ML Aggarwal Class 10 Maths Solutions Chapter 11 Section Formula

Question 1. Find the coordinates of the mid-points of the line segments joining the following pairs of points:
(i) (2, -3), (-6, 7)
(ii) (5, -11), (4, 3)
(iii) (a + 3, 5b), (2a - 1, 3b + 4)
Answer: The mid-point formula tells us that for any two points (x₁, y₁) and (x₂, y₂), the middle point is found by averaging the x-coordinates and y-coordinates separately.

(i) Using the formula with (2, -3) and (-6, 7):
\[ \left( \frac{2 + (-6)}{2}, \frac{(-3) + 7}{2} \right) = \left( \frac{-4}{2}, \frac{4}{2} \right) = (-2, 2) \]

(ii) For points (5, -11) and (4, 3):
\[ \left( \frac{5 + 4}{2}, \frac{-11 + 3}{2} \right) = \left( \frac{9}{2}, \frac{-8}{2} \right) = \left( \frac{9}{2}, -4 \right) \]

(iii) For the algebraic pair (a + 3, 5b) and (2a - 1, 3b + 4):
\[ \left( \frac{a + 3 + 2a - 1}{2}, \frac{5b + 3b + 4}{2} \right) = \left( \frac{3a + 2}{2}, \frac{8b + 4}{2} \right) = \left( \frac{3a + 2}{2}, 4b + 2 \right) \]
In simple words: The mid-point is halfway between two points. Add the x-values and divide by 2; add the y-values and divide by 2.

Exam Tip: Always apply the mid-point formula correctly - the x-coordinate of the midpoint is the average of the two x-coordinates, and similarly for y. Check your arithmetic carefully, especially with negative numbers.

Question 2. Find a point P which divides internally the line segment joining the points A(-3, 9) and B(1, -3) in the ratio 1 : 3.
Answer: When a point divides a line segment internally in a given ratio, we apply the section formula. For a point dividing the segment from A(x₁, y₁) to B(x₂, y₂) in the ratio m₁ : m₂, the coordinates are:
\[ P = \left( \frac{m_1 x_2 + m_2 x_1}{m_1 + m_2}, \frac{m_1 y_2 + m_2 y_1}{m_1 + m_2} \right) \]

Given: m₁ = 1, m₂ = 3, A(-3, 9), B(1, -3)
\[ P = \left( \frac{1(1) + 3(-3)}{1 + 3}, \frac{1(-3) + 3(9)}{1 + 3} \right) = \left( \frac{1 - 9}{4}, \frac{-3 + 27}{4} \right) = \left( \frac{-8}{4}, \frac{24}{4} \right) = (-2, 6) \]
In simple words: The section formula helps find where a point sits when it splits a line segment into two parts in a known ratio. Plug the numbers into the formula and simplify.

Exam Tip: Remember to place the m₁ term with the second point's coordinates and m₂ with the first point's coordinates in the numerator. The answer should always lie on the line segment itself.

Question 3(i). Find the coordinates of the points of trisection of the line segment joining the points (3, -3) and (6, 9).
Answer: Trisection means dividing a line segment into three equal parts using two points, P and Q. If AP = PQ = QB, then we need to find where these two division points lie.

Since the segment is divided equally, AP : PB = 1 : 2 (meaning P divides AB in the ratio 1 : 2), and AQ : QB = 2 : 1 (meaning Q divides AB in the ratio 2 : 1).

For point P (dividing in ratio 1 : 2):
\[ P = \left( \frac{1(6) + 2(3)}{1 + 2}, \frac{1(9) + 2(-3)}{1 + 2} \right) = \left( \frac{6 + 6}{3}, \frac{9 - 6}{3} \right) = \left( \frac{12}{3}, \frac{3}{3} \right) = (4, 1) \]

For point Q (dividing in ratio 2 : 1):
\[ Q = \left( \frac{2(6) + 1(3)}{2 + 1}, \frac{2(9) + 1(-3)}{2 + 1} \right) = \left( \frac{12 + 3}{3}, \frac{18 - 3}{3} \right) = \left( \frac{15}{3}, \frac{15}{3} \right) = (5, 5) \]
In simple words: Trisection splits a line into three equal pieces. The first trisecting point divides it 1 : 2, and the second divides it 2 : 1.

Exam Tip: For trisection problems, the ratios of the two points are always 1:2 and 2:1. Use the section formula twice, once for each point.

Question 3(ii). The line segment joining the points (3, -4) and (1, 2) is trisected at the points P and Q. If the coordinates of P and Q are (p, -2) and \( \left(\frac{5}{3}, q\right) \) respectively, find the values of p and q.
Answer: Since P and Q trisect the segment from A(3, -4) to B(1, 2), we know that P divides AB in the ratio 1 : 2 and Q divides AB in the ratio 2 : 1.

For point P (ratio 1 : 2):
\[ P = \left( \frac{1(1) + 2(3)}{1 + 2}, \frac{1(2) + 2(-4)}{1 + 2} \right) = \left( \frac{1 + 6}{3}, \frac{2 - 8}{3} \right) = \left( \frac{7}{3}, \frac{-6}{3} \right) = \left( \frac{7}{3}, -2 \right) \]

Comparing with (p, -2), we find \( p = \frac{7}{3} \).

For point Q (ratio 2 : 1):
\[ Q = \left( \frac{2(1) + 1(3)}{2 + 1}, \frac{2(2) + 1(-4)}{2 + 1} \right) = \left( \frac{2 + 3}{3}, \frac{4 - 4}{3} \right) = \left( \frac{5}{3}, 0 \right) \]

Comparing with \( \left(\frac{5}{3}, q\right) \), we find q = 0.
In simple words: Calculate the trisection points using the section formula, then match the calculated coordinates with the given forms to find the unknown values.

Exam Tip: Set up the section formula equations carefully, apply them correctly, and then compare x and y coordinates separately to solve for unknowns.

Question 4. The line segment joining the points A(3, 2) and B(5, 1) is divided at the point P in the ratio 1 : 2 and it lies on the line 3x - 18y + k = 0. Find the value of k.
Answer: First, find the coordinates of point P using the section formula. Since P divides AB in the ratio 1 : 2:
\[ P = \left( \frac{1(5) + 2(3)}{1 + 2}, \frac{1(1) + 2(2)}{1 + 2} \right) = \left( \frac{5 + 6}{3}, \frac{1 + 4}{3} \right) = \left( \frac{11}{3}, \frac{5}{3} \right) \]

Since P lies on the line 3x - 18y + k = 0, substitute the coordinates of P into the equation:
\[ 3\left(\frac{11}{3}\right) - 18\left(\frac{5}{3}\right) + k = 0 \]
\[ 11 - 30 + k = 0 \]
\[ -19 + k = 0 \]
\[ k = 19 \]
In simple words: Find where point P is located using the section formula. Then check that P satisfies the line equation by substituting, which helps you find the unknown k.

Exam Tip: Always verify that the point you found actually lies on the given line. Substitute both coordinates carefully into the equation.

Question 5. Find the coordinates of the point which is three-fourth of the way from A(3, 1) to B(-2, 5).
Answer: If point P is three-fourths of the way from A to B, then AP = \( \frac{3}{4} \) AB, which means AP : PB = 3 : 1. Using the section formula:
\[ P = \left( \frac{3(-2) + 1(3)}{3 + 1}, \frac{3(5) + 1(1)}{3 + 1} \right) = \left( \frac{-6 + 3}{4}, \frac{15 + 1}{4} \right) = \left( \frac{-3}{4}, \frac{16}{4} \right) = \left( -\frac{3}{4}, 4 \right) \]
In simple words: To find a point that is a certain fraction of the way along a segment, express that fraction as a ratio and apply the section formula.

Exam Tip: Converting a fraction like "three-fourths" to a ratio (3:1 in this case) makes it easy to use the section formula directly.

Question 6. The line segment joining A(-3, 1) and B(5, -4) is a diameter of a circle whose centre is C. Find the coordinates of the point C.
Answer: Since AB is a diameter of the circle, the centre C must be the midpoint of AB. Using the midpoint formula:
\[ C = \left( \frac{-3 + 5}{2}, \frac{1 + (-4)}{2} \right) = \left( \frac{2}{2}, \frac{-3}{2} \right) = \left( 1, -\frac{3}{2} \right) \]
In simple words: The centre of a circle is always at the midpoint of any diameter. Use the midpoint formula to find it.

Exam Tip: Remember that a diameter passes through the centre of the circle, so the centre is always the midpoint of any diameter.

