 0 are perpendicular to one another, find the value of a.
Answer: Express the first line in slope-intercept form:
\( 3x - 5y + 7 = 0 \)
\( \implies 5y = 3x + 7 \)
\( \implies y = \frac{3}{5}x + \frac{7}{5} \)
The slope is \( m_1 = \frac{3}{5} \). For the second line:
\( 4x + ay + 9 = 0 \)
\( \implies ay = -4x - 9 \)
\( \implies y = -\frac{4}{a}x - \frac{9}{a} \)
The slope is \( m_2 = -\frac{4}{a} \). Since the lines are perpendicular, the product of slopes equals -1:
\( m_1 \times m_2 = -1 \)
\( \implies \frac{3}{5} \times \left(-\frac{4}{a}\right) = -1 \)
\( \implies -\frac{12}{5a} = -1 \)
\( \implies \frac{12}{5a} = 1 \)
\( \implies a = \frac{12}{5} \)
In simple words: Perpendicular lines have slopes whose product is -1. Write out both slopes, multiply them, and set the result equal to -1 to find a.

Exam Tip: Remember the perpendicularity condition: \( m_1 \times m_2 = -1 \). Always cross-multiply and simplify carefully to avoid sign errors.

Question 4. If the lines 3x + by + 5 = 0 and ax - 5y + 7 = 0 are perpendicular to each other, find the relation connecting a and b.
Answer: Rearrange the first equation to find its slope:
\( 3x + by + 5 = 0 \)
\( \implies by = -3x - 5 \)
\( \implies y = -\frac{3}{b}x - \frac{5}{b} \)
The slope is \( m_1 = -\frac{3}{b} \). For the second line:
\( ax - 5y + 7 = 0 \)
\( \implies 5y = ax + 7 \)
\( \implies y = \frac{a}{5}x + \frac{7}{5} \)
The slope is \( m_2 = \frac{a}{5} \). Apply the perpendicularity condition:
\( m_1 \times m_2 = -1 \)
\( \implies -\frac{3}{b} \times \frac{a}{5} = -1 \)
\( \implies -\frac{3a}{5b} = -1 \)
\( \implies \frac{3a}{5b} = 1 \)
\( \implies 3a = 5b \)
In simple words: Extract the slope of each line by rearranging to y = mx + c form. Multiply the slopes and set their product equal to -1. Simplify to get the relation 3a = 5b.

Exam Tip: When the problem asks for a "relation" rather than a specific number, express your final answer as an equation connecting the two unknowns. Ensure all algebra steps are shown clearly.

Question 5. Is the line through (-2, 3) and (4, 1) perpendicular to the line 3x = y + 1? Does the line 3x = y + 1 bisect the join of (-2, 3) and (4, 1)?
Answer: Find the slope of the line passing through (-2, 3) and (4, 1) using the two-point form:
\( m_1 = \frac{1 - 3}{4 - (-2)} = \frac{-2}{6} = -\frac{1}{3} \)
The line through these points is:
\( y - 3 = -\frac{1}{3}(x + 2) \)
\( \implies y = -\frac{1}{3}x - \frac{11}{3} \)
For the line 3x = y + 1 or y = 3x - 1, the slope is \( m_2 = 3 \). Check perpendicularity:
\( m_1 \times m_2 = -\frac{1}{3} \times 3 = -1 \)
Yes, the lines are perpendicular. Now find the midpoint of (-2, 3) and (4, 1):
\( \text{Midpoint} = \left(\frac{-2 + 4}{2}, \frac{3 + 1}{2}\right) = (1, 2) \)
Check if (1, 2) lies on 3x = y + 1:
L.H.S. = 3(1) = 3
R.H.S. = 2 + 1 = 3
L.H.S. = R.H.S., so the point (1, 2) satisfies the equation. Therefore, the line 3x = y + 1 bisects the segment joining (-2, 3) and (4, 1).
In simple words: Two lines are perpendicular if their slopes multiply to give -1. A line bisects a segment if it passes through the segment's midpoint. Check both conditions using slopes and the midpoint formula.

Exam Tip: Always verify perpendicularity by computing the slope product explicitly. For bisection questions, find the midpoint and substitute it into the given line equation to confirm.

Question 6. The line through A(-2, 3) and B(4, b) is perpendicular to the line 2x - 4y = 5. Find the value of b.
Answer: Calculate the slope of the line through A and B:
\( m_1 = \frac{b - 3}{4 - (-2)} = \frac{b - 3}{6} \)
For the line 2x - 4y = 5, rearrange to find the slope:
\( 4y = 2x - 5 \)
\( \implies y = \frac{1}{2}x - \frac{5}{4} \)
The slope is \( m_2 = \frac{1}{2} \). Since the lines are perpendicular:
\( m_1 \times m_2 = -1 \)
\( \implies \frac{b - 3}{6} \times \frac{1}{2} = -1 \)
\( \implies \frac{b - 3}{12} = -1 \)
\( \implies b - 3 = -12 \)
\( \implies b = -9 \)
In simple words: Use the slope formula with the two points A and B, then apply the perpendicularity condition to solve for b. The product of the slopes must equal -1.

Exam Tip: When a coordinate contains an unknown, express the slope as a fraction with that unknown. Then use the perpendicularity condition to set up an equation and solve.

Question 7. If the lines 3x + y = 4, x - ay + 7 = 0 and bx + 2y + 5 = 0 form three consecutive sides of a rectangle, find the values of a and b.
Answer: In a rectangle, consecutive sides are perpendicular. This means lines (i) and (ii) are perpendicular, and lines (ii) and (iii) are perpendicular.
Find the slope of line (i): 3x + y = 4
\( \implies y = -3x + 4 \)
Slope \( m_1 = -3 \)
Find the slope of line (ii): x - ay + 7 = 0
\( \implies ay = x + 7 \)
\( \implies y = \frac{1}{a}x + \frac{7}{a} \)
Slope \( m_2 = \frac{1}{a} \)
Find the slope of line (iii): bx + 2y + 5 = 0
\( \implies 2y = -bx - 5 \)
\( \implies y = -\frac{b}{2}x - \frac{5}{2} \)
Slope \( m_3 = -\frac{b}{2} \)
For perpendicularity of lines (i) and (ii):
\( m_1 \times m_2 = -1 \)
\( \implies -3 \times \frac{1}{a} = -1 \)
\( \implies -\frac{3}{a} = -1 \)
\( \implies a = 3 \)
For perpendicularity of lines (ii) and (iii):
\( m_2 \times m_3 = -1 \)
\( \implies \frac{1}{a} \times \left(-\frac{b}{2}\right) = -1 \)
\( \implies -\frac{b}{2a} = -1 \)
\( \implies b = 2a = 2(3) = 6 \)
In simple words: Find each line's slope. Since the sides are consecutive sides of a rectangle, adjacent sides must be perpendicular. Set up two perpendicularity equations to find both a and b.

Exam Tip: Remember that in a rectangle (or any quadrilateral), consecutive sides meet at right angles. Use the perpendicularity condition systematically for each pair of adjacent sides.

Question 8. Find the value of 'p' if the lines 5x - 3y + 2 = 0 and 6x - py + 7 = 0 are perpendicular to each other. Hence find the equation of a line passing through (-2, -1) and parallel to 6x - py + 7 = 0.
Answer: Express the first line in slope form:
\( 5x - 3y + 2 = 0 \)
\( \implies 3y = 5x + 2 \)
\( \implies y = \frac{5}{3}x + \frac{2}{3} \)
The slope is \( m_1 = \frac{5}{3} \). For the second line:
\( 6x - py + 7 = 0 \)
\( \implies py = 6x + 7 \)
\( \implies y = \frac{6}{p}x + \frac{7}{p} \)
The slope is \( m_2 = \frac{6}{p} \). Apply the perpendicularity condition:
\( m_1 \times m_2 = -1 \)
\( \implies \frac{5}{3} \times \frac{6}{p} = -1 \)
\( \implies \frac{30}{3p} = -1 \)
\( \implies \frac{10}{p} = -1 \)
\( \implies p = -10 \)
Now substitute p = -10 into 6x - py + 7 = 0:
\( 6x - (-10)y + 7 = 0 \)
\( \implies 6x + 10y + 7 = 0 \)
The slope of this line is \( \frac{6}{10} = -\frac{3}{5} \) (after rearranging to y = mx + c form). Any line parallel to this has the same slope. Using the point-slope form with point (-2, -1):
\( y - (-1) = -\frac{3}{5}(x - (-2)) \)
\( \implies y + 1 = -\frac{3}{5}(x + 2) \)
\( \implies 5(y + 1) = -3(x + 2) \)
\( \implies 5y + 5 = -3x - 6 \)
\( \implies 3x + 5y + 11 = 0 \)
In simple words: Find p using the perpendicularity condition. Then substitute to get the second line's equation. Use the point-slope form to write the parallel line through the given point.

Exam Tip: This is a two-part question: first find the unknown coefficient using perpendicularity, then find a new line using the parallel condition. Write out both stages clearly to avoid missing marks.

Question 9. Find the equation of a line, which has the y-intercept 4, and is parallel to the line 2x - 3y - 7 = 0. Find the coordinates of the point where it cuts the x-axis.
Answer: Start with the given line 2x - 3y - 7 = 0. Rearrange it to slope-intercept form: 3y = 2x - 7, so y = (2/3)x - 7/3. Comparing with y = mx + c, the slope is 2/3. Since parallel lines share the same slope, the new line also has slope 2/3. The new line has y-intercept 4, so c = 4. Using y = mx + c with m = 2/3 and c = 4, we get y = (2/3)x + 4, which simplifies to 3y = 2x + 12, or 2x - 3y + 12 = 0. To find where this line meets the x-axis, set y = 0: 2x - 3(0) + 12 = 0, so 2x = -12, giving x = -6. Therefore, the line's equation is 2x - 3y + 12 = 0, and it intersects the x-axis at (-6, 0).
In simple words: Parallel lines have matching slopes. Find the slope of the first line, use it with the given y-intercept to build the new line, then plug in y = 0 to find where it crosses the x-axis.

Exam Tip: Always verify your final equation by confirming the slope matches the original line and the y-intercept is correct before finding the axis intersection.

