Exercise 7.2 Solutions
Coordinate Geometry
1. Find the coordinates of the point which divides the join of (–1, 7) and (4, –3) in the ratio 2 : 3.
Let the points be A(-1, 7) and B(4, -3). The ratio is m₁:m₂ = 2:3.
Using the Section Formula: P(x, y) = ( m₁x₂ + m₂x₁m₁ + m₂, m₁y₂ + m₂y₁m₁ + m₂ )
x = 2(4) + 3(-1)2 + 3 = 8 – 35 = 55 = 1.
y = 2(-3) + 3(7)2 + 3 = -6 + 215 = 155 = 3.
Answer: The coordinates of the point are (1, 3).
2. Find the coordinates of the points of trisection of the line segment joining (4, –1) and (–2, –3).
Let A=(4,-1) and B=(-2,-3). Trisection means two points P and Q divide the segment AB into three equal parts (AP = PQ = QB).
Point P divides AB in the ratio 1:2.
x₁ = 1(-2) + 2(4)1 + 2 = -2 + 83 = 63 = 2.
y₁ = 1(-3) + 2(-1)1 + 2 = -3 – 23 = –53.
So, P = (2, -5/3).
Point Q divides AB in the ratio 2:1.
x₂ = 2(-2) + 1(4)2 + 1 = -4 + 43 = 0.
y₂ = 2(-3) + 1(-1)2 + 1 = -6 – 13 = –73.
So, Q = (0, -7/3).
Answer: The points of trisection are (2, -5/3) and (0, -7/3).
3. To conduct Sports Day activities… Find the distance between both the flags. If Rashmi has to post a blue flag exactly halfway… where should she post her flag?

Step 1: Find the coordinates of the flags.
Niharika (Green flag): Runs on 2nd line, 1/4 of distance AD (100m).
Position G = (2, 1/4 × 100) = (2, 25).
Preet (Red flag): Runs on 8th line, 1/5 of distance AD.
Position R = (8, 1/5 × 100) = (8, 20).
Step 2: Find the distance between the flags.
Distance GR = √[(8 – 2)² + (20 – 25)²] = √[6² + (-5)²] = √36 + 25 = √61 m.
Step 3: Find the midpoint for Rashmi’s flag.
Rashmi’s position is the midpoint of GR.
Midpoint(x, y) = (
2 + 82,
25 + 202
) = (5, 45/2) = (5, 22.5).
Answer: Rashmi should post her flag on the 5th line at a distance of 22.5 m.
4. Find the ratio in which the line segment joining the points (–3, 10) and (6, –8) is divided by (–1, 6).
Let the ratio be k : 1. Let A=(-3,10), B=(6,-8), and P=(-1,6).
Using the section formula for the x-coordinate:
-1 = k(6) + 1(-3)k + 1
-(k + 1) = 6k – 3
-k – 1 = 6k – 3
2 = 7k => k = 2/7.
The ratio k : 1 is 2/7 : 1, which is 2 : 7.
Answer: The ratio is 2 : 7.
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.
Let the point of division on the x-axis be P(x, 0). Let the ratio be k : 1.
Using the section formula for the y-coordinate:
0 = k(5) + 1(-5)k + 1
0 = 5k – 5 => 5k = 5 => k = 1.
The ratio is 1 : 1. This means P is the midpoint.
Now find the x-coordinate:
x = 1(-4) + 1(1)1 + 1 = -32.
Answer: The ratio is 1 : 1 and the point of division is (-3/2, 0).
6. If (1, 2), (4, y), (x, 6) and (3, 5) are the vertices of a parallelogram taken in order, find x and y.
Let the vertices be A(1,2), B(4,y), C(x,6), D(3,5).
The diagonals of a parallelogram bisect each other. So, Midpoint of AC = Midpoint of BD.
Midpoint of AC = ( 1 + x2, 2 + 62 ) = ( 1 + x2, 4 ).
Midpoint of BD = ( 4 + 32, y + 52 ) = ( 72, y + 52 ).
Equating the coordinates:
1 + x2 = 72 => 1 + x = 7 => x = 6.
4 = y + 52 => 8 = y + 5 => y = 3.
Answer: x = 6, y = 3.
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).
The centre O(2, -3) is the midpoint of the diameter AB. Let A = (x, y).
Using the midpoint formula:
x + 12 = 2 => x + 1 = 4 => x = 3.
y + 42 = -3 => y + 4 = -6 => y = -10.
Answer: The coordinates of A are (3, -10).
8. If A and B are (–2, –2) and (2, –4), respectively, find the coordinates of P such that AP = 3/7 AB and P lies on the line segment AB.
The condition AP = (3/7)AB means that P divides the segment AB in the ratio 3:4.
(Because AP/AB = 3/7, so AP/PB = 3/(7-3) = 3/4).
So, m₁=3, m₂=4. A=(-2,-2), B=(2,-4).
Using the Section Formula:
x = 3(2) + 4(-2)3 + 4 = 6 – 87 = –27.
y = 3(-4) + 4(-2)3 + 4 = -12 – 87 = –207.
Answer: The coordinates of P are (-2/7, -20/7).
9. Find the coordinates of the points which divide the line segment joining A(–2, 2) and B(2, 8) into four equal parts.
Let the three points be P, Q, and R.
Point Q is the midpoint of AB (ratio 2:2 or 1:1).
Q = (
-2 + 22,
2 + 82
) = (0, 5).
Point P is the midpoint of AQ.
P = (
-2 + 02,
2 + 52
) = (-1, 7/2).
Point R is the midpoint of QB.
R = (
0 + 22,
5 + 82
) = (1, 13/2).
Answer: The points are (-1, 7/2), (0, 5), and (1, 13/2).
10. Find the area of a rhombus if its vertices are (3, 0), (4, 5), (–1, 4) and (–2, –1) taken in order.
Let the vertices be A(3,0), B(4,5), C(-1,4), D(-2,-1).
Area of a rhombus = 1/2 × (product of its diagonals).
Step 1: Find the length of diagonal AC.
AC = √[(-1 – 3)² + (4 – 0)²] = √[(-4)² + 4²] = √16 + 16 = √32 = 4√2.
Step 2: Find the length of diagonal BD.
BD = √[(-2 – 4)² + (-1 – 5)²] = √[(-6)² + (-6)²] = √36 + 36 = √72 = 6√2.
Step 3: Calculate the area.
Area = 1/2 × AC × BD = 1/2 × (4√2) × (6√2) = 1/2 × 24 × 2 = 24.
Answer: The area of the rhombus is 24 square units.