Question 7. The mid-point of the line segment joining the points (3m, 6) and (-4, 3n) is (1, 2m - 1). Find the values of m and n.
Answer: Applying the midpoint formula, the midpoint of (3m, 6) and (-4, 3n) is:
\[ \left( \frac{3m - 4}{2}, \frac{6 + 3n}{2} \right) \]

This equals (1, 2m - 1). Comparing x-coordinates:
\[ \frac{3m - 4}{2} = 1 \implies 3m - 4 = 2 \implies 3m = 6 \implies m = 2 \]

Comparing y-coordinates:
\[ \frac{6 + 3n}{2} = 2m - 1 \]

Substituting m = 2:
\[ \frac{6 + 3n}{2} = 2(2) - 1 = 3 \implies 6 + 3n = 6 \implies 3n = 0 \implies n = 0 \]
In simple words: Use the midpoint formula to get expressions for the x and y coordinates. Then set them equal to the given midpoint values and solve for the unknowns.

Exam Tip: Separate the x and y coordinate equations and solve them independently. This method prevents mistakes.

Question 8. The coordinates of the mid-point of the line segment PQ are (1, -2). The coordinates of P are (-3, 2). Find the coordinates of Q.
Answer: Let Q have coordinates (x, y). Since the midpoint of PQ is (1, -2), we have:
\[ \frac{-3 + x}{2} = 1 \quad \text{and} \quad \frac{2 + y}{2} = -2 \]

Solving the first equation:
\[ -3 + x = 2 \implies x = 5 \]

Solving the second equation:
\[ 2 + y = -4 \implies y = -6 \]

Therefore, Q = (5, -6).
In simple words: If you know the midpoint and one endpoint, you can work backwards to find the other endpoint by solving two simple equations.

Exam Tip: This is a reverse application of the midpoint formula. Set up equations by equating the midpoint components and solve carefully.

Question 9. AB is a diameter of a circle with centre C(-2, 5). If the point A is (3, -7). Find:
(i) The length of radius AC.
(ii) The coordinates of B.
Answer:
(i) To find the radius length, use the distance formula between C(-2, 5) and A(3, -7):
\[ AC = \sqrt{(3 - (-2))^2 + (-7 - 5)^2} = \sqrt{(5)^2 + (-12)^2} = \sqrt{25 + 144} = \sqrt{169} = 13 \]

The radius is 13 units.

(ii) Since C is the centre and AB is a diameter, C is the midpoint of AB. Let B have coordinates (x, y). Then:
\[ \frac{3 + x}{2} = -2 \quad \text{and} \quad \frac{-7 + y}{2} = 5 \]

From the first equation:
\[ 3 + x = -4 \implies x = -7 \]

From the second equation:
\[ -7 + y = 10 \implies y = 17 \]

Therefore, B = (-7, 17).
In simple words: The radius is the distance from the centre to any point on the circle. The centre is always the midpoint of any diameter.

Exam Tip: Use the distance formula to find the radius, and use the property that the centre is the midpoint of the diameter to locate the second endpoint.

Question 10. Find the reflection (image) of the point (5, -3) in the point (-1, 3).
Answer: The reflection of a point A in another point M means that M is the midpoint of the line segment joining A to its image A'. Let A' = (x, y). Then:
\[ \frac{5 + x}{2} = -1 \quad \text{and} \quad \frac{-3 + y}{2} = 3 \]

From the first equation:
\[ 5 + x = -2 \implies x = -7 \]

From the second equation:
\[ -3 + y = 6 \implies y = 9 \]

Therefore, A' = (-7, 9).
In simple words: To reflect a point in another point, the reflecting point becomes the midpoint between the original and its image. Solve for the image coordinates using this property.

Exam Tip: Always set the given point as the midpoint and solve the two equations (one for x, one for y) to find the reflected point.

Question 11. The line segment joining A \( \left( -1, \frac{5}{3} \right) \) and B(a, 5) is divided in the ratio 1 : 3 at P, the point where the line segment AB intersects y-axis. Calculate
(i) the value of a.
(ii) the coordinates of P.
Answer:
(i) Since P lies on the y-axis, its x-coordinate is 0. Using the section formula for the x-coordinate when the ratio is 1 : 3:
\[ 0 = \frac{1 \cdot a + 3 \cdot (-1)}{1 + 3} = \frac{a - 3}{4} \]
\[ a - 3 = 0 \implies a = 3 \]

(ii) Now with a = 3, find the coordinates of P using the section formula:
\[ P = \left( \frac{1(3) + 3(-1)}{1 + 3}, \frac{1(5) + 3 \cdot \frac{5}{3}}{1 + 3} \right) = \left( \frac{3 - 3}{4}, \frac{5 + 5}{4} \right) = \left( 0, \frac{10}{4} \right) = \left( 0, \frac{5}{2} \right) \]
In simple words: A point on the y-axis has x-coordinate 0. Use this fact with the section formula to find the unknown value a, then find P's coordinates.

Exam Tip: When a point lies on the y-axis, always remember that its x-coordinate is 0. Use this to set up an equation and solve for unknowns in division problems.

Question 11. Write the repeating decimal for each of the following and use a bar to show the repetend.
(i) \( \frac{1}{9} \)
(ii) \( \frac{-4}{3} \)
(iii) \( \frac{1}{6} \)
Answer:
(i) \( \frac{1}{9} = 0.\overline{1} \)
(ii) \( \frac{-4}{3} = -1.\overline{3} \)
(iii) \( \frac{1}{6} = 0.1\overline{6} \) — in this case, only the 6 repeats, not the 1 that comes before it.
In simple words: When you divide the numerator by the denominator, if the same digit (or set of digits) repeats infinitely, place a bar over those digits. The bar shows "this repeats forever."

Exam Tip: Place the bar only over the repeating digits, not over any non-repeating digits that appear first.

Question 12. The point P(-4, 1) divides the line segment joining the points A(2, -2) and B in the ratio 3 : 5. Find the point B.
Answer: Let the coordinates of B be (x, y). Using the section formula, when P divides AB in the ratio 3 : 5:
\( \left(\frac{3x + 5(2)}{3 + 5}, \frac{3y + 5(-2)}{3 + 5}\right) = (-4, 1) \)
\( \left(\frac{3x + 10}{8}, \frac{3y - 10}{8}\right) = (-4, 1) \)
From the x-coordinate: \( \frac{3x + 10}{8} = -4 \)
\( 3x + 10 = -32 \)
\( 3x = -42 \)
\( x = -14 \)
From the y-coordinate: \( \frac{3y - 10}{8} = 1 \)
\( 3y - 10 = 8 \)
\( 3y = 18 \)
\( y = 6 \)
Therefore, B = (-14, 6).
In simple words: The section formula helps us find an unknown endpoint when we know the dividing point and the ratio. Set up the formula, match the known coordinates, and solve for x and y.

Exam Tip: Always apply the section formula correctly: \( m_1x_2 + m_2x_1 \) goes in the numerator with the second point's x-coordinate, and check both x and y coordinates separately for accuracy.

Question 13(i). In what ratio does the point (5, 4) divide the line segment joining the points (2, 1) and (7, 6)?
Answer: Let point P(5, 4) divide the line segment joining (2, 1) and (7, 6) in the ratio \( m_1 : m_2 \). By the section formula:
\( x = \frac{m_1 \cdot 7 + m_2 \cdot 2}{m_1 + m_2} = 5 \)
\( 7m_1 + 2m_2 = 5(m_1 + m_2) \)
\( 7m_1 + 2m_2 = 5m_1 + 5m_2 \)
\( 2m_1 = 3m_2 \)
\( \frac{m_1}{m_2} = \frac{3}{2} \)
Therefore, the ratio is 3 : 2.
In simple words: To find a ratio, use the x-coordinate formula from the section formula. Plug in the known values and solve for the ratio between the two parts.

Exam Tip: Verify your answer by checking with the y-coordinate using the same ratio - both should give consistent results.

Question 13(ii). In what ratio does the point (-4, b) divide the line segment joining the points P(2, -2), Q(-14, 6)? Hence, find the value of b.
Answer: Let point P(-4, b) divide the segment joining (2, -2) and (-14, 6) in the ratio \( m_1 : m_2 \). Using the section formula for the x-coordinate:
\( -4 = \frac{m_1(-14) + m_2(2)}{m_1 + m_2} \)
\( -4(m_1 + m_2) = -14m_1 + 2m_2 \)
\( -4m_1 - 4m_2 = -14m_1 + 2m_2 \)
\( 10m_1 = 6m_2 \)
\( \frac{m_1}{m_2} = \frac{6}{10} = \frac{3}{5} \)
So \( m_1 : m_2 = 3 : 5 \). Now using the y-coordinate:
\( b = \frac{3(6) + 5(-2)}{3 + 5} = \frac{18 - 10}{8} = \frac{8}{8} = 1 \)
Therefore, b = 1.
In simple words: Find the ratio first using the x-coordinate, then substitute that ratio into the y-coordinate formula to find the unknown value.