Question 10. Find the equation of a straight line perpendicular to the line 2x + 5y + 7 = 0 and with y-intercept -3.
Answer: Begin with the given line 2x + 5y + 7 = 0. Convert to slope-intercept form: 5y = -2x - 7, so y = (-2/5)x - 7/5. The slope is -2/5. For perpendicular lines, the product of their slopes equals -1. If m₁ = -2/5, then m₁ × m₂ = -1 gives m₂ = 5/2. The new line has slope 5/2 and y-intercept -3. Using y = mx + c: y = (5/2)x + (-3), which becomes y = (5x - 6)/2. Multiplying by 2: 2y = 5x - 6, or 5x - 2y - 6 = 0.
In simple words: When two lines are perpendicular, flip the slope upside down and change its sign. Then use the given y-intercept to write the equation.

Exam Tip: Remember the perpendicularity condition: m₁ × m₂ = -1. If the original slope is a fraction, the perpendicular slope is its negative reciprocal.

Question 11. Find the equation of a straight line perpendicular to the line 3x - 4y + 12 = 0 and having same y-intercept as 2x - y + 5 = 0.
Answer: First, find the slope of the given line 3x - 4y + 12 = 0. Rearrange: 4y = 3x + 12, so y = (3/4)x + 3. The slope is 3/4. For a perpendicular line, m₂ = -4/3. Next, determine the y-intercept from 2x - y + 5 = 0. Rewrite as y = 2x + 5, so the y-intercept is 5. The new line has slope -4/3 and y-intercept 5. Substituting into y = mx + c: y = (-4/3)x + 5, which becomes y = (-4x + 15)/3. Multiplying by 3: 3y = -4x + 15, or 4x + 3y - 15 = 0.
In simple words: Find the perpendicular slope from the first line, then find the y-intercept from the second line. Combine them to form the equation.

Exam Tip: This question requires extracting information from two different lines - one provides the perpendicular condition, the other provides the y-intercept. Keep these roles distinct.

Question 12. Find the equation of the line passing through (0, 4) and parallel to the line 3x + 5y + 15 = 0.
Answer: Start with the given line 3x + 5y + 15 = 0. Convert to slope-intercept form: 5y = -3x - 15, so y = (-3/5)x - 3. The slope is -3/5. Since parallel lines have equal slopes, the new line also has slope -3/5 and passes through (0, 4). Using the point-slope form y - y₁ = m(x - x₁): y - 4 = (-3/5)(x - 0), which gives 5(y - 4) = -3x. Expanding: 5y - 20 = -3x, so 3x + 5y - 20 = 0.
In simple words: Extract the slope from the original line. Since the new line is parallel, it has the same slope. Use the given point and slope to write the equation.

Exam Tip: When a point is given, use the point-slope form. When only a y-intercept is given, use the slope-intercept form. Choose the method that fits your information.

Question 13. (i) The line 4x - 3y + 12 = 0 meets the x-axis at A. Write down the coordinates of A. (ii) Determine the equation of the line passing through A and perpendicular to 4x - 3y + 12 = 0.
Answer: (i) On the x-axis, the y-coordinate is 0. Substituting y = 0 into 4x - 3y + 12 = 0: 4x - 3(0) + 12 = 0, so 4x = -12, giving x = -3. Therefore, A = (-3, 0). (ii) Convert the given line to slope-intercept form: 4x - 3y + 12 = 0 becomes 3y = 4x + 12, so y = (4/3)x + 4. The slope is 4/3. For a perpendicular line, m₂ = -3/4. Using the point-slope form with point (-3, 0): y - 0 = (-3/4)(x - (-3)), so 4y = -3(x + 3), which gives 4y = -3x - 9, or 3x + 4y + 9 = 0.
In simple words: On the x-axis, y is always 0. Find where the line crosses by setting y = 0. Then use the perpendicular slope condition with that point to form the new line.

Exam Tip: Always remember that on the x-axis, y = 0, and on the y-axis, x = 0. These are key facts for finding axis intercepts.

Question 14. Find the equation of the line that is parallel to 2x + 5y - 7 = 0 and passes through the mid-point of the line segment joining the points (2, 7) and (-4, 1).
Answer: First, find the slope of 2x + 5y - 7 = 0. Rearrange: 5y = -2x + 7, so y = (-2/5)x + 7/5. The slope is -2/5. A parallel line has the same slope -2/5. Next, find the mid-point of the segment joining (2, 7) and (-4, 1) using the mid-point formula: M = ((2 + (-4))/2, (7 + 1)/2) = (-1, 4). Now use point-slope form with slope -2/5 and point (-1, 4): y - 4 = (-2/5)(x - (-1)), so 5(y - 4) = -2(x + 1), giving 5y - 20 = -2x - 2, which simplifies to 5y + 2x = 18, or 2x + 5y - 18 = 0.
In simple words: Find the slope of the given line. Calculate the mid-point of the two given points. Then write the equation using the slope and mid-point.

Exam Tip: The mid-point formula is \( \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right) \). Always simplify this before using it in the point-slope form.

Question 15. Find the equation of the line that is perpendicular to 3x + 2y - 8 = 0 and passes through the mid-point of the line segment joining the points (5, -2) and (2, 2).
Answer: Convert the given line 3x + 2y - 8 = 0 to slope-intercept form: 2y = -3x + 8, so y = (-3/2)x + 4. The slope is -3/2. For a perpendicular line, m₂ = 2/3. Calculate the mid-point of (5, -2) and (2, 2): M = ((5 + 2)/2, (-2 + 2)/2) = (7/2, 0). Using point-slope form with slope 2/3 and point (7/2, 0): y - 0 = (2/3)(x - 7/2), so 3y = 2(x - 7/2), which gives 3y = 2x - 7, or 2x - 3y - 7 = 0.
In simple words: Find the perpendicular slope by flipping and changing the sign of the original slope. Calculate the mid-point. Then write the equation using the perpendicular slope and mid-point.

Exam Tip: When finding the perpendicular slope, use m₁ × m₂ = -1. Check your work by confirming the perpendicularity condition holds.

Question 16. Find the equation of a straight line passing through the intersection of 2x + 5y - 4 = 0 with x-axis and parallel to the line 3x - 7y + 8 = 0.
Answer: To find where 2x + 5y - 4 = 0 meets the x-axis, set y = 0: 2x + 5(0) - 4 = 0, so 2x = 4, giving x = 2. The intersection point is (2, 0). Next, find the slope of 3x - 7y + 8 = 0. Rearrange: 7y = 3x + 8, so y = (3/7)x + 8/7. The slope is 3/7. The parallel line has slope 3/7 and passes through (2, 0). Using point-slope form: y - 0 = (3/7)(x - 2), so 7y = 3(x - 2), which gives 7y = 3x - 6, or 3x - 7y - 6 = 0.
In simple words: Find the intersection point by setting y = 0. Extract the slope from the reference line. Write the equation using that slope and the intersection point.

Exam Tip: On the x-axis, y is always 0. This is the quickest way to find axis intersection points without solving complex systems.

Question 17. Line AB is perpendicular to line CD. Coordinates of B, C and D are (4, 0), (0, -1) and (4, 3) respectively. Find (i) the slope of CD (ii) the equation of line AB
Answer: (i) Use the slope formula with points C(0, -1) and D(4, 3): slope = (3 - (-1))/(4 - 0) = 4/4 = 1. So the slope of CD is 1. (ii) Since AB is perpendicular to CD, and the slope of CD is 1, the slope of AB satisfies: (slope of AB) × 1 = -1, so slope of AB = -1. The line AB passes through B(4, 0). Using point-slope form: y - 0 = -1(x - 4), which gives y = -x + 4, or x + y = 4.
In simple words: Use the two points on CD to find its slope. Then use the perpendicularity rule to get the slope of AB. Finally, use the perpendicular slope with point B to write the equation.

Exam Tip: When two lines are perpendicular, their slopes multiply to give -1. This is the defining relationship to remember.

Question 18. Find the equation of a line parallel to the line 2x + y - 7 = 0 and passing through the point of intersection of the lines x + y - 4 = 0 and 2x - y = 8.
Answer: Solve the system x + y - 4 = 0 and 2x - y = 8 simultaneously. From the first equation: y = 4 - x. Substitute into the second: 2x - (4 - x) = 8, so 2x - 4 + x = 8, giving 3x = 12, hence x = 4. Then y = 4 - 4 = 0. The intersection point is (4, 0). Find the slope of 2x + y - 7 = 0: rearrange to y = -2x + 7, so slope = -2. A parallel line has slope -2 and passes through (4, 0). Using point-slope form: y - 0 = -2(x - 4), which gives y = -2x + 8, or 2x + y - 8 = 0.
In simple words: Solve both equations together to find their intersection point. Extract the slope from the reference line. Use that slope with the intersection point to build the final equation.

Exam Tip: When finding the intersection of two lines, solve them as a simultaneous system. Substitution or elimination methods both work - choose whichever seems simpler for the given equations.

Question 19. The equation of a line is 3x + 4y - 7 = 0. Find (i) slope of the line. (ii) the equation of a line perpendicular to the given line and passing through the intersection of the lines x - y + 2 = 0 and 3x + y - 10 = 0.
Answer:
(i) Start with 3x + 4y - 7 = 0. Rearrange to express it as y = mx + c:
\( 4y = -3x + 7 \)
\( y = -\frac{3}{4}x + \frac{7}{4} \)
Comparing with y = mx + c, the slope (m₁) = -3/4.

(ii) For a line perpendicular to the given line, let the slope be m₂. Using the perpendicularity condition:
\( m_1 \times m_2 = -1 \)
\( -\frac{3}{4} \times m_2 = -1 \)
\( m_2 = \frac{4}{3} \)

Now find where x - y + 2 = 0 and 3x + y - 10 = 0 meet. Add the two equations:
\( x - y + 2 + 3x + y - 10 = 0 \)
\( 4x - 8 = 0 \)
\( x = 2 \)

Put x = 2 in the first equation:
\( 2 - y + 2 = 0 \)
\( y = 4 \)

So the intersection point is (2, 4). Using the point-slope form with slope 4/3 and point (2, 4):
\( y - 4 = \frac{4}{3}(x - 2) \)
\( 3(y - 4) = 4(x - 2) \)
\( 3y - 12 = 4x - 8 \)
\( 4x - 3y + 4 = 0 \)

The equation of the required line is 4x - 3y + 4 = 0.
In simple words: To find the slope, rewrite the equation in the form y = mx + c. For a perpendicular line, use the rule that multiplying the two slopes gives -1. Find where two lines cross, then write the equation of the new line through that point.

