CBSE Class 10 Mathematics Coordinate Geometry Assignment Set 23

Exercise II: Section Formula

Question 1. Find the coordinates of the point which divides the join of \( (-1, 7) \) and \( (4, -3) \) in the ratio \( 2 : 3 \).

Answer:
Let \( A(-1, 7),\ B(4, -3) \) and C divide AB internally in the ratio \( 2 : 3 \).
\( x = \frac{2(4) + 3(-1)}{5} = \frac{5}{5} = 1 \) and \( y = \frac{2(-3) + 3(7)}{5} = \frac{15}{5} = 3 \).
Hence the point is \( C(1, 3) \).
In simple words: Use the section formula with m = 2 and n = 3.

Teacher's Note:
1. Use the section formula for an internal division: multiply each coordinate of the second point by m and of the first point by n, then divide by m + n.
2. Check that the point you obtain lies between the two given points.

Question 2. Find the coordinates of the points of trisection of the line segment joining \( (4, -1) \) and \( (-2, -3) \).
Answer:
Let \( P(4, -1) \) and \( Q(-2, -3) \), and let A and B be the points of trisection, so \( PA = AB = BQ \).
A divides PQ in the ratio \( 1 : 2 \): \( A = \left( \frac{1(-2) + 2(4)}{3},\ \frac{1(-3) + 2(-1)}{3} \right) = \left( 2,\ -\frac{5}{3} \right) \).
B divides PQ in the ratio \( 2 : 1 \): \( B = \left( \frac{2(-2) + 1(4)}{3},\ \frac{2(-3) + 1(-1)}{3} \right) = \left( 0,\ -\frac{7}{3} \right) \).
Hence the points of trisection are \( \left( 2, -\frac{5}{3} \right) \) and \( \left( 0, -\frac{7}{3} \right) \).
In simple words: Two points of trisection divide the segment in the ratios 1 : 2 and 2 : 1.

Teacher's Note:
1. Take the ratio from the end you start measuring from.
2. Check that B is the mid-point of AQ.

Question 3. To conduct Sports Day activities, in your rectangular shaped school ground ABCD, lines have been drawn with chalk powder at a distance of 1 m each. 100 flower pots have been placed at a distance of 1 m from each other along AD. Niharika runs \( \frac{1}{4} \)th the distance AD on the 2nd line and posts a green flag. Preet runs \( \frac{1}{5} \)th the distance AD on the eighth line and posts a red flag. What is the distance between both the flags? If Rashmi has to post a blue flag exactly halfway between the line segment joining the two flags, where should she post her flag?
Answer:
Taking A as the origin, AD along the y-axis and the lines as \( x = 1, 2, \ldots, 10 \): \( AD = 100 \) m.
Green flag: \( \frac{1}{4} \times 100 = 25 \) m along the 2nd line, so its coordinates are \( (2, 25) \). Red flag: \( \frac{1}{5} \times 100 = 20 \) m along the 8th line, so its coordinates are \( (8, 20) \).
Distance between the flags \( = \sqrt{(8 - 2)^2 + (20 - 25)^2} = \sqrt{36 + 25} = \sqrt{61} \) m.
The blue flag is at the mid-point: \( \left( \frac{2 + 8}{2},\ \frac{25 + 20}{2} \right) = \left( 5,\ \frac{45}{2} \right) \).
Hence the distance between the flags is \( \sqrt{61} \) m, and Rashmi should post the blue flag on the 5th line at 22.5 m.
In simple words: Treat the ground as a coordinate plane with the lines as x-values and the distance along AD as y-values.

Teacher's Note:
1. Convert the word description into coordinates first.
2. The point halfway between two points is their mid-point.

Question 4. Find the ratio in which the line segment joining \( (-3, 10) \) and \( (6, -8) \) is divided by \( (-1, 6) \).

Answer:
Let the ratio be \( k : 1 \). Then \( -1 = \frac{k(6) + 1(-3)}{k + 1} \Rightarrow -k - 1 = 6k - 3 \Rightarrow 7k = 2 \Rightarrow k = \frac{2}{7} \).
Check with the y-coordinate: \( \frac{k(-8) + 10}{k + 1} = \frac{-16/7 + 10}{9/7} = \frac{54/7}{9/7} = 6 \), as required.
Hence the point divides the segment in the ratio \( 2 : 7 \).
In simple words: Assume the ratio k : 1, apply the section formula and match the x-coordinate; then confirm with the y-coordinate.

Teacher's Note:
1. Use the x-coordinate (or the y-coordinate) to find k.
2. Verify with the other coordinate to be sure the point lies on the segment.

Question 5. Find the ratio in which the line segment joining \( A(1, -5) \) and \( B(-4, 5) \) is divided by the x-axis. Also find the coordinates of the point of division.
Answer:
Let the x-axis divide AB in the ratio \( k : 1 \) at \( P(x, 0) \). The y-coordinate: \( \frac{5k - 5}{k + 1} = 0 \Rightarrow k = 1 \).
The x-coordinate: \( x = \frac{-4k + 1}{k + 1} = \frac{-3}{2} \).
Hence the x-axis divides AB in the ratio \( 1 : 1 \) at the point \( \left( -\frac{3}{2}, 0 \right) \).
In simple words: A point on the x-axis has y-coordinate 0, which gives the ratio.