Exam Tip: Keep the ratio as a fraction (like 6/10) until you finish finding x, then simplify it to its lowest terms (3/5) for the final answer.

Question 14. The line segment joining A(2, 3) and B(6, -5) is intersected by x-axis at a point K. Write down the ordinate of the point K. Hence, find the ratio in which K divides AB. Also find the coordinates of point K.
Answer: Since K lies on the x-axis, its y-coordinate (ordinate) is 0. Let K divide AB in the ratio \( m_1 : m_2 \). Using the section formula for the y-coordinate:
\( 0 = \frac{m_1(-5) + m_2(3)}{m_1 + m_2} \)
\( 0 = -5m_1 + 3m_2 \)
\( 5m_1 = 3m_2 \)
\( \frac{m_1}{m_2} = \frac{3}{5} \)
So the ratio is 3 : 5. Now find the x-coordinate:
\( x = \frac{3(6) + 5(2)}{3 + 5} = \frac{18 + 10}{8} = \frac{28}{8} = \frac{7}{2} \)
Therefore, K = \( \left(\frac{7}{2}, 0\right) \).
In simple words: Any point on the x-axis has a y-coordinate of 0. Set the y-coordinate formula equal to 0, solve for the ratio, then use that ratio to find the x-coordinate.

Exam Tip: Remember that the x-axis is where y = 0, and the y-axis is where x = 0 - this helps determine coordinates immediately without extra work.

Question 15. If A = (-4, 3) and B = (8, -6),
(i) find the length of AB.
(ii) in what ratio is the line segment joining AB, divided by the x-axis?
Answer:
(i) Using the distance formula:
\( AB = \sqrt{(8 - (-4))^2 + (-6 - 3)^2} = \sqrt{12^2 + (-9)^2} = \sqrt{144 + 81} = \sqrt{225} = 15 \) units
(ii) From the graph, the origin O(0, 0) lies on segment AB. Let O divide AB in the ratio \( m_1 : m_2 \). Using the x-coordinate from the section formula:
\( 0 = \frac{m_1(8) + m_2(-4)}{m_1 + m_2} \)
\( 8m_1 - 4m_2 = 0 \)
\( 8m_1 = 4m_2 \)
\( \frac{m_1}{m_2} = \frac{4}{8} = \frac{1}{2} \)
Therefore, the x-axis divides AB in the ratio 1 : 2.
In simple words: The distance formula uses the differences of x and y coordinates squared, then takes the square root. To find where an axis cuts a segment, set the appropriate coordinate to zero and solve for the ratio.

Exam Tip: When working with axes intersections, remember x-axis has y = 0 and y-axis has x = 0 - use this to simplify the section formula immediately.

Question 16. In what ratio does the line x - y - 2 = 0 divide the line segment joining the points (3, -1) and (8, 9)? Also, find the coordinates of the point of division.
Answer: Let the point P(x, y) where the line cuts the segment have coordinates from the section formula. If the line divides the segment joining (3, -1) and (8, 9) in the ratio \( m_1 : m_2 \), then:
\( x = \frac{m_1(8) + m_2(3)}{m_1 + m_2} = \frac{8m_1 + 3m_2}{m_1 + m_2} \)
\( y = \frac{m_1(9) + m_2(-1)}{m_1 + m_2} = \frac{9m_1 - m_2}{m_1 + m_2} \)
Since P lies on the line x - y - 2 = 0, we have x - y - 2 = 0. Substituting:
\( \frac{8m_1 + 3m_2}{m_1 + m_2} - \frac{9m_1 - m_2}{m_1 + m_2} - 2 = 0 \)
\( 8m_1 + 3m_2 - 9m_1 + m_2 - 2(m_1 + m_2) = 0 \)
\( -3m_1 + 2m_2 = 0 \)
\( 3m_1 = 2m_2 \)
\( \frac{m_1}{m_2} = \frac{2}{3} \)
So the ratio is 2 : 3. Now find the coordinates:
\( x = \frac{8(2) + 3(3)}{2 + 3} = \frac{16 + 9}{5} = \frac{25}{5} = 5 \)
\( y = \frac{9(2) - 3}{2 + 3} = \frac{18 - 3}{5} = \frac{15}{5} = 3 \)
Therefore, the line divides the segment in the ratio 2 : 3 and the coordinates of the point are (5, 3).
In simple words: When a line cuts a segment, the cutting point satisfies both the line's equation and the section formula. Substitute the formula expressions into the line equation and solve for the ratio.

Exam Tip: Always verify your final coordinates by substituting them back into the original line equation - they should satisfy it exactly.

Question 17. Given, a line segment AB joining the points A(-4, 6) and B(8, -3). Find:
(i) The ratio in which AB is divided by the y-axis.
(ii) The coordinates of the point of intersection.
(iii) The length of AB.
Answer:
(i) The y-axis intersects where x = 0. Let the y-axis divide AB in the ratio \( m_1 : m_2 \). Using the section formula:
\( 0 = \frac{m_1(8) + m_2(-4)}{m_1 + m_2} \)
\( 8m_1 - 4m_2 = 0 \)
\( 8m_1 = 4m_2 \)
\( \frac{m_1}{m_2} = \frac{4}{8} = \frac{1}{2} \)
Therefore, the required ratio is 1 : 2.
(ii) Using the y-coordinate from the section formula:
\( y = \frac{1(-3) + 2(6)}{1 + 2} = \frac{-3 + 12}{3} = \frac{9}{3} = 3 \)
The coordinates of the intersection point are (0, 3).
(iii) Using the distance formula:
\( AB = \sqrt{(8 - (-4))^2 + (-3 - 6)^2} = \sqrt{12^2 + (-9)^2} = \sqrt{144 + 81} = \sqrt{225} = 15 \) units
In simple words: On the y-axis, x is always 0. Set the x-coordinate formula to 0 to get the ratio. Then use that ratio in the y-coordinate formula to find where the segment meets the y-axis.

Exam Tip: For axes problems, immediately set the relevant coordinate to zero - this step saves time and reduces errors significantly.

Question 18. Calculate the length of the median through the vertex A of the triangle ABC with vertices A(7, -3), B(5, 3) and C(3, -1).
Answer: A median from a vertex reaches the midpoint of the opposite side. Let D be the midpoint of BC. Using the midpoint formula:
\( D = \left(\frac{5 + 3}{2}, \frac{3 + (-1)}{2}\right) = \left(\frac{8}{2}, \frac{2}{2}\right) = (4, 1) \)
The median AD connects A(7, -3) to D(4, 1). Using the distance formula:
\( AD = \sqrt{(7 - 4)^2 + (-3 - 1)^2} = \sqrt{3^2 + (-4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \) units
In simple words: A median starts from a vertex and goes to the midpoint of the opposite side. First find the midpoint using the midpoint formula, then calculate the distance from the vertex to that midpoint.

Exam Tip: Always identify which side is opposite to the given vertex - this determines which two points' midpoint you need to find first.

Question 19. Three consecutive vertices of a parallelogram ABCD are A(1, 2), B(1, 0) and C(4, 0). Find the fourth vertex D.
Answer: In a parallelogram, the diagonals bisect each other. Let O be the midpoint of diagonal AC. Using the midpoint formula:
\( O = \left(\frac{1 + 4}{2}, \frac{2 + 0}{2}\right) = \left(\frac{5}{2}, 1\right) \)
Since the diagonals bisect each other, O is also the midpoint of diagonal BD. If D = (x, y), then:
\( \left(\frac{5}{2}, 1\right) = \left(\frac{1 + x}{2}, \frac{0 + y}{2}\right) \)
From the x-coordinate: \( \frac{5}{2} = \frac{1 + x}{2} \)
\( 5 = 1 + x \)
\( x = 4 \)
From the y-coordinate: \( 1 = \frac{0 + y}{2} \)
\( 2 = y \)
Therefore, D = (4, 2).
In simple words: In a parallelogram, the two diagonals cut each other in the middle at the same point. Find where one diagonal's midpoint is, then use that to locate the fourth vertex.

Exam Tip: The key property of a parallelogram is that its diagonals bisect each other - use this property instead of trying to work with parallel sides, which is more calculation-heavy.