Exam Tip: Always convert to y = mx + c form to read off the slope directly. Remember that perpendicular lines have slopes whose product is -1, not that the slopes are negative reciprocals - they mean the same thing, but the multiplication rule is quicker to apply.

Question 20. Find the equation of the perpendicular from the point (1, -2) on the line 4x - 3y - 5 = 0. Also find the coordinates of the foot of perpendicular.
Answer:
First, convert 4x - 3y - 5 = 0 to y = mx + c form:
\( 3y = 4x - 5 \)
\( y = \frac{4}{3}x - \frac{5}{3} \)

The slope of the given line (m₁) = 4/3. Let m₂ be the slope of the perpendicular line:
\( m_1 \times m_2 = -1 \)
\( \frac{4}{3} \times m_2 = -1 \)
\( m_2 = -\frac{3}{4} \)

The perpendicular line passes through (1, -2) with slope -3/4:
\( y - (-2) = -\frac{3}{4}(x - 1) \)
\( 4(y + 2) = -3(x - 1) \)
\( 4y + 8 = -3x + 3 \)
\( 3x + 4y + 5 = 0 \)

To find the foot of the perpendicular (intersection of 4x - 3y - 5 = 0 and 3x + 4y + 5 = 0), multiply the first equation by 4 and the second by 3:
\( 16x - 12y - 20 = 0 \) ... (i)
\( 9x + 12y + 15 = 0 \) ... (ii)

Add (i) and (ii):
\( 25x - 5 = 0 \)
\( x = \frac{1}{5} \)

Substitute x = 1/5 into 4x - 3y - 5 = 0:
\( 4 \cdot \frac{1}{5} - 3y - 5 = 0 \)
\( \frac{4}{5} - 3y - 5 = 0 \)
\( 3y = \frac{4}{5} - 5 = \frac{4 - 25}{5} = -\frac{21}{5} \)
\( y = -\frac{7}{5} \)

The equation of the perpendicular is 3x + 4y + 5 = 0, and the foot of the perpendicular is (1/5, -7/5).
In simple words: To drop a perpendicular from a point to a line, find a line through the point that is perpendicular to the given line. The foot is where these two lines meet.

Exam Tip: Always verify your foot of perpendicular by checking it satisfies both line equations. Use elimination method (multiply equations strategically) to find intersection points cleanly.

Question 21. Prove that the line through (0, 0) and (2, 3) is parallel to the line through (2, -2) and (6, 4).
Answer:
The slope formula for a line through two points (x₁, y₁) and (x₂, y₂) is:
\( \text{Slope} = \frac{y_2 - y_1}{x_2 - x_1} \)

For the line through (0, 0) and (2, 3):
\( m_1 = \frac{3 - 0}{2 - 0} = \frac{3}{2} \)

For the line through (2, -2) and (6, 4):
\( m_2 = \frac{4 - (-2)}{6 - 2} = \frac{6}{4} = \frac{3}{2} \)

Since m₁ = m₂ = 3/2, the two lines have equal slopes. Therefore, the lines are parallel to each other.
In simple words: Lines are parallel when they have the same slope. We calculated the slope of each line and found both equal 3/2, so they must be parallel.

Exam Tip: To prove two lines are parallel, always show their slopes are equal using the slope formula. Never assume from coordinates alone.

Question 22(i). Prove that the line through (-2, 6) and (4, 8) is perpendicular to the line through (8, 12) and (4, 24).
Answer:
Using the slope formula:

Slope of the line through (-2, 6) and (4, 8):
\( m_1 = \frac{8 - 6}{4 - (-2)} = \frac{2}{6} = \frac{1}{3} \)

Slope of the line through (8, 12) and (4, 24):
\( m_2 = \frac{24 - 12}{4 - 8} = \frac{12}{-4} = -3 \)

Check the perpendicularity condition:
\( m_1 \times m_2 = \frac{1}{3} \times (-3) = -1 \)

Since the product of the slopes equals -1, the two lines are perpendicular to each other.
In simple words: Two lines are perpendicular when multiplying their slopes gives -1. We found m₁ = 1/3 and m₂ = -3, and their product is indeed -1.

Exam Tip: To prove perpendicularity, always compute the product of slopes. If it equals exactly -1, the lines are perpendicular. This is quicker than checking angles.

Question 22(ii). Show that the triangle formed by the points A(1, 3), B(3, -1) and C(-5, -5) is a right angled triangle (by using slopes).
Answer:
Using the slope formula for each side:

Slope of AB (line from A to B):
\( m_1 = \frac{-1 - 3}{3 - 1} = \frac{-4}{2} = -2 \)

Slope of BC (line from B to C):
\( m_2 = \frac{-5 - (-1)}{-5 - 3} = \frac{-4}{-8} = \frac{1}{2} \)

Check if AB and BC are perpendicular:
\( m_1 \times m_2 = (-2) \times \frac{1}{2} = -1 \)

Since the product of slopes equals -1, sides AB and BC are perpendicular to each other. Therefore, the angle at B is a right angle (90°), making triangle ABC a right-angled triangle.
In simple words: A triangle is right-angled if two of its sides meet at a right angle. We found that AB and BC are perpendicular (their slopes multiply to -1), so the angle at B is 90°.

Exam Tip: Always calculate slopes of all three sides to identify which two are perpendicular. Label your answer clearly with the angle at which the right angle occurs.

Question 23. Find the equation of the line through the point (-1, 3) and parallel to the line joining the points (0, -2) and (4, 5).
Answer:
First, find the slope of the line joining (0, -2) and (4, 5):
\( m = \frac{5 - (-2)}{4 - 0} = \frac{7}{4} \)

A line parallel to this has the same slope, so m = 7/4. Use the point-slope form with the point (-1, 3):
\( y - 3 = \frac{7}{4}(x - (-1)) \)
\( y - 3 = \frac{7}{4}(x + 1) \)
\( 4(y - 3) = 7(x + 1) \)
\( 4y - 12 = 7x + 7 \)
\( 7x - 4y + 19 = 0 \)

The equation of the required line is 7x - 4y + 19 = 0.
In simple words: Parallel lines have the same slope. Find the slope of the given line, then use that slope and the given point to write the equation of the new line.

Exam Tip: Always simplify the equation to standard form ax + by + c = 0. Check by substituting the given point - it must satisfy the equation.

Question 24. A(-1, 3), B(4, 2), C(3, -2) are the vertices of a triangle. (i) Find the coordinates of the centroid G of the triangle. (ii) Find the equation of the line through G and parallel to AC.
Answer:
(i) The centroid of a triangle with vertices (x₁, y₁), (x₂, y₂), and (x₃, y₃) is:
\( G(x, y) = \left(\frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3}\right) \)

Substitute A(-1, 3), B(4, 2), C(3, -2):
\( G = \left(\frac{-1 + 4 + 3}{3}, \frac{3 + 2 + (-2)}{3}\right) \)
\( G = \left(\frac{6}{3}, \frac{3}{3}\right) \)
\( G = (2, 1) \)

(ii) Find the slope of AC:
\( m = \frac{-2 - 3}{3 - (-1)} = \frac{-5}{4} \)

A line parallel to AC also has slope -5/4. Using point-slope form with G(2, 1):
\( y - 1 = -\frac{5}{4}(x - 2) \)
\( 4(y - 1) = -5(x - 2) \)
\( 4y - 4 = -5x + 10 \)
\( 5x + 4y - 14 = 0 \)

The equation of the required line is 5x + 4y - 14 = 0.
In simple words: The centroid is the point where the three medians of a triangle meet - it divides each median in a 2:1 ratio. Find it by averaging the x-coordinates and y-coordinates. Then use its coordinates and the slope to write the parallel line.

Exam Tip: The centroid formula is essential - remember it divides each coordinate by 3, not by 2. Double-check by verifying each vertex coordinate is used exactly once.

Question 25. Find the equation of the line through (0, -3) and perpendicular to the line joining the points (-3, 2) and (9, 1).
Answer:
Find the slope of the line through (-3, 2) and (9, 1):
\( m_1 = \frac{1 - 2}{9 - (-3)} = \frac{-1}{12} = -\frac{1}{12} \)

For a line perpendicular to this, use m₁ × m₂ = -1:
\( -\frac{1}{12} \times m_2 = -1 \)
\( m_2 = 12 \)

The perpendicular line passes through (0, -3) with slope 12. Using point-slope form:
\( y - (-3) = 12(x - 0) \)
\( y + 3 = 12x \)
\( 12x - y - 3 = 0 \)

The equation of the required line is 12x - y - 3 = 0.
In simple words: Find the slope of the given line, then use the perpendicularity rule to find the slope of the new line. Pass the new line through the given point and write its equation.

Exam Tip: When a given slope is a fraction like -1/12, its perpendicular slope is the negative reciprocal: 12/1 = 12. Always multiply to check: (-1/12) × 12 = -1. ✓

Question 26. The vertices of a △ABC are A(3, 8), B(-1, 2) and C(6, -6). Find: (i) slope of BC. (ii) equation of a line perpendicular to BC and passing through A.
Answer:
(i) Slope of BC using points B(-1, 2) and C(6, -6):
\( m_1 = \frac{-6 - 2}{6 - (-1)} = \frac{-8}{7} = -\frac{8}{7} \)

The slope of BC is -8/7.

(ii) Let m₂ be the slope of the perpendicular line. Using m₁ × m₂ = -1:
\( -\frac{8}{7} \times m_2 = -1 \)
\( m_2 = \frac{7}{8} \)

The line with slope 7/8 passing through A(3, 8) has equation:
\( y - 8 = \frac{7}{8}(x - 3) \)
\( 8(y - 8) = 7(x - 3) \)
\( 8y - 64 = 7x - 21 \)
\( 7x - 8y + 43 = 0 \)

The equation of the required line is 7x - 8y + 43 = 0.
In simple words: Calculate the slope of BC from the two given points. Use the perpendicularity rule to find the slope of the line through A that is perpendicular to BC. Write the equation using the point-slope form.

