Class 10 Maths (NCERT)
Chapter 7: Coordinate Geometry
Exercise 7.2 – Stepwise Solutions
Important Formulas (Concept)
Section Formula (Internal Division):
If P divides A(x1,y1) and B(x2,y2) in ratio m:n, then
P = ( (mx2+nx1)/(m+n) , (my2+ny1)/(m+n) )
Midpoint Formula:
M = ( (x1+x2)/2 , (y1+y2)/2 )
Distance Formula:
AB = √[(x2−x1)² + (y2−y1)²]
1. Point dividing (-1, 7) and (4, -3) in ratio 2:3
A(-1,7), B(4,-3), m:n = 2:3
x = (2·4 + 3·(-1)) / (2+3) = (8 – 3)/5 = 1
y = (2·(-3) + 3·7) / 5 = (-6 + 21)/5 = 3
Answer: (1, 3)
2. Points of trisection of segment joining (4, -1) and (-2, -3)
A(4,-1), B(-2,-3)
Trisection points divide AB internally in ratios 1:2 and 2:1
Point P (ratio 1:2)
x = (1·(-2) + 2·4)/3 = (-2 + 8)/3 = 2
y = (1·(-3) + 2·(-1))/3 = (-3 – 2)/3 = -5/3
P = (2, -5/3)
Point Q (ratio 2:1)
x = (2·(-2) + 1·4)/3 = (-4 + 4)/3 = 0
y = (2·(-3) + 1·(-1))/3 = (-6 – 1)/3 = -7/3
Q = (0, -7/3)
3. Sports Day flags (Fig. 7.12)
Lines are 1 m apart and pots on AD are also 1 m apart.
Take A as origin (0,0). Along AB = x-axis (in metres), along AD = y-axis (in metres).
Niharika runs 1/4 of AD on 2nd line ⇒ x = 2, y = 100/4 = 25
So green flag G = (2, 25)
Preet runs 1/5 of AD on 8th line ⇒ x = 8, y = 100/5 = 20
So red flag R = (8, 20)
(i) Distance between flags
GR = √[(8−2)² + (20−25)²]
= √[6² + (−5)²] = √[36 + 25] = √61
Distance = √61 m
(ii) Blue flag exactly halfway between them
Midpoint M = ( (2+8)/2 , (25+20)/2 )
M = (5, 45/2) = (5, 22.5)
Blue flag position = (5, 22.5)
4. Ratio in which (-1, 6) divides segment joining (-3,10) and (6,-8)
A(-3,10), B(6,-8), P(-1,6)
Let AP:PB = m:n
Using x-coordinate:
-1 = (m·6 + n·(-3))/(m+n)
-1(m+n) = 6m – 3n
-m – n = 6m – 3n
2n = 7m
m:n = 2:7
Answer: The ratio is 2:7 (internal division)
5. Ratio in which x-axis divides the segment joining A(1,-5) and B(-4,5)
Let P be point where it meets x-axis ⇒ y = 0
AP:PB = m:n
Using y-coordinate section formula:
0 = (m·5 + n·(-5))/(m+n)
0 = 5m – 5n ⇒ m = n
So ratio = 1:1
Coordinates of P (midpoint):
x = (1 + (-4))/2 = -3/2
y = 0
Answer: Ratio 1:1 and point P = (-3/2, 0)
6. If (1,2), (4,y), (x,6), (3,5) are vertices of a parallelogram in order, find x and y
Let A(1,2), B(4,y), C(x,6), D(3,5)
Diagonals of a parallelogram bisect each other.
Midpoint of AC = Midpoint of BD
Midpoint of AC = ((1+x)/2 , (2+6)/2) = ((1+x)/2 , 4)
Midpoint of BD = ((4+3)/2 , (y+5)/2) = (7/2 , (y+5)/2)
Equate x-coordinates:
(1+x)/2 = 7/2 ⇒ 1 + x = 7 ⇒ x = 6
Equate y-coordinates:
4 = (y+5)/2 ⇒ 8 = y + 5 ⇒ y = 3
Answer: x = 6, y = 3
7. Find coordinates of point A if AB is diameter of circle with centre (2,-3) and B is (1,4)
Centre is midpoint of diameter AB.
Let A(x,y), B(1,4), midpoint = (2,-3)
(x+1)/2 = 2 ⇒ x+1 = 4 ⇒ x = 3
(y+4)/2 = -3 ⇒ y+4 = -6 ⇒ y = -10
Answer: A = (3, -10)
8. If A(-2,-2) and B(2,-4), find P such that AP = (3/7)AB and P lies on segment AB
AP = (3/7)AB ⇒ AP:PB = 3:4
Using section formula (m:n = 3:4):
x = (3·2 + 4·(-2))/7 = (6 – 8)/7 = -2/7
y = (3·(-4) + 4·(-2))/7 = (-12 – 8)/7 = -20/7
Answer: P = (-2/7, -20/7)
9. Points dividing segment joining A(-2,2) and B(2,8) into four equal parts
We need 3 points that divide AB in ratios 1:3, 2:2, 3:1
A(-2,2), B(2,8)
P1 (1:3)
x = (1·2 + 3·(-2))/4 = (2 – 6)/4 = -1
y = (1·8 + 3·2)/4 = (8 + 6)/4 = 7/2
P1 = (-1, 7/2)
P2 (2:2) (midpoint)
x = (2·2 + 2·(-2))/4 = 0
y = (2·8 + 2·2)/4 = (16+4)/4 = 5
P2 = (0, 5)
P3 (3:1)
x = (3·2 + 1·(-2))/4 = (6 – 2)/4 = 1
y = (3·8 + 1·2)/4 = (24+2)/4 = 13/2
P3 = (1, 13/2)
10. Find area of rhombus with vertices (3,0), (4,5), (-1,4), (-2,-1) taken in order
Let A(3,0), B(4,5), C(-1,4), D(-2,-1)
Diagonals are AC and BD.
AC = √[(-1−3)² + (4−0)²] = √[(-4)² + 4²] = √(16+16) = √32 = 4√2
BD = √[(-2−4)² + (-1−5)²] = √[(-6)² + (-6)²] = √(36+36) = √72 = 6√2
Area of rhombus = 1/2 × (AC × BD)
= 1/2 × (4√2 × 6√2)
= 1/2 × (24 × 2) = 24
Answer: Area = 24 square units
Final Answers (Quick)
1) (1,3)
2) (2, -5/3) and (0, -7/3)
3) Distance = √61 m, Blue flag at (5, 22.5)
4) 2:7
5) Ratio 1:1, point (-3/2, 0)
6) x=6, y=3
7) A = (3, -10)
8) P = (-2/7, -20/7)
9) (-1,7/2), (0,5), (1,13/2)
10) 24 sq units