Question 20. If the points A(-2, -1), B(1, 0), C(p, 3) and D(1, q) form a parallelogram ABCD, find the values of p and q.
Answer: In a parallelogram, the diagonals cut each other in half. This means the midpoint of diagonal AC equals the midpoint of diagonal BD. Using the midpoint formula on AC with A(-2, -1) and C(p, 3):
\( x = \frac{p + (-2)}{2} = \frac{p - 2}{2} \) and \( y = \frac{3 + (-1)}{2} = 1 \)
Using the midpoint formula on BD with B(1, 0) and D(1, q):
\( x = \frac{1 + 1}{2} = 1 \) and \( y = \frac{0 + q}{2} = \frac{q}{2} \)
Since both midpoints are the same, equating the coordinates gives:
\( \frac{p - 2}{2} = 1 \) and \( \frac{q}{2} = 1 \)
\( \Rightarrow p - 2 = 2 \) and \( q = 2 \)
\( \Rightarrow p = 4 \) and \( q = 2 \)
In simple words: In a parallelogram, the two diagonals always cross at their middle points. Set the midpoints of both diagonals equal to find p and q.

Exam Tip: Always use the property that diagonals bisect each other in a parallelogram - this is the key to finding unknown coordinates quickly.

Question 21. If two vertices of a parallelogram are (3, 2), (-1, 0) and its diagonals meet at (2, -5), find the other two vertices of the parallelogram.
Answer: First, calculate the midpoint of the segment joining (3, 2) and (-1, 0):
\( \left( \frac{3 + (-1)}{2}, \frac{2 + 0}{2} \right) = (1, 1) \)
Since (1, 1) is not the same as (2, -5), these two points cannot be opposite vertices. They must be adjacent vertices. Let the given points be A(3, 2) and B(-1, 0), and the unknown vertices be C(x, y) and D(m, n). The diagonals are AC and BD, and they meet at (2, -5).
Using the midpoint formula with A(3, 2) and C(x, y) having midpoint (2, -5):
\( 2 = \frac{x + 3}{2} \) and \( -5 = \frac{y + 2}{2} \)
\( \Rightarrow x + 3 = 4 \) and \( y + 2 = -10 \)
\( \Rightarrow x = 1 \) and \( y = -12 \)
Now for D(m, n) using B(-1, 0) and midpoint (2, -5):
\( 2 = \frac{m + (-1)}{2} \) and \( -5 = \frac{n + 0}{2} \)
\( \Rightarrow m - 1 = 4 \) and \( n = -10 \)
\( \Rightarrow m = 5 \) and \( n = -10 \)
In simple words: When the diagonals' meeting point is given, apply the midpoint formula twice - once for each diagonal - to find the two missing vertices.

Exam Tip: Check that the given points are not opposite by finding their midpoint first - if it differs from the given intersection point, they are adjacent, not opposite.

Question 22. Find the third vertex of a triangle if its two vertices are (-1, 4) and (5, 2) and mid-point of one side is (0, 3).
Answer: Let A(-1, 4) and B(5, 2) be the two known vertices, and C(x, y) be the unknown third vertex. First, find the midpoint of AB:
\( \left( \frac{-1 + 5}{2}, \frac{4 + 2}{2} \right) = (2, 3) \)
Since (0, 3) is not the midpoint of AB, it must be the midpoint of either AC or BC. Assume (0, 3) is the midpoint of AC:
\( 0 = \frac{x + (-1)}{2} \) and \( 3 = \frac{y + 4}{2} \)
\( \Rightarrow x - 1 = 0 \) and \( y + 4 = 6 \)
\( \Rightarrow x = 1 \) and \( y = 2 \)
Now assume (0, 3) is the midpoint of BC:
\( 0 = \frac{5 + x}{2} \) and \( 3 = \frac{2 + y}{2} \)
\( \Rightarrow x + 5 = 0 \) and \( y + 2 = 6 \)
\( \Rightarrow x = -5 \) and \( y = 4 \)
In simple words: If one side's midpoint is given and you know two vertices, the midpoint formula can determine the third vertex - try both possibilities until one works.

Exam Tip: Remember that a triangle has three sides; if the given midpoint doesn't match one side's, it belongs to another - always check both cases.

Question 23. Find the coordinates of the vertices of the triangle, the middle points of whose sides are \( \left(0, \frac{1}{2}\right) \), \( \left(\frac{1}{2}, \frac{1}{2}\right) \) and \( \left(\frac{1}{2}, 0\right) \).
Answer: Let ABC be a triangle with vertices A(x₁, y₁), B(x₂, y₂), and C(x₃, y₃). Let D, E, and F be the midpoints of AB, BC, and CA respectively. From the given information:
D is the midpoint of AB: \( 0 = \frac{x_1 + x_2}{2} \) and \( \frac{1}{2} = \frac{y_1 + y_2}{2} \)
\( \Rightarrow x_1 + x_2 = 0 \) ... [Eq 1] and \( y_1 + y_2 = 1 \) ... [Eq 2]
E is the midpoint of BC: \( \frac{1}{2} = \frac{x_2 + x_3}{2} \) and \( \frac{1}{2} = \frac{y_2 + y_3}{2} \)
\( \Rightarrow x_2 + x_3 = 1 \) ... [Eq 3] and \( y_2 + y_3 = 1 \) ... [Eq 4]
F is the midpoint of CA: \( \frac{1}{2} = \frac{x_3 + x_1}{2} \) and \( 0 = \frac{y_3 + y_1}{2} \)
\( \Rightarrow x_3 + x_1 = 1 \) ... [Eq 5] and \( y_3 + y_1 = 0 \) ... [Eq 6]
Adding Equations 1, 3, and 5:
\( 2(x_1 + x_2 + x_3) = 2 \Rightarrow x_1 + x_2 + x_3 = 1 \) ... [Eq 7]
Subtracting Eq 3 from Eq 7: \( x_1 = 0 \)
Subtracting Eq 5 from Eq 7: \( x_2 = 0 \)
Subtracting Eq 1 from Eq 7: \( x_3 = 1 \)
Adding Equations 2, 4, and 6:
\( 2(y_1 + y_2 + y_3) = 2 \Rightarrow y_1 + y_2 + y_3 = 1 \) ... [Eq 8]
Subtracting Eq 4 from Eq 8: \( y_1 = 0 \)
Subtracting Eq 6 from Eq 8: \( y_2 = 1 \)
Subtracting Eq 2 from Eq 8: \( y_3 = 0 \)
In simple words: When midpoints of all three sides are known, add the equations from each midpoint and solve the system to recover the three vertices.

Exam Tip: Solving this type of problem requires setting up three equations from three midpoints and using addition/subtraction strategically to isolate each coordinate.

Question 24. Show by section formula that the points (3, -2), (5, 2) and (8, 8) are collinear.
Answer: Let the point (5, 2) divide the line joining (3, -2) and (8, 8) in the ratio m₁ : m₂. Using the section formula for the x-coordinate:
\( 5 = \frac{m_1 \times 8 + m_2 \times 3}{m_1 + m_2} \)
\( \Rightarrow 5(m_1 + m_2) = 8m_1 + 3m_2 \)
\( \Rightarrow 5m_1 + 5m_2 = 8m_1 + 3m_2 \)
\( \Rightarrow 2m_2 = 3m_1 \Rightarrow \frac{m_1}{m_2} = \frac{2}{3} \) ... [Eq 1]
Using the section formula for the y-coordinate:
\( 2 = \frac{m_1 \times 8 + m_2 \times (-2)}{m_1 + m_2} \)
\( \Rightarrow 2(m_1 + m_2) = 8m_1 - 2m_2 \)
\( \Rightarrow 2m_1 + 2m_2 = 8m_1 - 2m_2 \)
\( \Rightarrow 4m_2 = 6m_1 \Rightarrow \frac{m_1}{m_2} = \frac{2}{3} \) ... [Eq 2]
Since both equations yield the same ratio \( \frac{m_1}{m_2} = \frac{2}{3} \), the point (5, 2) lies on the line segment joining (3, -2) and (8, 8). This proves the three points are collinear.
In simple words: If a point lies on the line joining two other points at a specific ratio, all three points are collinear - verify by checking that both coordinates give the same dividing ratio.

Exam Tip: When proving collinearity using the section formula, obtain the ratio from both x and y coordinates separately - they must match exactly for the points to be collinear.