Exam Tip: After writing the equation, expand and collect terms carefully. Check your arithmetic by verifying the given point (A in this case) satisfies your final equation.

Question 27. The vertices of a triangle are A(10, 4), B(4, -9) and C(-2, -1). Find the equation of the altitude through A.
Answer: Given: Triangle vertices are A(10, 4), B(4, -9), and C(-2, -1).

First, find the slope of line BC using the slope formula:
\( m_1 = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-1 - (-9)}{-2 - 4} = \frac{8}{-6} = -\frac{4}{3} \)

Since the altitude from A is perpendicular to BC, the product of their slopes equals -1:
\( m_1 \times m_2 = -1 \)
\( -\frac{4}{3} \times m_2 = -1 \)
\( m_2 = \frac{3}{4} \)

Using the point-slope formula with point A(10, 4) and slope \( m_2 = \frac{3}{4} \):
\( y - 4 = \frac{3}{4}(x - 10) \)
\( 4(y - 4) = 3(x - 10) \)
\( 4y - 16 = 3x - 30 \)
\( 3x - 4y - 14 = 0 \)

In simple words: An altitude is a line drawn from one corner of a triangle straight down to the opposite side, hitting it at a right angle. To find it, we first calculate how steep the opposite side is, then draw a line from point A that crosses it at 90 degrees.

Exam Tip: Remember that perpendicular lines have slopes that multiply to give -1. Always use this relationship when finding altitudes or perpendicular bisectors.

Question 28. A(2, -4), B(3, 3) and C(-1, 5) are the vertices of triangle ABC. Find the equation of: (i) the median of the triangle through A. (ii) the altitude of the triangle through B.
Answer:
(i) Median through A:

A median connects a vertex to the midpoint of the opposite side. First, find the midpoint D of BC:
\( D = \left(\frac{3 + (-1)}{2}, \frac{3 + 5}{2}\right) = \left(\frac{2}{2}, \frac{8}{2}\right) = (1, 4) \)

Now use the two-point formula with A(2, -4) and D(1, 4):
\( y - (-4) = \frac{4 - (-4)}{1 - 2}(x - 2) \)
\( y + 4 = \frac{8}{-1}(x - 2) \)
\( y + 4 = -8(x - 2) \)
\( y + 4 = -8x + 16 \)
\( 8x + y - 12 = 0 \)

(ii) Altitude through B:

An altitude from B is perpendicular to side AC. First, find the slope of AC:
\( m_1 = \frac{5 - (-4)}{-1 - 2} = \frac{9}{-3} = -3 \)

The slope of the altitude (perpendicular to AC) is:
\( m_1 \times m_2 = -1 \)
\( -3 \times m_2 = -1 \)
\( m_2 = \frac{1}{3} \)

Using point-slope form with B(3, 3):
\( y - 3 = \frac{1}{3}(x - 3) \)
\( 3(y - 3) = x - 3 \)
\( 3y - 9 = x - 3 \)
\( x - 3y + 6 = 0 \)

In simple words: A median joins a corner to the middle of the opposite side. An altitude goes from a corner straight across to the opposite side at a 90-degree angle, not to its middle.

Exam Tip: For medians, always find the midpoint first using the midpoint formula. For altitudes, use the perpendicularity condition - slopes of perpendicular lines multiply to -1.

Question 29. Find the equation of the right bisector of the line segment joining the points (1, 2) and (5, -6).
Answer: A right bisector is a line that cuts a segment exactly in half and crosses it at a 90-degree angle.

Step 1: Find the slope of the segment joining (1, 2) and (5, -6):
\( m_1 = \frac{-6 - 2}{5 - 1} = \frac{-8}{4} = -2 \)

Step 2: The slope of the right bisector (perpendicular to the segment) is:
\( m_1 \times m_2 = -1 \)
\( -2 \times m_2 = -1 \)
\( m_2 = \frac{1}{2} \)

Step 3: Find the midpoint of the segment:
\( M = \left(\frac{1 + 5}{2}, \frac{2 + (-6)}{2}\right) = \left(\frac{6}{2}, \frac{-4}{2}\right) = (3, -2) \)

Step 4: Apply point-slope form with M(3, -2) and slope \( m_2 = \frac{1}{2} \):
\( y - (-2) = \frac{1}{2}(x - 3) \)
\( 2(y + 2) = x - 3 \)
\( 2y + 4 = x - 3 \)
\( x - 2y - 7 = 0 \)

In simple words: The right bisector cuts a line segment in half and stands at a right angle to it. Find the middle point, calculate the perpendicular slope, then write the equation.

Exam Tip: The right bisector always passes through the midpoint of the segment. Verify your answer by checking that the midpoint satisfies your equation.

Question 30. Points A and B have coordinates (7, -3) and (1, 9) respectively. Find (i) the slope of AB. (ii) the equation of the perpendicular bisector of the line segment AB. (iii) the value of p if (-2, p) lies on it.
Answer:
(i) Slope of AB:
\( m_1 = \frac{y_2 - y_1}{x_2 - x_1} = \frac{9 - (-3)}{1 - 7} = \frac{12}{-6} = -2 \)

(ii) Equation of the perpendicular bisector:

Since the perpendicular bisector is perpendicular to AB:
\( m_1 \times m_2 = -1 \)
\( -2 \times m_2 = -1 \)
\( m_2 = \frac{1}{2} \)

The midpoint M of AB is:
\( M = \left(\frac{7 + 1}{2}, \frac{-3 + 9}{2}\right) = \left(\frac{8}{2}, \frac{6}{2}\right) = (4, 3) \)

Using point-slope form:
\( y - 3 = \frac{1}{2}(x - 4) \)
\( 2(y - 3) = x - 4 \)
\( 2y - 6 = x - 4 \)
\( x - 2y + 2 = 0 \)

(iii) Find p if (-2, p) lies on the perpendicular bisector:

Substitute x = -2 and y = p into the equation:
\( -2 - 2p + 2 = 0 \)
\( -2p = 0 \)
\( p = 0 \)

In simple words: The perpendicular bisector of a segment passes through its centre point and stands at a 90-degree angle. Any point on this line is equally far from both endpoints of the original segment.

Exam Tip: Always check that the given point satisfies the perpendicular bisector equation by substitution. This is a quick verification method.

Question 31. The points B(1, 3) and D(6, 8) are two opposite vertices of a square ABCD. Find the equation of the diagonal AC.
Answer: In a square, the two diagonals are perpendicular to each other and bisect one another at their common midpoint.

Step 1: Find the slope of diagonal BD:
\( m_1 = \frac{8 - 3}{6 - 1} = \frac{5}{5} = 1 \)

Step 2: Since diagonal AC is perpendicular to diagonal BD:
\( m_1 \times m_2 = -1 \)
\( 1 \times m_2 = -1 \)
\( m_2 = -1 \)

Step 3: Find the midpoint of BD (this is also the midpoint of AC, since diagonals of a square bisect each other):
\( \text{Midpoint} = \left(\frac{1 + 6}{2}, \frac{3 + 8}{2}\right) = \left(\frac{7}{2}, \frac{11}{2}\right) \)

Step 4: Use point-slope form with slope -1 and the midpoint:
\( y - \frac{11}{2} = -1\left(x - \frac{7}{2}\right) \)
\( y - \frac{11}{2} = -x + \frac{7}{2} \)
\( y + x = \frac{7}{2} + \frac{11}{2} \)
\( x + y = 9 \)
\( x + y - 9 = 0 \)

In simple words: In a square, the two diagonals cross each other at right angles at the centre. If you know one diagonal and one vertex, you can find the equation of the other diagonal.

Exam Tip: The key property of a square is that its diagonals bisect each other at right angles and are equal in length. Use this to find slopes and midpoints quickly.

Question 32. ABCD is a rhombus. The coordinates of A and C are (3, 6) and (-1, 2) respectively. Write down the equation of BD.
Answer: In a rhombus, the diagonals bisect each other at right angles.

Step 1: Find the slope of diagonal AC:
\( m_1 = \frac{2 - 6}{-1 - 3} = \frac{-4}{-4} = 1 \)

Step 2: Since diagonal BD is perpendicular to diagonal AC:
\( m_1 \times m_2 = -1 \)
\( 1 \times m_2 = -1 \)
\( m_2 = -1 \)

Step 3: Find the midpoint of AC (this is also the midpoint of BD, since the diagonals bisect each other):
\( \text{Midpoint} = \left(\frac{3 + (-1)}{2}, \frac{6 + 2}{2}\right) = \left(\frac{2}{2}, \frac{8}{2}\right) = (1, 4) \)

Step 4: Use point-slope form with slope -1 and the midpoint (1, 4):
\( y - 4 = -1(x - 1) \)
\( y - 4 = -x + 1 \)
\( x + y - 5 = 0 \)

In simple words: A rhombus is a four-sided shape with all sides equal. Its diagonals cross each other at right angles and split each other exactly in half at their meeting point.

Exam Tip: The property that diagonals of a rhombus are perpendicular and bisect each other is essential. Always use the perpendicularity condition and the midpoint to write the equation of the second diagonal.

Question 33. Find the image of the point (1, 2) in the line x - 2y - 7 = 0.
Answer: The image of a point in a line is the reflection of that point across the line. The line itself becomes the perpendicular bisector of the segment joining the original point and its image.

Step 1: Rewrite the line equation to find its slope:
\( x - 2y - 7 = 0 \)
\( 2y = x - 7 \)
\( y = \frac{1}{2}x - \frac{7}{2} \)

The slope of the given line is \( m_1 = \frac{1}{2} \)

Step 2: Let P = (1, 2) be the original point and P' = (a, b) be its image.