Teacher's Note:
1. Use the y-coordinate to find the ratio.
2. Substitute the ratio in the x-coordinate formula to get the point.

Question 6. If \( (1, 2),\ (4, y),\ (x, 6) \) and \( (3, 5) \) are the vertices of a parallelogram taken in order, find x and y.

Answer:
Let \( P(1, 2),\ Q(4, y),\ R(x, 6),\ S(3, 5) \). The diagonals PR and QS of a parallelogram bisect each other, so they have the same mid-point.
Mid-point of \( PR = \left( \frac{1 + x}{2},\ 4 \right) \); mid-point of \( QS = \left( \frac{7}{2},\ \frac{5 + y}{2} \right) \).
\( \frac{1 + x}{2} = \frac{7}{2} \Rightarrow x = 6 \) and \( 4 = \frac{5 + y}{2} \Rightarrow y = 3 \).
Hence \( x = 6 \) and \( y = 3 \).
In simple words: In a parallelogram the diagonals have the same mid-point.

Teacher's Note:
1. Join the opposite vertices to get the two diagonals.
2. Equate the mid-points coordinate by coordinate.

Question 7. Find the coordinates of a point A, where AB is the diameter of a circle whose centre is \( (2, -3) \) and B is \( (1, 4) \).

Answer:
The centre C(2, -3) is the mid-point of the diameter AB. Let \( A = (x, y) \).
\( \frac{x + 1}{2} = 2 \Rightarrow x = 3 \) and \( \frac{y + 4}{2} = -3 \Rightarrow y = -10 \).
Hence the coordinates of A are \( (3, -10) \).
In simple words: The centre of a circle is the mid-point of every diameter.

Teacher's Note:
1. Use the mid-point formula with the centre as the mid-point.
2. Solve for x and y separately.

Question 8. If A and B are \( (-2, -2) \) and \( (2, -4) \), respectively, find the coordinates of P such that \( AP = \frac{3}{7} AB \) and P lies on the line segment AB.
Answer:
\( AP = \frac{3}{7}AB \Rightarrow 7AP = 3(AP + PB) \Rightarrow 4AP = 3PB \Rightarrow \frac{AP}{PB} = \frac{3}{4} \). So P divides AB in the ratio \( 3 : 4 \).
\( P = \left( \frac{3(2) + 4(-2)}{7},\ \frac{3(-4) + 4(-2)}{7} \right) = \left( -\frac{2}{7},\ -\frac{20}{7} \right) \).
Hence \( P\left( -\frac{2}{7}, -\frac{20}{7} \right) \).
In simple words: Convert AP = (3/7)AB into the ratio AP : PB = 3 : 4 and apply the section formula.

Teacher's Note:
1. Since P is on AB, PB = AB − AP.
2. The ratio is part to part, not part to whole.

Question 9. Find the coordinates of the points which divide the line segment joining \( A(-2, 2) \) and \( B(2, 8) \) into four equal parts.
Answer:
Let P, Q, R divide AB into four equal parts. Q is the mid-point of AB: \( Q = \left( \frac{-2 + 2}{2},\ \frac{2 + 8}{2} \right) = (0, 5) \).
P is the mid-point of AQ: \( P = \left( \frac{-2 + 0}{2},\ \frac{2 + 5}{2} \right) = \left( -1, \frac{7}{2} \right) \). R is the mid-point of QB: \( R = \left( \frac{0 + 2}{2},\ \frac{5 + 8}{2} \right) = \left( 1, \frac{13}{2} \right) \).
Hence the points are \( P\left( -1, \frac{7}{2} \right),\ Q(0, 5),\ R\left( 1, \frac{13}{2} \right) \).
In simple words: Find the middle point first, then the mid-points of the two halves.

Teacher's Note:
1. Three points are needed for four equal parts.
2. Each new point is a mid-point of two known points.

Question 10. Find the area of a rhombus if its vertices are \( (3, 0),\ (4, 5),\ (-1, 4) \) and \( (-2, -1) \) taken in order. [Hint: Area of a rhombus \( = \frac{1}{2} \) (product of the lengths of its diagonals)]

Answer:
Let \( P(3, 0),\ Q(4, 5),\ R(-1, 4),\ S(-2, -1) \). The diagonals are \( PR = \sqrt{(3 + 1)^2 + (0 - 4)^2} = \sqrt{32} = 4\sqrt{2} \) and \( QS = \sqrt{(4 + 2)^2 + (5 + 1)^2} = \sqrt{72} = 6\sqrt{2} \).
Area \( = \frac{1}{2} \times 4\sqrt{2} \times 6\sqrt{2} = 24 \).
Hence the area of the rhombus is 24 square units.
In simple words: The area of a rhombus is half the product of its diagonals.

Teacher's Note:
1. Find the lengths of both diagonals with the distance formula.
2. Apply the given hint.