Question 25. Find the value of p for which the points (-5, 1), (1, p) and (4, -2) are collinear.
Answer: Since A(-5, 1), B(1, p), and C(4, -2) are collinear, point A divides BC in some ratio m₁ : m₂. Using the section formula for the x-coordinate:
\( -5 = \frac{m_1 \times 4 + m_2 \times 1}{m_1 + m_2} \)
\( \Rightarrow -5(m_1 + m_2) = 4m_1 + m_2 \)
\( \Rightarrow -5m_1 - 5m_2 = 4m_1 + m_2 \)
\( \Rightarrow -9m_1 = 6m_2 \Rightarrow \frac{m_1}{m_2} = -\frac{2}{3} \) ... [Eq 1]
Using the section formula for the y-coordinate:
\( 1 = \frac{m_1 \times (-2) + m_2 \times p}{m_1 + m_2} \)
\( \Rightarrow 1(m_1 + m_2) = -2m_1 + m_2 p \)
\( \Rightarrow m_1 + m_2 = -2m_1 + m_2 p \)
\( \Rightarrow 3m_1 = m_2(p - 1) \Rightarrow \frac{m_1}{m_2} = \frac{p - 1}{3} \) ... [Eq 2]
Comparing Eq 1 and Eq 2:
\( -\frac{2}{3} = \frac{p - 1}{3} \)
\( \Rightarrow -2 = p - 1 \Rightarrow p = -1 \)
In simple words: For three points to be collinear, the ratio found from the x-coordinates must equal the ratio from the y-coordinates - use this to solve for the unknown coordinate.

Exam Tip: Set up the section formula equations for both coordinates, obtain the ratio from each, then equate them to find the unknown parameter.

Question 26. The mid-point of the line segment AB shown in the adjoining diagram is (4, -3). Write down the coordinates of A and B.
Answer: From the diagram, point A lies on the x-axis and point B lies on the y-axis. This means A has coordinates (x, 0) and B has coordinates (0, y). The midpoint P of AB is (4, -3). Using the midpoint formula:
\( 4 = \frac{x + 0}{2} \) and \( -3 = \frac{0 + y}{2} \)
\( \Rightarrow x = 8 \) and \( y = -6 \)
In simple words: When a point is on the x-axis, its y-coordinate is 0. When on the y-axis, its x-coordinate is 0. Use the midpoint formula to find the missing values.

Exam Tip: Always identify from the diagram which axes the endpoints lie on - this greatly simplifies the midpoint formula application.

Question 27. Find the coordinates of the centroid of a triangle whose vertices are A(-1, 3), B(1, -1) and C(5, 1).
Answer: The centroid of a triangle with vertices (x₁, y₁), (x₂, y₂), and (x₃, y₃) is located at:
\( \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) \)
Substituting A(-1, 3), B(1, -1), and C(5, 1):
\( \left( \frac{-1 + 1 + 5}{3}, \frac{3 + (-1) + 1}{3} \right) = \left( \frac{5}{3}, \frac{3}{3} \right) = \left( \frac{5}{3}, 1 \right) \)
In simple words: The centroid is found by adding all three x-coordinates and dividing by 3, then doing the same for the y-coordinates.

Exam Tip: The centroid always divides each median in the ratio 2:1 from vertex to midpoint - remember this property for verification.

Question 28. Two vertices of a triangle are (3, -5) and (-7, 4). Find the third vertex given that the centroid is (2, -1).
Answer: Let the unknown third vertex be (x, y). The centroid formula gives:
\( 2 = \frac{3 + (-7) + x}{3} \) and \( -1 = \frac{-5 + 4 + y}{3} \)
\( \Rightarrow 2 = \frac{-4 + x}{3} \) and \( -1 = \frac{-1 + y}{3} \)
\( \Rightarrow 6 = -4 + x \) and \( -3 = -1 + y \)
\( \Rightarrow x = 10 \) and \( y = -2 \)
In simple words: When the centroid and two vertices are known, reverse the centroid formula by multiplying by 3 to find the sum, then solve for the missing vertex.

Exam Tip: Rearrange the centroid formula to isolate the unknown coordinates - multiply the centroid value by 3, then subtract the known coordinates.

Question 29. The vertices of a triangle are A(-5, 3), B(p, -1) and C(6, q). Find the values of p and q if the centroid of the triangle ABC is the point (1, -1).
Answer: Using the centroid formula:
\( 1 = \frac{-5 + p + 6}{3} \) and \( -1 = \frac{3 + (-1) + q}{3} \)
\( \Rightarrow 1 = \frac{1 + p}{3} \) and \( -1 = \frac{2 + q}{3} \)
\( \Rightarrow 3 = 1 + p \) and \( -3 = 2 + q \)
\( \Rightarrow p = 2 \) and \( q = -5 \)
In simple words: Substitute the centroid coordinates into the centroid formula and the given vertex coordinates to set up two equations, then solve for the unknowns p and q.

Exam Tip: Be careful with signs when substituting negative coordinates - double-check your arithmetic in the addition before dividing by 3.

Question 1. The points A(9, 0), B(9, 6), C(-9, 6) and D(-9, 0) are the vertices of a
(1) rectangle
(2) square
(3) rhombus
(4) trapezium
Answer: (1) rectangle
In simple words: When you calculate the distances between the four points using the distance formula, you find that opposite sides are equal in length. This means the shape is a rectangle.

Exam Tip: Always check if opposite sides are equal (AB = CD and BC = DA) - this property alone identifies a rectangle. Don't forget to also verify that the diagonals are equal, which further confirms a rectangle.

Question 2. If P(a/3, 4) is the mid-point of the line segment joining the points Q(-6, 5) and R(-2, 3), then the value of a is
(1) -4
(2) -6
(3) 12
(4) -12
Answer: (4) -12
In simple words: The midpoint formula tells us the x-coordinate of the midpoint equals the sum of the two x-coordinates divided by 2. Setting this equal to a/3 and solving gives us a = -12.

Exam Tip: Write out the midpoint formula completely - don't skip steps. Examiners want to see how you set up the equation and solve it, step by step.

Question 3. If the end points of a diameter of a circle are A(-2, 3) and B(4, -5) then the coordinates of its centre are
(1) (2, -2)
(2) (1, -1)
(3) (-1, 1)
(4) (-2, 2)
Answer: (2) (1, -1)
In simple words: The centre of a circle is always at the midpoint of any diameter. Using the midpoint formula with the two endpoints gives you the centre coordinates.

Exam Tip: Remember this key fact - the centre of a circle lies exactly at the midpoint of its diameter. This concept links two important formulas together.

Question 4. If one end of a diameter of a circle is (2, 3) and the centre is (-2, 5), then the other end is
(1) (-6, 7)
(2) (6, -7)
(3) (0, 8)
(4) (0, 4)
Answer: (1) (-6, 7)
In simple words: Since the centre is the midpoint of the diameter, you can work backwards using the midpoint formula. If you know one endpoint and the midpoint, you can find the other endpoint.

Exam Tip: Use the midpoint formula rearranged: if the centre is (h, k) and one endpoint is (x₁, y₁), then the other endpoint (x₂, y₂) satisfies h = (x₁ + x₂)/2 and k = (y₁ + y₂)/2. Solve for x₂ and y₂.

Question 5. If the mid-point of the line segment joining the points P(a, b - 2) and Q(-2, 4) is R(2, -3), then the values of a and b are
(1) a = 4, b = -5
(2) a = 6, b = 8
(3) a = 6, b = -8
(4) a = -6, b = 8
Answer: (3) a = 6, b = -8
In simple words: Apply the midpoint formula to both the x and y coordinates separately. This gives you two equations that you can solve to find the values of a and b.

Exam Tip: Always split the midpoint calculation into two parts - one for the x-coordinate and one for the y-coordinate. This makes it easier to track your work and reduces mistakes.

Question 6. The point which lies on the perpendicular bisector of the line segment joining the points A(-2, -5) and B(2, 5) is
(1) (0, 0)
(2) (0, 2)
(3) (2, 0)
(4) (-2, 0)
Answer: (1) (0, 0)
In simple words: Any point on the perpendicular bisector of a line segment must be equidistant from both endpoints. The perpendicular bisector passes through the midpoint of the segment.

Exam Tip: The midpoint of a segment always lies on its perpendicular bisector. Find the midpoint using the standard formula, and check which given option matches it.

Question 7. The coordinates of the point which is equidistant from the three vertices of △AOB (shown in the adjoining figure) are
(1) (x, y)
(2) (y, x)
(3) (x/2, y/2)
(4) (y/2, x/2)
Answer: (1) (x, y)
In simple words: For a right-angled triangle, the point equidistant from all three vertices is located at the midpoint of the hypotenuse. In this case, AB is the hypotenuse, so apply the midpoint formula to find the answer.