The slope of PP' (the line joining P and P') must be perpendicular to the given line:
\( m_1 \times m_2 = -1 \)
\( \frac{1}{2} \times m_2 = -1 \)
\( m_2 = -2 \)

The slope of PP' is \( \frac{b - 2}{a - 1} = -2 \)
\( b - 2 = -2(a - 1) \)
\( b - 2 = -2a + 2 \)
\( 2a + b = 4 \) ... (equation i)

Step 3: The midpoint of PP' must lie on the given line:
\( \text{Midpoint M} = \left(\frac{a + 1}{2}, \frac{b + 2}{2}\right) \)

Substitute into the line equation:
\( \frac{a + 1}{2} - 2\left(\frac{b + 2}{2}\right) - 7 = 0 \)
\( \frac{a + 1}{2} - (b + 2) - 7 = 0 \)
\( \frac{a + 1}{2} - b - 2 - 7 = 0 \)
\( \frac{a + 1}{2} - b - 9 = 0 \)
\( a + 1 - 2b - 18 = 0 \)
\( a - 2b = 17 \) ... (equation ii)

Step 4: Solve the system of equations:
From equation (i): \( 2a + b = 4 \)
From equation (ii): \( a - 2b = 17 \)

Multiply equation (ii) by 2:
\( 2a - 4b = 34 \)

Subtract from equation (i):
\( (2a + b) - (2a - 4b) = 4 - 34 \)
\( 5b = -30 \)
\( b = -6 \)

Substitute back into equation (i):
\( 2a + (-6) = 4 \)
\( 2a = 10 \)
\( a = 5 \)

Therefore, the image P' = (5, -6)

In simple words: The image of a point in a line is like its reflection in a mirror. To find it, we use two key ideas: the line joining the point and its image must be perpendicular to the mirror line, and the mirror line must pass through the midpoint between them.

Exam Tip: Always set up two equations when finding an image - one from the perpendicularity condition and one from the midpoint lying on the line. Solve simultaneously to get the coordinates.

Question 34. If the line x - 4y - 6 = 0 is the perpendicular bisector of the line segment PQ and the coordinates of P are (1, 3), find the coordinates of Q.
Answer: Since the line x - 4y - 6 = 0 is the perpendicular bisector of segment PQ, it means this line passes through the midpoint of PQ and is perpendicular to PQ.

Step 1: Find the slope of the given line:
\( x - 4y - 6 = 0 \)
\( 4y = x - 6 \)
\( y = \frac{1}{4}x - \frac{6}{4} \)

The slope of the perpendicular bisector is \( \frac{1}{4} \)

Step 2: Since PQ is perpendicular to this line, the slope of PQ is:
\( m_1 \times m_2 = -1 \)
\( \frac{1}{4} \times m_1 = -1 \)
\( m_1 = -4 \)

Step 3: Write the equation of line PQ using point P(1, 3) and slope -4:
\( y - 3 = -4(x - 1) \)
\( y - 3 = -4x + 4 \)
\( 4x + y - 7 = 0 \)

Step 4: Find the intersection of PQ and the perpendicular bisector:
Solve simultaneously:
\( x - 4y - 6 = 0 \) ... (1)
\( 4x + y - 7 = 0 \) ... (2)

From (1): \( x = 4y + 6 \)

Substitute into (2):
\( 4(4y + 6) + y - 7 = 0 \)
\( 16y + 24 + y - 7 = 0 \)
\( 17y + 17 = 0 \)
\( y = -1 \)

Substitute back:\( x = 4(-1) + 6 = 2 \)

The midpoint M of PQ is (2, -1)

Step 5: Let Q = (a, b). Using the midpoint formula:
\( \frac{1 + a}{2} = 2 \)
\( 1 + a = 4 \)
\( a = 3 \)

\( \frac{3 + b}{2} = -1 \)
\( 3 + b = -2 \)
\( b = -5 \)

Therefore, Q = (3, -5)

In simple words: If a line is the perpendicular bisector of a segment, it cuts the segment in half and stands at a right angle to it. Find where this line crosses the segment (the midpoint), then use this to calculate the other endpoint.

Exam Tip: When a perpendicular bisector is given, always find the intersection point with the line through the known endpoint. This intersection gives you the midpoint, from which you can find the other endpoint using the midpoint formula.

Question 1. The slope of a line parallel to y-axis is
(a) 0
(b) 1
(c) -1
(d) not defined
Answer: (d) not defined
In simple words: A vertical line (parallel to the y-axis) has a slope that cannot be defined because the line goes straight up and down with no horizontal change.

Exam Tip: Remember that parallel lines share the same slope, and a vertical line's slope is always undefined because division by zero occurs in the slope formula.

Question 2. The slope of a line which makes an angle of 30° with the positive direction of x-axis is
(a) 1
(b) \( \frac{1}{\sqrt{3}} \)
(c) \( \sqrt{3} \)
(d) \( -\frac{1}{\sqrt{3}} \)
Answer: (b) \( \frac{1}{\sqrt{3}} \)
In simple words: When a line tilts at 30° from the horizontal, its slope equals the tangent of that angle. Tan 30° gives us \( \frac{1}{\sqrt{3}} \).

Exam Tip: For any line, slope equals tan(angle with x-axis). Learn the tangent values of common angles: 30°, 45°, 60°, and 90°.

Question 3. The slope of the line passing through the points (0, -4) and (-6, 2) is
(a) 0
(b) 1
(c) -1
(d) 6
Answer: (c) -1
In simple words: Using the slope formula with these two points: we take the change in y (from -4 to 2, which is 6) and divide it by the change in x (from 0 to -6, which is -6). This gives us -1.

Exam Tip: Always use the slope formula \( m = \frac{y_2 - y_1}{x_2 - x_1} \) carefully, watching signs on both the numerator and denominator.

Question 4. The slope of the line passing through the points (3, -2) and (-7, -2) is
(a) 0
(b) 1
(c) \( -\frac{1}{10} \)
(d) not defined
Answer: (a) 0
In simple words: When both points have the same y-coordinate (-2), the line is horizontal. Horizontal lines always have a slope of zero.

Exam Tip: If the y-values are identical for both points, the slope is always 0 - this is a horizontal line.

Question 5. The slope of the line passing through the points (3, -2) and (3, -4) is
(a) -2
(b) 0
(c) 1
(d) not defined
Answer: (d) not defined
In simple words: Both points share the same x-coordinate (3), so the line is vertical. Vertical lines have undefined slopes because we cannot divide by zero.

Exam Tip: When x-values are the same for both points, the line is vertical and the slope is undefined.

Question 6. The inclination of the line \( y = \sqrt{3}x - 5 \) is
(a) 30°
(b) 60°
(c) 45°
(d) 0°
Answer: (b) 60°
In simple words: The slope of this line is \( \sqrt{3} \), which equals tan 60°. So the angle the line makes with the x-axis is 60°.

Exam Tip: Identify the slope from the equation y = mx + c, then find the angle using tan θ = m. Recall that tan 60° = \( \sqrt{3} \).

Question 7. If the slope of the line passing through the points (2, 5) and (k, 3) is 2, then the value of k is
(a) -2
(b) -1
(c) 1
(d) 2
Answer: (c) 1
In simple words: We know the slope is 2 and we plug the two points into the slope formula. Solving for k: (3 - 5)/(k - 2) = 2 gives us k = 1.

Exam Tip: When given a slope and coordinates with an unknown variable, substitute into the slope formula and solve algebraically for the unknown.

Question 8. The slope of a line parallel to the line passing through the points (0, 6) and (7, 3) is
(a) \( \frac{3}{7} \)
(b) \( -\frac{3}{7} \)
(c) \( \frac{7}{3} \)
(d) \( -\frac{7}{3} \)
Answer: (b) \( -\frac{3}{7} \)
In simple words: First, find the slope of the given line by using its two points: (3 - 6)/(7 - 0) = -3/7. Since parallel lines have equal slopes, any line parallel to it also has slope -3/7.

Exam Tip: Parallel lines always have identical slopes. Find the slope of the given line and that is the answer.

Question 9. The slope of a line perpendicular to the line passing through the points (2, 5) and (-3, 6) is
(a) \( -\frac{1}{5} \)
(b) \( \frac{1}{5} \)
(c) -5
(d) 5
Answer: (d) 5
In simple words: The slope of the first line is (6 - 5)/(-3 - 2) = -1/5. For a perpendicular line, multiply slopes to get -1: (-1/5) × m = -1, so m = 5.

Exam Tip: For perpendicular lines, their slopes multiply to give -1. If one slope is m, the perpendicular slope is -1/m.

Question 10. The slope of a line parallel to the line 2x + 3y - 7 = 0 is
(a) \( -\frac{3}{2} \)
(b) \( \frac{3}{2} \)
(c) \( -\frac{2}{3} \)
(d) \( \frac{2}{3} \)
Answer: (a) \( -\frac{3}{2} \)
In simple words: Rearrange the equation to y = mx + c form: 3y = -2x + 7, so y = -2/3 x + 7/3. The slope is -2/3. Parallel lines share this slope.

Exam Tip: Convert the equation to slope-intercept form by solving for y, then read off the slope coefficient directly.

Question 11. The slope of a line perpendicular to the line 3x = 4y + 11 is
(a) \( \frac{3}{4} \)
(b) \( -\frac{3}{4} \)
(c) \( \frac{4}{3} \)
(d) \( -\frac{4}{3} \)
Answer: (d) \( -\frac{4}{3} \)
In simple words: Rearrange: 4y = 3x - 11, so y = 3/4 x - 11/4. The slope is 3/4. For a perpendicular line, the slope is -4/3 (the negative reciprocal).

Exam Tip: Find the slope from the equation, then take its negative reciprocal for the perpendicular slope.

Question 12. If the lines 2x + 3y = 5 and kx - 6y = 7 are parallel, then the value of k is
(a) 4
(b) -4
(c) \( \frac{1}{4} \)
(d) \( -\frac{1}{4} \)
Answer: (b) -4
In simple words: Rewrite both equations in y = mx + c form. First line: y = -2/3 x + 5/3 (slope = -2/3). Second line: y = k/6 x - 7/6 (slope = k/6). For parallel lines, set the slopes equal: -2/3 = k/6, so k = -4.

Exam Tip: For two lines to be parallel, their slopes must be equal. Convert both equations to find the slopes and equate them.