Exam Tip: Recognise when a triangle is right-angled - this is key to solving these problems quickly. The circumcentre of a right-angled triangle always sits at the midpoint of the hypotenuse.

Question 8. The fourth vertex D of a parallelogram ABCD whose vertices are A(-2, 3), B(6, 7) and C(8, 3) is
(1) (0, 1)
(2) (0, -1)
(3) (-1, 0)
(4) (1, 0)
Answer: (2) (0, -1)
In simple words: In a parallelogram, the two diagonals cut each other exactly in half. Find the midpoint of one diagonal, then use that same midpoint to find the fourth vertex on the other diagonal.

Exam Tip: The key property is that diagonals of a parallelogram bisect each other - use this to set up equations for the unknown coordinates and solve.

Question 9. The point which divides the line segment joining the points (7, -6) and (3, 4) in the ratio 1 : 2 internally lies in the
(1) Ist quadrant
(2) IInd quadrant
(3) IIIrd quadrant
(4) IVth quadrant
Answer: (4) IVth quadrant
In simple words: Use the section formula to find the exact coordinates of the point that divides the segment in the given ratio. Then check the signs of the x and y coordinates to determine which quadrant the point is in.

Exam Tip: Apply the section formula carefully - remember that a positive x-coordinate and negative y-coordinate means the point lies in the fourth quadrant.

Question 10. The centroid of the triangle whose vertices are (-4, -2), (6, 2) and (4, 6) is
(1) (2, 2)
(2) (2, 3)
(3) (3, 3)
(4) (0, -1)
Answer: (1) (2, 2)
In simple words: The centroid of a triangle is found by adding all three x-coordinates and dividing by 3, then doing the same for the y-coordinates. This single point represents the triangle's centre of gravity.

Exam Tip: The centroid formula is straightforward - add the three coordinates of each axis and divide by 3. Don't mix up the x and y coordinates during your calculation.

Question 11. A(1, 4), B(4, 1) and C(x, 4) are the vertices of △ABC. If the centroid of the triangles is G(4, 3), then x is equal to
(1) 2
(2) 1
(3) 7
(4) 4
Answer: (3) 7
In simple words: You know the centroid and two of the three vertices. Substitute these values into the centroid formula and solve the resulting equation to find the unknown x-coordinate.

Exam Tip: Set up the centroid formula with the known values and treat x as an unknown. Working through the formula step by step will isolate and reveal the value of x.

Question. Assertion (A): The point P(3, -1) divides the line segment joining the points A(1, -3) and B(6, 2) internally in the ratio 2 : 3.
Reason (R): The coordinates of the point which divides the line segment joining the points A(x₁, y₁) and B(x₂, y₂) internally in the ratio m₁ : m₂ are ((m₁x₂ + m₂x₁)/(m₁ + m₂), (m₁y₂ + m₂y₁)/(m₁ + m₂)).
(1) Assertion (A) is true, Reason (R) is false.
(2) Assertion (A) is false, Reason (R) is true.
(3) Both Assertion (A) and Reason (R) are correct, and Reason (R) is the correct reason for Assertion (A).
(4) Both Assertion (A) and Reason (R) are correct, and Reason (R) is incorrect reason for Assertion (A).
Answer: (3) Both Assertion (A) and Reason (R) are correct, and Reason (R) is the correct reason for Assertion (A).
In simple words: The Reason provides the section formula needed to verify the Assertion. When you use this formula with the given points and ratio, you indeed get the point P(3, -1), proving both statements are true and connected.

Exam Tip: For assertion-reason questions, always verify both statements independently. Then confirm whether the reason actually explains why the assertion is true - they must be logically connected, not just both correct.

Question. Assertion (A): If the coordinates of the mid-points of the sides AB and AC of △ABC are D(3, 5) and E(-3, 5) respectively, then BC = 12 units.
Reason (R): The line joining the mid-points of two sides of a triangle is parallel to the third side and equal to half of it.
(1) Assertion (A) is true, Reason (R) is false.
(2) Assertion (A) is false, Reason (R) is true.
(3) Both Assertion (A) and Reason (R) are correct, and Reason (R) is the correct reason for Assertion (A).
(4) Both Assertion (A) and Reason (R) are correct, and Reason (R) is incorrect reason for Assertion (A).
Answer: (3) Both Assertion (A) and Reason (R) are correct, and Reason (R) is the correct reason for Assertion (A).
In simple words: The segment connecting the two midpoints (D and E) has length 6. By the midpoint theorem, this segment is half the length of BC, so BC must be 12 units. The Reason directly explains why the Assertion is true.

Exam Tip: The midpoint theorem is essential - when a line joins the midpoints of two sides of a triangle, it is always half the length of and parallel to the third side. Use this to set up distance calculations quickly.

Question 3. Assertion (A): The point (-5, 0) lies on the x-axis. Reason (R): The y-coordinates of a point on x-axis is zero.
(1) Assertion (A) is true, Reason (R) is false.
(2) Assertion (A) is false, Reason (R) is true.
(3) Both Assertion (A) and Reason (R) are correct, and Reason (R) is the correct reason for Assertion (A).
(4) Both Assertion (A) and Reason (R) are correct, and Reason (R) is incorrect reason for Assertion (A).
Answer: (3) Both Assertion (A) and Reason (R) are correct, and Reason (R) is the correct reason for Assertion (A).
In simple words: A point stays on the x-axis when its y-coordinate equals zero. Since (-5, 0) has y-coordinate 0, it is on the x-axis. The reason explains why the assertion is true.

Exam Tip: For Assertion-Reason questions, always check if both statements are correct first, then verify whether the Reason actually explains the Assertion before selecting your answer.

Chapter Test

Question 1. The base BC of an equilateral triangle ABC lies on y-axis. The coordinates of the point C are (0, -3). If origin is the mid-point of the base BC, find the coordinates of the points A and B.
Answer: Since the origin is the mid-point of BC and C is at (0, -3), using the mid-point formula we get:
\( 0 = \frac{x + 0}{2} \) and \( 0 = \frac{y - 3}{2} \)
Solving, we get x = 0 and y = 3, so B is at (0, 3).
From the diagram, BC = 6 units. Since triangle ABC is equilateral, AB = BC = AC = 6 units.
Since A lies on the x-axis, let A be at (a, 0). Using the distance formula:
\( AB = \sqrt{(a - 0)^2 + (0 - 3)^2} = 6 \)
\( \sqrt{a^2 + 9} = 6 \)
\( a^2 + 9 = 36 \)
\( a^2 = 27 \)
\( a = \pm 3\sqrt{3} \)
Therefore, A is at \( (\pm 3\sqrt{3}, 0) \) and B is at (0, 3).
In simple words: The mid-point formula helps find B at (0, 3). Since all sides of an equilateral triangle are equal and measure 6 units, point A must be \( (\pm 3\sqrt{3}, 0) \) on the x-axis.

Exam Tip: When working with equilateral triangles, remember that all three sides are equal. Use this property along with the distance formula to find unknown vertices.

Question 2. Find the coordinates of the point that divides the line segment joining the points P(5, -2) and Q(9, 6) internally in the ratio of 3 : 1.
Answer: Let R be the point with coordinates (x, y) that divides PQ in the ratio 3 : 1.
Using the section formula:
\( x = \frac{m_1 x_2 + m_2 x_1}{m_1 + m_2} = \frac{3 \times 9 + 1 \times 5}{3 + 1} = \frac{27 + 5}{4} = \frac{32}{4} = 8 \)
\( y = \frac{m_1 y_2 + m_2 y_1}{m_1 + m_2} = \frac{3 \times 6 + 1 \times (-2)}{3 + 1} = \frac{18 - 2}{4} = \frac{16}{4} = 4 \)
Therefore, R = (8, 4).
In simple words: The section formula helps find the point that splits a line segment in a given ratio. Plug in the coordinates and the ratio to get the answer.

Exam Tip: Always label m₁ and m₂ clearly from the ratio given (m₁ belongs to the first point and m₂ to the second). This prevents sign and calculation errors.

Question 3. Find the coordinates of the point P which is three-fourth of the way from A(3, -1) to B(-2, 5).
Answer: If P is three-fourths of the way from A to B, then AP = (3/4) AB.
This means AP = (3/4)(AP + PB), which gives AP : PB = 3 : 1.
Using the section formula with m₁ : m₂ = 3 : 1:
\( x = \frac{3 \times (-2) + 1 \times 3}{3 + 1} = \frac{-6 + 3}{4} = -\frac{3}{4} \)
\( y = \frac{3 \times 5 + 1 \times (-1)}{3 + 1} = \frac{15 - 1}{4} = \frac{16}{4} = 4 \)
Therefore, P = \( \left( -\frac{3}{4}, 4 \right) \).
In simple words: When a point is three-fourths of the way along a segment, it divides that segment in the ratio 3 : 1. Use this ratio with the section formula to find the coordinates.