Question 13. If the line 3x - 4y + 7 = 0 and 2x + ky + 5 = 0 are perpendicular to each other, then the value of k is
(a) \( \frac{3}{2} \)
Answer: From the first equation: 4y = 3x + 7, so y = 3/4 x + 7/4 (slope = 3/4). From the second equation: ky = -2x - 5, so y = -2/k x - 5/k (slope = -2/k). For perpendicular lines, the product of slopes equals -1:
\( \frac{3}{4} \times \left(-\frac{2}{k}\right) = -1 \)
\( -\frac{6}{4k} = -1 \)
\( \frac{6}{4k} = 1 \)
\( 6 = 4k \)
\( k = \frac{3}{2} \)
In simple words: Find both slopes by rearranging the equations. Then use the perpendicular condition: the product of the two slopes must be -1. Solve for k.

Exam Tip: For perpendicular lines given as general linear equations, convert to slope-intercept form and apply m₁ × m₂ = -1.

Question. If 3x - 4y + 7 = 0 and 2x + ky + 5 = 0, then find k if both lines are perpendicular to each other.
Answer: Start by rearranging the equations into slope-intercept form. From 3x - 4y + 7 = 0, you get \( y = \frac{3}{4}x + \frac{7}{4} \). From 2x + ky + 5 = 0, you get \( y = -\frac{2}{k}x - \frac{5}{k} \). When you compare these with \( y = mx + c \), the slope of the first line is \( \frac{3}{4} \) and the slope of the second line is \( -\frac{2}{k} \). For two lines to be perpendicular, the product of their slopes must equal -1. So, \( \frac{3}{4} \times \left(-\frac{2}{k}\right) = -1 \), which simplifies to \( k = \frac{4 \times -2}{3 \times -1} = \frac{-8}{-3} = \frac{8}{3} \). Wait, let me recalculate: \( \frac{3}{4} \times \left(-\frac{2}{k}\right) = -1 \) gives \( -\frac{6}{4k} = -1 \), so \( k = \frac{6}{4} = \frac{3}{2} \).
In simple words: When two lines are perpendicular, multiply their slopes and you should get -1. Use this rule to find k.

Exam Tip: Always convert line equations to slope-intercept form first. Remember the perpendicularity condition: \( m_1 \times m_2 = -1 \).

Question 14. Which of the following equations represents a line passing through origin?
(a) 3x - 2y + 5 = 0
(b) 2x - 3y = 0
(c) x = 5
(d) y = -6
Answer: (b) 2x - 3y = 0
In simple words: A line passes through the origin when putting x = 0 and y = 0 into the equation gives 0 = 0. Only option (b) satisfies this test.

Exam Tip: Always substitute x = 0 and y = 0 into each equation. If the result equals 0, that line goes through the origin.

Question 15. Points A(x, y), B(3, -2) and C(4, -5) are collinear. The value of y in terms of x is:
(a) 3x - 11
(b) 11 - 3x
(c) 3x - 7
(d) 7 - 3x
Answer: (d) 7 - 3x
In simple words: When three points lie on the same line, they all have the same slope. Find the slope using points B and C, then use this slope with either point to get the relationship between x and y.

Exam Tip: For collinear points, slope of AB = slope of BC. Set them equal and simplify to find the required relationship.

Question 16. Which of the following equation represents a line equally inclined to the axes?
(a) 2x - 3y + 7 = 0
(b) x - y = 7
(c) x = 7
(d) y = -7
Answer: (b) x - y = 7
In simple words: A line is equally inclined to both axes when its slope equals 1 or -1. Rewrite x - y = 7 as y = x - 7, which has slope 1.

Exam Tip: A line equally inclined to the axes makes a 45-degree angle with each axis. Its slope is always \( \pm 1 \).

Assertion-Reason Type Questions

Question. y = -2 is the equation of a line.
Assertion (A): Its x-intercept is zero.
Reason (R): It does not intersect x-axis.
(a) Assertion (A) is true, Reason (R) is false.
(b) Assertion (A) is false, Reason (R) is true.
(c) Both Assertion (A) and Reason (R) are correct, and Reason (R) is the correct reason for Assertion (A).
(d) Both Assertion (A) and Reason (R) are correct, and Reason (R) is incorrect reason for Assertion (A).
Answer: (b) Assertion (A) is false, Reason (R) is true
In simple words: The line y = -2 is horizontal and stays 2 units below the x-axis. It never touches the x-axis, so it has no x-intercept. The reason is correct, but the assertion is wrong.

Exam Tip: Remember: x-intercept exists only when a line crosses the x-axis. A horizontal line parallel to the x-axis has no x-intercept.

Question. The slope of a line passing through (-1, 0) is 1.
Assertion (A): Its x-intercept and y-intercept are equal.
Reason (R): It makes an isosceles triangle with the coordinate axes.
(a) Assertion (A) is true, but Reason (R) is false.
(b) Assertion (A) is false, but Reason (R) is true.
(c) Both Assertion (A) and Reason (R) are correct, and Reason (R) is the correct reason for Assertion (A).
(d) Both Assertion (A) and Reason (R) are correct, and Reason (R) is incorrect reason for Assertion (A).
Answer: (b) Assertion (A) is false, but Reason (R) is true
In simple words: Using the point-slope form, the line equation is y = x + 1. The x-intercept is -1 and the y-intercept is 1 - they are not equal. However, the triangle formed with both axes has two equal sides of length 1 each, making it isosceles.

Exam Tip: When a line makes an isosceles triangle with the axes, two of the triangle's sides are equal in length. Use the distance formula to verify this fact.

Question. Given below are the equation of two lines: y = 2x + 8 and y = \( \frac{1}{2}x - 7 \)
Assertion (A): The two lines are perpendicular to each other.
Reason (R): Their slopes are reciprocal of each other.
(a) Assertion (A) is true, Reason (R) is false.
(b) Assertion (A) is false, Reason (R) is true.
(c) Both Assertion (A) and Reason (R) are correct, and Reason (R) is the correct reason for Assertion (A).
(d) Both Assertion (A) and Reason (R) are correct, and Reason (R) is incorrect reason for Assertion (A).
Answer: (b) Assertion (A) is false, Reason (R) is true
In simple words: The slopes are 2 and \( \frac{1}{2} \) - they are reciprocals of each other, so the reason is true. However, for perpendicular lines, the product of slopes must be -1, not 1. Since \( 2 \times \frac{1}{2} = 1 \), the lines are not perpendicular.

Exam Tip: Two slopes being reciprocals is not enough - they must be negative reciprocals for the lines to be perpendicular.

Question. l and m are two lines in the figure given below:
Assertion (A): They have equal y-intercept.
Reason (R): Their inclination is same.
(a) Assertion (A) is true, Reason (R) is false.
(b) Assertion (A) is false, Reason (R) is true.
(c) Both Assertion (A) and Reason (R) are correct, and Reason (R) is the correct reason for Assertion (A).
(d) Both Assertion (A) and Reason (R) are correct, and Reason (R) is incorrect reason for Assertion (A).
Answer: (a) Assertion (A) is true, Reason (R) is false
In simple words: Both lines cross the y-axis at the same point, so they share the same y-intercept. However, from the figure, the x-intercepts are equal in distance but opposite in sign, which means the lines have different angles of inclination.

Exam Tip: Look carefully at where lines intersect the axes. Equal y-intercepts mean they cross the y-axis at the same height, not that their slopes are equal.

Question. Assertion (A): The line 3x + 3y + 5 = 0 crosses the x-axis at the point \( \left(0, -\frac{5}{3}\right) \).
Reason (R): The ordinate of every point on x-axis is zero.
(a) Assertion (A) is true, Reason (R) is false.
(b) Assertion (A) is false, Reason (R) is true.
(c) Both Assertion (A) and Reason (R) are correct, and Reason (R) is the correct reason for Assertion (A).
(d) Both Assertion (A) and Reason (R) are correct, and Reason (R) is incorrect reason for Assertion (A).
Answer: (b) Assertion (A) is false, Reason (R) is true
In simple words: The reason is correct - all points on the x-axis have y-coordinate equal to 0. Setting y = 0 in 3x + 3y + 5 = 0 gives \( x = -\frac{5}{3} \), so the crossing point is \( \left(-\frac{5}{3}, 0\right) \), not \( \left(0, -\frac{5}{3}\right) \). The assertion is wrong.

Exam Tip: When finding where a line crosses the x-axis, substitute y = 0, not x = 0. The x-intercept point has the form (a, 0), not (0, a).

Chapter Test

Question 1. Find the equation of a line whose inclination is 60° and y-intercept is -4.
Answer: Given that the angle of inclination is 60° and the y-intercept is -4. The slope of the line is \( m = \tan 60° = \sqrt{3} \). Using the slope-intercept form y = mx + c, substitute m = \( \sqrt{3} \) and c = -4 to get \( y = \sqrt{3}x - 4 \).
In simple words: The inclination angle tells you the slope through the tangent function. Then plug the slope and y-intercept into y = mx + c.

Exam Tip: Always convert inclination angle to slope using \( m = \tan \theta \) before writing the equation.

Question 2. Write down the gradient and the intercept on the y-axis of the line 3y + 2x = 12.
Answer: Rearrange the equation 3y + 2x = 12 into slope-intercept form. Dividing by 3 gives \( y = -\frac{2}{3}x + 4 \). Comparing with y = mx + c, the gradient (slope) is \( -\frac{2}{3} \) and the y-intercept is 4.
In simple words: Rewrite the equation as y = ... to find the slope and y-intercept directly.

Exam Tip: The gradient is the coefficient of x after rearranging to y = mx + c form. The intercept on the y-axis is the constant term c.

Question 3. If the equation of a line is y = \( \sqrt{3}x + 1 \), find its inclination.
Answer: Comparing y = \( \sqrt{3}x + 1 \) with y = mx + c, the slope is m = \( \sqrt{3} \). Since m = tan θ, you have tan θ = \( \sqrt{3} \). This means θ = 60°.
In simple words: Extract the slope from the equation and use tan θ = m to find the angle.

Exam Tip: Remember key angle values: tan 30° = \( \frac{1}{\sqrt{3}} \), tan 45° = 1, tan 60° = \( \sqrt{3} \).