Exam Tip: Always convert fractional distances (like "three-fourths of the way") into a ratio form before applying the section formula.

Question 4. P and Q are the points on the line segment joining the points A(3, -1) and B(-6, 5) such that AP = PQ = QB. Find the coordinates of P and Q.
Answer: Since AP = PQ = QB, the segment AB is divided into three equal parts.
P divides AB in the ratio 1 : 2, and Q divides it in the ratio 2 : 1.
For P, using the section formula:
\( a = \frac{1 \times (-6) + 2 \times 3}{1 + 2} = \frac{-6 + 6}{3} = 0 \)
\( b = \frac{1 \times 5 + 2 \times (-1)}{1 + 2} = \frac{5 - 2}{3} = 1 \)
So P = (0, 1).
For Q, using the section formula:
\( c = \frac{2 \times (-6) + 1 \times 3}{2 + 1} = \frac{-12 + 3}{3} = -3 \)
\( d = \frac{2 \times 5 + 1 \times (-1)}{2 + 1} = \frac{10 - 1}{3} = 3 \)
So Q = (-3, 3).
Therefore, P = (0, 1) and Q = (-3, 3).
In simple words: When three equal segments are marked on a line, the first point divides the line in ratio 1 : 2 and the second in ratio 2 : 1. Use the section formula for each.

Exam Tip: Recognize when a segment is divided into equal parts and convert this into the corresponding ratio before using the section formula.

Question 5. The center of a circle is (α + 2, α - 5). Find the value of α, given that the circle passes through points (2, -2) and (8, -2).
Answer: Since both points lie on the circle, they are equidistant from the center.
Using the distance formula, let OA and OB be the distances from center O(α + 2, α - 5) to points A(2, -2) and B(8, -2) respectively.
\( OA = \sqrt{(2 - α - 2)^2 + (-2 - α + 5)^2} = \sqrt{(-α)^2 + (-α + 3)^2} = \sqrt{2α^2 - 6α + 9} \)
\( OB = \sqrt{(8 - α - 2)^2 + (-2 - α + 5)^2} = \sqrt{(6 - α)^2 + (3 - α)^2} = \sqrt{2α^2 + 45 - 18α} \)
Since OA = OB, we have:
\( 2α^2 - 6α + 9 = 2α^2 + 45 - 18α \)
\( -6α + 18α = 45 - 9 \)
\( 12α = 36 \)
\( α = 3 \)
In simple words: Both points are on the circle, so they must be the same distance from the center. Set these distances equal and solve for the unknown.

Exam Tip: When a circle's center has unknown coordinates, use the property that all points on the circle are equidistant from the center to form an equation.

Question 6. The mid-point of the line segment joining A(2, p) and B(q, 4) is (3, 5). Calculate the values of p and q.
Answer: Using the mid-point formula, the mid-point of A(2, p) and B(q, 4) is:
\( \left( \frac{2 + q}{2}, \frac{p + 4}{2} \right) = (3, 5) \)
From the x-coordinate: \( \frac{2 + q}{2} = 3 \) gives \( 2 + q = 6 \), so q = 4.
From the y-coordinate: \( \frac{p + 4}{2} = 5 \) gives \( p + 4 = 10 \), so p = 6.
Therefore, p = 6 and q = 4.
In simple words: The mid-point formula gives the average of the two endpoints. Set each coordinate equal to the given mid-point values and solve for the unknowns.

Exam Tip: Always separate the x and y coordinates when using the mid-point formula to avoid mixing up your calculations.

Question 7. The ends of a diameter of a circle have the coordinates (3, 0) and (-5, 6). PQ is another diameter where Q has the coordinates (-1, -2). Find the coordinates of P and the radius of the circle.
Answer: The center of the circle is the mid-point of the diameter AB:
\( \text{Center} = \left( \frac{3 + (-5)}{2}, \frac{0 + 6}{2} \right) = (-1, 3) \)
Since PQ is also a diameter, the center is also the mid-point of PQ. With Q = (-1, -2) and center = (-1, 3):
\( (-1, 3) = \left( \frac{-1 + x}{2}, \frac{-2 + y}{2} \right) \)
From the x-coordinate: \( -1 = \frac{-1 + x}{2} \) gives x = -1.
From the y-coordinate: \( 3 = \frac{-2 + y}{2} \) gives y = 8.
So P = (-1, 8).
The radius is the distance from the center (-1, 3) to any point on the circle, say (3, 0):
\( r = \sqrt{(3 - (-1))^2 + (0 - 3)^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \) units.
In simple words: Both diameters pass through the center. Find the center from one diameter, then use it to find the other endpoint. The radius is the distance from center to any point on the circle.

Exam Tip: Remember that the center of a circle is always the mid-point of any diameter. This property is key to solving diameter-related problems.

Question 8. In what ratio does the point (-4, 6) divide the line segment joining the points A(-6, 10) and B(3, -8)?
Answer: Let the point (-4, 6) divide AB in the ratio m : n.
Using the section formula for the x-coordinate:
\( -4 = \frac{m \times 3 + n \times (-6)}{m + n} = \frac{3m - 6n}{m + n} \)
\( -4(m + n) = 3m - 6n \)
\( -4m - 4n = 3m - 6n \)
\( -7m = -2n \)
\( \frac{m}{n} = \frac{2}{7} \)
Therefore, m : n = 2 : 7.
In simple words: Use the section formula with the x-coordinate of the given point. Set up an equation and solve for the ratio of m to n.

Exam Tip: When finding a ratio, work with one coordinate (usually x) that gives a clean equation. Verify with the y-coordinate if needed.

Question 9. Find the ratio in which the point P(-3, p) divides the line segment joining the points (-5, -4) and (-2, 3). Hence, find the value of p.
Answer: Let P(-3, p) divide the segment in the ratio m : n.
Using the section formula for the x-coordinate:
\( -3 = \frac{m \times (-2) + n \times (-5)}{m + n} = \frac{-2m - 5n}{m + n} \)
\( -3(m + n) = -2m - 5n \)
\( -3m - 3n = -2m - 5n \)
\( -m = -2n \)
\( \frac{m}{n} = \frac{2}{1} \)
So m : n = 2 : 1.
Now using the y-coordinate:
\( p = \frac{2 \times 3 + 1 \times (-4)}{2 + 1} = \frac{6 - 4}{3} = \frac{2}{3} \)
In simple words: Use the x-coordinate to find the ratio, then use that ratio with the y-coordinate formula to find p.

Exam Tip: Once you find the ratio from one coordinate, always use it to calculate the unknown coordinate. This ensures your answer is consistent.

Question 10. In what ratio is the line joining the points (4, 2) and (3, -5) divided by the x-axis? Also find the coordinates of the point of division.
Answer: Let point P on the x-axis divide the line segment joining points A(4, 2) and B(3, -5) in the ratio m : n. Let P have coordinates (x, 0).

Using the section formula for the y-coordinate:
\( 0 = \frac{m \times (-5) + n \times 2}{m + n} \)

\( \implies 0 = -5m + 2n \)

\( \implies 5m = 2n \)

\( \implies m : n = 2 : 5 \)

Now, using the section formula for the x-coordinate with m : n = 2 : 5:
\( x = \frac{2 \times 3 + 5 \times 4}{2 + 5} = \frac{6 + 20}{7} = \frac{26}{7} \)

Therefore, the coordinates of the division point are \( \left(\frac{26}{7}, 0\right) \), and the line is divided in the ratio 2 : 5 by the x-axis.
In simple words: When a line meets the x-axis, the y-value at that point is always zero. We use this fact to find what ratio the point divides the line segment.

Exam Tip: On the x-axis, always set y = 0. This single condition lets you find the ratio directly without extra calculation.

Question 11. If the abscissa of a point P is 2, find the ratio in which it divides the line segment joining the points (-4, 3) and (6, 3). Hence, find the coordinates of P.
Answer: Let A(-4, 3) and B(6, 3) be the two endpoints, and P(2, y) divide segment AB in the ratio m : n.