Question 4. If the line y = mx + c passes through the points (2, -4) and (-3, 1), determine the values of m and c.
Answer: Use the two-point form of a line. With points (2, -4) and (-3, 1), the slope is \( m = \frac{1 - (-4)}{-3 - 2} = \frac{5}{-5} = -1 \). Substituting m = -1 and point (2, -4) into y = mx + c gives -4 = -1(2) + c, so c = -2. Therefore, m = -1 and c = -2.
In simple words: Find slope using the two points, then use one point to find the y-intercept.

Exam Tip: Always verify your equation by checking that both points satisfy y = mx + c after you find m and c.

Question 5. If the points (1, 4), (3, -2) and (p, -5) lie on a line, find the value of p.
Answer: First, find the equation of the line through (1, 4) and (3, -2) using the two-point form. The slope is \( m = \frac{-2 - 4}{3 - 1} = \frac{-6}{2} = -3 \). Substituting into y - 4 = -3(x - 1) and simplifying gives y = -3x + 7. Now, since (p, -5) also lies on this line, substitute y = -5: -5 = -3p + 7, which gives 3p = 12, so p = 4.
In simple words: Get the line equation from two points, then use the third point to solve for the unknown coordinate.

Exam Tip: For collinear points, all three must satisfy the same line equation. Set up an equation using the third point and solve for the unknown variable.

Question 6. Find the inclination of the line joining the points P(4, 0) and Q(7, 3).
Answer: To find the slope of the line joining two points, we use the formula \( m = \frac{y_2 - y_1}{x_2 - x_1} \). Substituting P(4, 0) and Q(7, 3), we get \( m = \frac{3 - 0}{7 - 4} = \frac{3}{3} = 1 \). Since \( m = \tan \theta \) and \( m = 1 \), we have \( \tan \theta = \tan 45° \), which gives \( \theta = 45° \). Therefore, the inclination of the line is 45°.
In simple words: The slope is found by dividing the change in y by the change in x. A slope of 1 means the line makes a 45-degree angle with the x-axis.

Exam Tip: Remember that inclination is the angle a line makes with the positive x-axis. Use the slope formula carefully and match it to a known angle.

Question 7. Find the equation of the line passing through the point of intersection of the lines 2x + y = 5 and x - 2y = 5 and having y-intercept equal to \( -\frac{7}{3} \).
Answer: First, we find the point where the two lines meet. Multiplying the first equation by 2 gives \( 4x + 2y = 10 \). Adding this to the second equation: \( 4x + 2y + x - 2y = 10 + 5 \), so \( 5x = 15 \) and \( x = 3 \). Substituting into the first equation: \( 2(3) + y = 5 \), giving \( y = -1 \). The point of intersection is (3, -1). Since the y-intercept is \( -\frac{7}{3} \), we have \( c = -\frac{7}{3} \). Using \( y = mx + c \) with the point (3, -1): \( -1 = 3m - \frac{7}{3} \), so \( -7 = 21m - 3 \), leading to \( m = -\frac{4}{21} \). Therefore, \( y = -\frac{4}{21}x - \frac{7}{3} \), which simplifies to \( 4x + 21y + 9 = 0 \).
In simple words: Find where the two lines cross, then use that point along with the given y-intercept to write the equation of the new line.

Exam Tip: Always verify your intersection point by substituting into both original equations. Check that your final equation passes through the intersection point.

Question 8. If the lines \( \frac{x}{3} + \frac{y}{4} = 7 \) and 3x + ky = 11 are perpendicular to each other, find the value of k.
Answer: We convert both equations to the form \( y = mx + c \) to find their slopes. For the first line: \( \frac{4x + 3y}{12} = 7 \), giving \( 4x + 3y = 84 \) and \( y = -\frac{4}{3}x + 28 \). The slope is \( m_1 = -\frac{4}{3} \). For the second line: \( 3x + ky = 11 \) becomes \( y = -\frac{3}{k}x + \frac{11}{k} \), so \( m_2 = -\frac{3}{k} \). For perpendicular lines, \( m_1 \times m_2 = -1 \). Thus: \( -\frac{4}{3} \times -\frac{3}{k} = -1 \), giving \( \frac{4}{k} = -1 \), so \( k = -4 \).
In simple words: Two lines are perpendicular when the product of their slopes equals -1. Find each slope, set up the product equation, and solve for k.

Exam Tip: Always convert equations to slope-intercept form first. The perpendicularity condition \( m_1 \times m_2 = -1 \) is key for these problems.

Question 9. Write down the equation of a line parallel to x - 2y + 8 = 0 and passing through the point (1, 2).
Answer: First, we convert the given line to slope-intercept form: \( x - 2y + 8 = 0 \) becomes \( 2y = x + 8 \), so \( y = \frac{1}{2}x + 4 \). The slope is \( \frac{1}{2} \). A parallel line has the same slope, so we seek a line with slope \( \frac{1}{2} \) passing through (1, 2). Using point-slope form: \( y - 2 = \frac{1}{2}(x - 1) \), which gives \( 2(y - 2) = x - 1 \), so \( 2y - 4 = x - 1 \), and therefore \( x - 2y + 3 = 0 \).
In simple words: Parallel lines have the same slope. Find the slope of the given line, then use it with the given point to write the new line's equation.

Exam Tip: Remember that parallel lines never meet and always have equal slopes. Use the point-slope form for quick calculation.

Question 10. Write down the equation of the line passing through (-3, 2) and perpendicular to the line 3y = 5 - x.
Answer: We convert the given equation to slope-intercept form: \( 3y = 5 - x \) becomes \( y = -\frac{1}{3}x + \frac{5}{3} \). The slope is \( m_1 = -\frac{1}{3} \). For a perpendicular line, we use the condition \( m_1 \times m_2 = -1 \), giving \( -\frac{1}{3} \times m_2 = -1 \), so \( m_2 = 3 \). Now we apply point-slope form with slope 3 and point (-3, 2): \( y - 2 = 3(x - (-3)) \), which gives \( y - 2 = 3(x + 3) \), so \( y - 2 = 3x + 9 \), and therefore \( 3x - y + 11 = 0 \).
In simple words: Find the slope of the given line. Use the perpendicularity rule to get the new slope. Then apply the point-slope formula with the given point.

Exam Tip: Check that your perpendicular slope is the negative reciprocal of the original slope. Verify the final equation by substituting the given point.

Question 11. Find the equation of the line perpendicular to the line joining the points A(1, 2) and B(6, 7) and passing through the point which divides the line segment AB in the ratio 3 : 2.
Answer: The slope of line AB is \( s_1 = \frac{7 - 2}{6 - 1} = \frac{5}{5} = 1 \). For a perpendicular line, \( s_1 \times s_2 = -1 \), so \( s_2 = -1 \). Using the section formula, the point dividing AB in ratio 3:2 is \( \left( \frac{3 \times 6 + 2 \times 1}{3 + 2}, \frac{3 \times 7 + 2 \times 2}{3 + 2} \right) = \left( \frac{20}{5}, \frac{25}{5} \right) = (4, 5) \). Using point-slope form with slope -1 and point (4, 5): \( y - 5 = -1(x - 4) \), giving \( y - 5 = -x + 4 \), so \( x + y - 9 = 0 \).
In simple words: Find the slope of the line joining the two points. Use the section formula to locate the dividing point. Then write the perpendicular line through that point.

Exam Tip: Apply the section formula carefully with the correct order of points and ratio. Double-check the perpendicular slope calculation before using point-slope form.

Question 12. The points A(7, 3) and C(0, -4) are two opposite vertices of a rhombus ABCD. Find the equation of the diagonal BD.
Answer: The slope of diagonal AC is \( m_1 = \frac{-4 - 3}{0 - 7} = \frac{-7}{-7} = 1 \). In a rhombus, the diagonals bisect each other at right angles. Therefore, BD is perpendicular to AC, so \( m_1 \times m_2 = -1 \), giving \( m_2 = -1 \). The midpoint of AC (the center of the rhombus) is \( \left( \frac{7 + 0}{2}, \frac{3 + (-4)}{2} \right) = \left( \frac{7}{2}, -\frac{1}{2} \right) \). Using point-slope form: \( y - \left( -\frac{1}{2} \right) = -1 \left( x - \frac{7}{2} \right) \), which simplifies to \( y + \frac{1}{2} = -x + \frac{7}{2} \), so \( 2y + 1 = -2x + 7 \), giving \( x + y - 3 = 0 \).
In simple words: In a rhombus, diagonals cross at right angles. Find the slope of one diagonal, then use the perpendicularity property to get the slope of the other. The midpoint becomes the point through which the second diagonal passes.

Exam Tip: Remember that rhombus diagonals bisect each other at right angles - this is a key geometric property. The intersection point serves as the pivotal point for the perpendicular diagonal.

Question 13. A and B are two points on the x-axis and y-axis respectively.
(a) Write down the co-ordinates of A and B.
(b) P is a point on AB such that AP : PB = 3 : 1. Using section formula find the coordinates of point P.
(c) Find the equation of a line passing through P and perpendicular to AB.
Answer:
(a) From the figure, point A lies on the x-axis at (4, 0) and point B lies on the y-axis at (0, 4).

(b) Let P have coordinates (x, y). By the section formula: \( (x, y) = \left( \frac{3 \times 0 + 1 \times 4}{3 + 1}, \frac{3 \times 4 + 1 \times 0}{3 + 1} \right) = \left( \frac{4}{4}, \frac{12}{4} \right) = (1, 3) \). So the coordinates of P are (1, 3).

(c) The slope of AB is \( \frac{4 - 0}{0 - 4} = -1 \). For a perpendicular line, the slope becomes 1 (since \( (-1) \times 1 = -1 \)). Using point-slope form through P(1, 3): \( y - 3 = 1(x - 1) \), giving \( y - 3 = x - 1 \), so \( y = x + 2 \).
In simple words: A lies where the line crosses the x-axis, and B where it crosses the y-axis. The section formula divides the segment in the given ratio. The perpendicular line has a slope that is the negative reciprocal.

Exam Tip: Always read coordinates from the figure carefully. Apply the section formula with the correct order and ratio. For perpendicular lines, multiply the slopes to check they equal -1.

Question 14. A straight line passes through P(2, 1) and cuts the axes in points A, B. If BP : PA = 3 : 1.
Find :
(i) the coordinates of A and B.
(ii) the equation of the line AB.
Answer:
(i) Since A lies on the x-axis, let A = (x, 0). Since B lies on the y-axis, let B = (0, y). Point P(2, 1) divides BA in the ratio 3:1. Using the section formula: \( 2 = \frac{3x + 0}{3 + 1} = \frac{3x}{4} \), so \( x = \frac{8}{3} \). Similarly, \( 1 = \frac{3 \times 0 + 1 \times y}{3 + 1} = \frac{y}{4} \), so \( y = 4 \). Therefore, A = \( \left( \frac{8}{3}, 0 \right) \) and B = (0, 4).

(ii) The slope of line AB is \( \frac{4 - 0}{0 - \frac{8}{3}} = \frac{4}{-\frac{8}{3}} = -\frac{3}{2} \). Using the intercept form: \( \frac{x}{\frac{8}{3}} + \frac{y}{4} = 1 \), which simplifies to \( \frac{3x}{8} + \frac{y}{4} = 1 \). Multiplying by 8: \( 3x + 2y = 8 \).
In simple words: Use the section formula to find where the line crosses each axis. Then write the equation using those intercepts, or find the slope and use point-slope form.

Exam Tip: When a point divides a line segment joining the axes, apply the section formula carefully. Verify both intercepts by substituting them back into your final equation.

Question 15. A straight line makes on the coordinate axes positive intercepts whose sum is 7. If the line passes through the point (-3, 8), find its equation.
Answer: Let the line make intercept a and b with the x-axis and y-axis respectively. The line meets the x-axis at A(a, 0) and the y-axis at B(0, b).

Given that a + b = 7, so b = 7 - a.

Using the two point form, the equation of line AB is:

\( y - 0 = \frac{b - 0}{0 - a}(x - a) \)

\( y = \frac{b}{-a}(x - a) \)

\( -ay = bx - ba \)

\( bx + ay - ab = 0 \) ...(i)

Since the line passes through (-3, 8), substitute these values:

\( (7 - a)(-3) + 8a - a(7 - a) = 0 \)

\( -21 + 3a + 8a - 7a + a^2 = 0 \)

\( a^2 + 4a - 21 = 0 \)

\( a^2 + 7a - 3a - 21 = 0 \)

\( a(a + 7) - 3(a + 7) = 0 \)

\( (a - 3)(a + 7) = 0 \)

\( a = 3 \text{ or } a = -7 \)

Since only positive intercepts are made, a ≠ -7. Therefore, a = 3.

b = 7 - 3 = 4

Substituting a and b into equation (i):

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

The equation of the required line is \( 4x + 3y = 12 \).
In simple words: A line meeting both axes with positive values adds up to 7. By using the fact that it passes through (-3, 8), we find the two intercepts are 3 and 4, giving the line equation \( 4x + 3y = 12 \).

Exam Tip: Always use the intercept form and remember to reject negative solutions when the problem specifies positive intercepts. Verify your answer by checking that the line passes through the given point.

Question 16. If the coordinates of the vertex A of a square ABCD are (3, -2) and the equation of diagonal BD is 3x - 7y + 6 = 0, find the equation of the diagonal AC. Also find the coordinates of the centre of the square.
Answer: In a square, the diagonals bisect each other at right angles at the centre O.

From the equation of BD: 3x - 7y + 6 = 0

\( 7y = 3x + 6 \)

\( y = \frac{3}{7}x + \frac{6}{7} \)

Slope of BD: \( m_1 = \frac{3}{7} \)

Since AC is perpendicular to BD:

\( m_1 \times m_2 = -1 \)

\( \frac{3}{7} \times m_2 = -1 \)

\( m_2 = -\frac{7}{3} \)

Equation of AC using point-slope form with A(3, -2):

\( y - (-2) = -\frac{7}{3}(x - 3) \)

\( 3(y + 2) = -7(x - 3) \)

\( 3y + 6 = -7x + 21 \)

\( 7x + 3y = 15 \)

To find the centre O, solve the equations of AC and BD:

BD: \( 3x - 7y = -6 \) ...(i)

AC: \( 7x + 3y = 15 \) ...(ii)

Multiply (i) by 3 and (ii) by 7:

\( 9x - 21y = -18 \) ...(iii)

\( 49x + 21y = 105 \) ...(iv)

Adding (iii) and (iv):

\( 58x = 87 \)

\( x = \frac{3}{2} \)

Substituting into (i):

\( 3 \times \frac{3}{2} - 7y = -6 \)

\( \frac{9}{2} - 7y = -6 \)

\( 7y = \frac{9}{2} + 6 = \frac{21}{2} \)

\( y = \frac{3}{2} \)

The equation of AC is \( 7x + 3y - 15 = 0 \) and the coordinates of the centre are \( \left(\frac{3}{2}, \frac{3}{2}\right) \).
In simple words: The diagonals of a square cross each other at right angles. Using this fact along with the given diagonal BD and vertex A, we find the other diagonal and the centre point.

Exam Tip: Remember that in a square, diagonals are perpendicular and bisect each other. Use the perpendicularity condition \( m_1 \times m_2 = -1 \) to find the slope of the other diagonal quickly.

Question 17. If the lines kx - y + 4 = 0 and 2y = 6x + 7 are perpendicular to each other, find the value of k.
Answer: From the first equation: kx - y + 4 = 0

\( y = kx + 4 \)

Slope: \( m_1 = k \)

From the second equation: 2y = 6x + 7

\( y = 3x + \frac{7}{2} \)

Slope: \( m_2 = 3 \)

For perpendicular lines, the product of their slopes equals -1:

\( m_1 \times m_2 = -1 \)

\( k \times 3 = -1 \)

\( k = -\frac{1}{3} \)
In simple words: Two lines are perpendicular when the product of their slopes is -1. Find each slope, then use this rule to get k = -1/3.

Exam Tip: Always rewrite line equations in slope-intercept form (y = mx + c) to identify the slope clearly before applying the perpendicularity condition.

Question 18. Find the equation of a line parallel to 2y = 6x + 7 and passing through (-1, 1).
Answer: Parallel lines have equal slopes. From 2y = 6x + 7:

\( y = 3x + \frac{7}{2} \)

The slope is 3. Since the required line is parallel, it also has slope 3.

Using point-slope form with the point (-1, 1):

\( y - 1 = 3[x - (-1)] \)

\( y - 1 = 3(x + 1) \)

\( y - 1 = 3x + 3 \)

\( y = 3x + 4 \)

The equation of the line parallel to 2y = 6x + 7 and passing through (-1, 1) is \( y = 3x + 4 \).
In simple words: Parallel lines have the same slope. Extract the slope from the given line, then use the given point to write the new line's equation.

Exam Tip: For parallel lines, always extract the slope from the given line first. Then apply the point-slope form carefully with the new point to avoid sign errors.

Question 19. A line segment joining P (2, -3) and Q (0, -1) is cut by the x-axis at the point R. A line AB cuts the y-axis at T(0, 6) and is perpendicular to PQ at S. Find the: (a) equation of line PQ (b) equation of line AB (c) coordinates of points R and S.
Answer: (a) Equation of line PQ:

Slope of PQ:

\( m = \frac{-1 - (-3)}{0 - 2} = \frac{2}{-2} = -1 \)

Using point-slope form with P(2, -3):

\( y - (-3) = -1(x - 2) \)

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

\( x + y + 1 = 0 \)

The equation of line PQ is \( x + y + 1 = 0 \).

(b) Equation of line AB:

For perpendicular lines: \( m_1 \times m_2 = -1 \)

\( -1 \times m_2 = -1 \)

\( m_2 = 1 \)

Line AB passes through T(0, 6) with slope 1:

\( y - 6 = 1(x - 0) \)

\( y = x + 6 \)

\( x - y + 6 = 0 \)

The equation of line AB is \( x - y + 6 = 0 \).

(c) Coordinates of R and S:

Point R lies on PQ and on the x-axis, so y = 0:

\( x + 0 + 1 = 0 \)

\( x = -1 \)

Therefore, R = (-1, 0).

Point S is the intersection of AB and PQ:

From AB: \( y = x + 6 \)

Substitute into PQ:

\( x + (x + 6) + 1 = 0 \)

\( 2x + 7 = 0 \)

\( x = -\frac{7}{2} \)

\( y = -\frac{7}{2} + 6 = \frac{5}{2} \)

Therefore, S = \( \left(-\frac{7}{2}, \frac{5}{2}\right) \).

The coordinates of R = (-1, 0) and S = \( \left(-\frac{7}{2}, \frac{5}{2}\right) \).
In simple words: Find the slope of PQ, use perpendicularity to get AB's slope, then find where each line crosses the axes or meets the other line.

Exam Tip: When finding the x-intercept, set y = 0 in the line equation. When finding the intersection of two lines, solve them simultaneously. Always verify your coordinates satisfy both original equations.

Question 20. Find the coordinates of the centroid P of the triangle ABC, whose vertices are A(-1, 3), B(3, -1) and C(0, 0). Hence, find the equation of a line passing through P and parallel to AB.
Answer: The centroid of a triangle with vertices \( (x_1, y_1) \), \( (x_2, y_2) \), and \( (x_3, y_3) \) is:

\( \left(\frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3}\right) \)

Substituting A(-1, 3), B(3, -1), and C(0, 0):

\( P = \left(\frac{-1 + 3 + 0}{3}, \frac{3 + (-1) + 0}{3}\right) = \left(\frac{2}{3}, \frac{2}{3}\right) \)

Slope of AB:

\( m = \frac{-1 - 3}{3 - (-1)} = \frac{-4}{4} = -1 \)

Since the required line is parallel to AB, it has slope -1.

Using point-slope form with P\( \left(\frac{2}{3}, \frac{2}{3}\right) \):

\( y - \frac{2}{3} = -1\left(x - \frac{2}{3}\right) \)

\( 3y - 2 = -1(3x - 2) \)

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

\( 3x + 3y = 4 \)

The required equation is \( 3x + 3y = 4 \).
In simple words: The centroid is found by averaging the x-coordinates and y-coordinates of all three vertices. Then use the slope of AB to write a parallel line through this centroid.

Exam Tip: The centroid formula is a standard result - memorize it to save time. When writing the parallel line equation, clear fractions early by multiplying the point-slope form to avoid arithmetic errors.