Using the section formula for the x-coordinate:
\( 2 = \frac{m \times 6 + n \times (-4)}{m + n} \)

\( \implies 2(m + n) = 6m - 4n \)

\( \implies 2m + 2n = 6m - 4n \)

\( \implies 6n = 4m \)

\( \implies m : n = 3 : 2 \)

Using the section formula for the y-coordinate with m : n = 3 : 2:
\( y = \frac{3 \times 3 + 2 \times 3}{3 + 2} = \frac{9 + 6}{5} = \frac{15}{5} = 3 \)

Therefore, the coordinates of P are (2, 3), and it divides the line in the ratio 3 : 2.
In simple words: The abscissa is just the x-coordinate. We use the x-value to find the ratio, then use that ratio to find the y-value.

Exam Tip: When one coordinate is given (here, x = 2), always substitute it into the section formula first to find the ratio before finding the other coordinate.

Question 12. Determine the ratio in which the line 2x + y - 4 = 0 divide the line segment joining the points A(2, -2) and B(3, 7). Also find the coordinates of the point of the division.
Answer: Suppose the line 2x + y - 4 = 0 divides segment AB in the ratio m : n at point P. Using the section formula, the coordinates of P are:

\( x = \frac{3m + 2n}{m + n}, \quad y = \frac{7m - 2n}{m + n} \)

Since P lies on the line 2x + y - 4 = 0, substitute these values:
\( 2 \cdot \frac{3m + 2n}{m + n} + \frac{7m - 2n}{m + n} - 4 = 0 \)

\( \implies \frac{6m + 4n + 7m - 2n - 4(m + n)}{m + n} = 0 \)

\( \implies 6m + 4n + 7m - 2n - 4m - 4n = 0 \)

\( \implies 9m - 2n = 0 \)

\( \implies m : n = 2 : 9 \)

Substituting m : n = 2 : 9 back:
\( x = \frac{3 \times 2 + 2 \times 9}{2 + 9} = \frac{6 + 18}{11} = \frac{24}{11} \)

\( y = \frac{7 \times 2 - 2 \times 9}{2 + 9} = \frac{14 - 18}{11} = -\frac{4}{11} \)

Therefore, the coordinates of the division point are \( \left(\frac{24}{11}, -\frac{4}{11}\right) \), and the ratio is 2 : 9.
In simple words: A line cuts through a segment at one point. We use the fact that the point must satisfy both the line equation and the section formula to find where it cuts.

Exam Tip: The key trick is to substitute the section formula expressions for x and y into the line equation - this gives you an equation in m and n only, which you can solve for the ratio.

Question 13. ABCD is a parallelogram. If the coordinates of A, B and D are (10, -6), (2, -6) and (4, -2) respectively, find the coordinates of C.
Answer: In parallelogram ABCD, the diagonals AC and BD cut each other in half. Let O be the point where they meet.

First, find O as the midpoint of BD:
\( O = \left(\frac{2 + 4}{2}, \frac{-6 + (-2)}{2}\right) = (3, -4) \)

Now, O is also the midpoint of AC. If C has coordinates (x, y):
\( (3, -4) = \left(\frac{10 + x}{2}, \frac{-6 + y}{2}\right) \)

Comparing both coordinates:
\( \frac{10 + x}{2} = 3 \implies 10 + x = 6 \implies x = -4 \)

\( \frac{-6 + y}{2} = -4 \implies -6 + y = -8 \implies y = -2 \)

Therefore, the coordinates of C are (-4, -2).
In simple words: In any parallelogram, the two diagonals always cross exactly at their midpoints. This property helps us find any missing corner.

Exam Tip: Remember that the key property of a parallelogram is that its diagonals bisect each other - use this as your starting point whenever a vertex is missing.

Question 14. ABCD is a parallelogram whose vertices A and B have coordinates (2, -3) and (-1, -1) respectively. If the diagonals of the parallelogram meet at the point M(1, -4), find the coordinates of C and D. Hence, find the perimeter of the parallelogram.
Answer: Since M(1, -4) is where the diagonals cross, it is the midpoint of both AC and BD.

For AC: If C has coordinates (x₁, y₁), then:
\( 1 = \frac{2 + x_1}{2} \implies x_1 = 0 \)

\( -4 = \frac{-3 + y_1}{2} \implies y_1 = -5 \)

So C = (0, -5).

For BD: If D has coordinates (x₂, y₂), then:
\( 1 = \frac{-1 + x_2}{2} \implies x_2 = 3 \)

\( -4 = \frac{-1 + y_2}{2} \implies y_2 = -7 \)

So D = (3, -7).

Now find the side lengths. Using the distance formula for AB:
\( AB = \sqrt{(2 - (-1))^2 + (-3 - (-1))^2} = \sqrt{3^2 + (-2)^2} = \sqrt{9 + 4} = \sqrt{13} \)

For BC:
\( BC = \sqrt{(0 - (-1))^2 + (-5 - (-1))^2} = \sqrt{1^2 + (-4)^2} = \sqrt{1 + 16} = \sqrt{17} \)

The perimeter of parallelogram ABCD is:
\( 2(AB + BC) = 2(\sqrt{13} + \sqrt{17}) \text{ units} \)

Therefore, C = (0, -5), D = (3, -7), and the perimeter is \( 2(\sqrt{13} + \sqrt{17}) \) units.
In simple words: Once you know where the diagonals cross, you can find all four corners. Then measure two neighbouring sides and double their sum to get the perimeter.

Exam Tip: In a parallelogram, opposite sides are equal, so you only need to find the lengths of two adjacent sides and then multiply the sum by 2.

Question 15. Given O (0, 0), P(1, 2), S(-3, 0). P divides OQ in the ratio 2 : 3 and OPRS is a parallelogram.
Find:
(i) the coordinates of Q.
(ii) the coordinates of R.
(iii) the ratio in which RQ is divided by the x-axis.
Answer:
(i) Let Q have coordinates (a, b). Point P(1, 2) divides segment OQ in the ratio 2 : 3. Using the section formula:

\( 1 = \frac{2 \times a + 3 \times 0}{2 + 3} = \frac{2a}{5} \implies a = \frac{5}{2} \)

\( 2 = \frac{2 \times b + 3 \times 0}{2 + 3} = \frac{2b}{5} \implies b = 5 \)

Therefore, Q = \( \left(\frac{5}{2}, 5\right) \).

(ii) In parallelogram OPRS, diagonals OR and PS bisect each other. Let M be their meeting point. Since M is the midpoint of PS:

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

Since M is also the midpoint of OR, and O = (0, 0), if R = (c, d):

\( -1 = \frac{0 + c}{2} \implies c = -2 \)

\( 1 = \frac{0 + d}{2} \implies d = 2 \)

Therefore, R = (-2, 2).

(iii) Let N be the point on the x-axis that divides RQ in the ratio m₁ : m₂. At this point, the y-coordinate is 0. Using the section formula:

\( 0 = \frac{m_1 \times 5 + m_2 \times (-2)}{m_1 + m_2} \)

\( \implies \frac{5m_1}{2} = 2m_2 \)

\( \implies m_1 : m_2 = 4 : 5 \)

Therefore, RQ is divided by the x-axis in the ratio 4 : 5.
In simple words: Use the section formula three times - once to find Q, once to find the crossing point of diagonals, and once more to find where the segment touches the x-axis.

Exam Tip: For multi-part questions about parallelograms, always start with the diagonal property (diagonals bisect each other) - it is your most powerful tool for finding missing vertices.

Question 16. If A(5, -1), B(-3, -2) and C(-1, 8) are the vertices of a triangle ABC, find the length of the median through A and the coordinates of the centroid of triangle ABC.
Answer: Let D be the midpoint of side BC. Using the midpoint formula:

\( D = \left(\frac{-3 + (-1)}{2}, \frac{-2 + 8}{2}\right) = (-2, 3) \)

The median from vertex A goes to point D. Using the distance formula:

\( AD = \sqrt{(-2 - 5)^2 + (3 - (-1))^2} = \sqrt{(-7)^2 + 4^2} = \sqrt{49 + 16} = \sqrt{65} \text{ units} \)

The centroid of a triangle is found by averaging all three vertex coordinates:

\( G = \left(\frac{5 + (-3) + (-1)}{3}, \frac{-1 + (-2) + 8}{3}\right) = \left(\frac{1}{3}, \frac{5}{3}\right) \)

Therefore, the centroid is at \( \left(\frac{1}{3}, \frac{5}{3}\right) \), and the median through A has length \( \sqrt{65} \) units.
In simple words: A median of a triangle goes from one corner to the middle of the opposite side. The centroid is the point where all three medians meet - you can find it by averaging the x-values and y-values of all three corners separately.

Exam Tip: The centroid formula is straightforward: add all x-coordinates and divide by 3; add all y-coordinates and divide by 3. This is faster and less error-prone than finding individual medians.