7

Coordinate
Geometry

NCERT SOLUTIONS

What's inside
– In-Chapter Q's (solved)
– Textbook Exercise Q's (solved)
 Exercise – 7.1
1. Find the distance between the following pairs of points :
(i) (2, 3) and (4, 1) (ii) (– 5, 7) and (– 1, 3)
(iii) (a, b) and (– a, – b)
Sol. : (i) Let the given points are A(2, 3) and B(4, 1)
Here, x1 = 2, y1 = 3, x2 = 4, y2 = 1
\ AB = (x2 - x1) 2 + (y2 - y1) 2
= (4 - 2) 2 + (1 - 3) 2
= 2 2 + (- 2) 2
= 4 + 4 = 8 = 2 2 units
(ii) (– 5, 7) and (– 1, 3)
Let, the given points are A(– 5, 7) and B(– 1, 3)
Here, x1 = – 5, y1 = 7, x2 = – 1, y2 = 3
\ AB = (x2 - x1) 2 + (y2 - y1) 2
= {(- 1) - (- 5)} 2 + (3 - 7) 2
= 42 + (- 4) 2 = 16 + 16 = 32
\ AB = 4 2 units
(iii) (a, b) and (– a, – b)
Let, the given points are A(a, b) and B(– a, – b)
Here, x1 = a, y1 = b, x2 = – a, y2 = – b
\ AB = (x2 - x1) 2 + (y2 - y1) 2
= ( - a - a) 2 + ( - b - b) 2
= (- 2a) 2 + (- 2b) 2
= 4a 2 + 4b 2 = 4 ( a 2 + b 2)
\ AB = 2 a2 + b2 units
2. Find the distance between the points (0, 0) and (36, 15). Can you now find the distance
between the two towns A and B? If A town B is located 36 cm east and 15 km north of
the town A.
Sol. : Let given poitns are A(0, 0) and B(36, 15)
Here, x1 = 0, y1 = 0, x2 = 36, y2 = 15
\ AB = (x2 - x1) 2 + (y2 - y1) 2
= (36 - 0) 2 + (15 - 0) 2
= 362 + 152 = 1296 + 225
= 1521 = 39 units

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 North

B (36, 15)

15 km

West East
A 36 km
(0, 0)

South

Given a town B is located 36 km east and 15 km north of town A.

Let, origin is A (0, 0) and point is B (36, 15)
Now, we have found the distance between points (0, 0) and (36, 15).
Hence, distance between town A and B = 39 km.
3. Determine if the points (1, 5), (2, 3) and (–2, –11) are collinear.
Sol. : Let, the given points are A(1, 5), B(2, 3) and C(– 2, – 11)
then, AB = (2 - 1) 2 + (3 - 5) 2
= 12 + (- 2) 2 = 1 + 4 = 5 units
BC = (- 2 - 2) 2 + (- 11 - 3) 2
= (- 4) 2 + (- 14) 2
= 16 + 196 = 212 units
CA = (1 + 2) 2 + (5 + 11) 2
= 32 + 162 = 9 + 256 = 265 units
Now,
AB + BC = 5 + 212
= 5 + 2 53
and AC = 265
Hence, AB + BC ≠ AC
Hence, given points are not collinear.
4. Check whether (5, –2), (6, 4) and (7, –2) are the vertices of an isosceles triangle. 
Sol. : Let, the given points A(5, – 2), B(6, 4) and C(7, – 2) then,
AB = (6 - 5) 2 + [4 - (- 2)] 2

= 12 + 62 = 1 + 36 = 37

BC = (7 - 6) 2 + (- 2 - 4) 2

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 = 1 2 + (- 6) 2 = 1 + 36 = 37

AC = (7 - 5) 2 + (- 2 + 2) 2

= 22 + 02 = 4 =2

Here, AB = BC = 37 units
Hence, DABC is an isosceles triangle.
5. In a classroom, 4 friends are seated at the points A, B, C and D as shown in figure below.
Champa and Chameli walk into the class and after observing for a few minutes Champa
asks Chameli "Don't you think ABCD is square ?" Chameli disagrees. Using distance
formula, find which of them is correct. 

Sol. : In the given figure, the points A, B, C and D are (3, 4), (6, 7), (9, 4) and (6, 1) respectively.
\ AB = (6 - 3) 2 + (7 - 4) 2 = 32 + 32

= 9+9 = 18 = 3 2 units
BC = (9 - 6) 2 + (4 - 7) 2

= 3 2 + (- 3) 2 = 9+9

= 18 = 3 2 units
CD = (6 - 9) 2 + (1 - 4) 2

= (- 3) 2 + (- 3) 2 = 9+9

= 18 = 3 2 units
DA = (6 - 3) 2 + (1 - 4) 2

= (3) 2 + (–3) 2 = 9+9

= 18 = 3 2 units
\ AB = BC = CD = DA
Now, AC = (9 - 3) 2 + (4 - 4) 2 = 62 + 02

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 = 36 + 0 = 36 = 6 units
and BD = (6 - 6) 2 + (1 - 7) 2
= 0 + (- 6) 2 = 36 = 6 units
\ AC = BD
\ Four sides are equal and diagonals are also equal.
So, ABCD is a square.
Hence, Champa is correct.
6. Name the type of quadrilateral formed, if any, by the following points, and give reasons
for your answer :
(i) (– 1, – 2), (1, 0), (– 1, 2) and (– 3, 0)
(ii) (–3, 5), (3, 1), (0, 3) and (–1, 4)
(iii) (4, 5), (7, 6), (4, 3), and (1, 2)
Sol. : (i) Let, the given points are A(–1, –2), B(1, 0), C(–1, 2) and D(–3, 0), then,
AB = (1 + 1) 2 + (0 + 2) 2
= 22 + 22 = 4 + 4
= 8 = 2 2 units
BC = (- 1 - 1) 2 + (2 - 0) 2
= (- 2) 2 + 22 = 4 + 4
= 8 = 2 2 units
CD = (- 3 + 1) 2 + (0 - 2) 2
= (- 2) 2 + (- 2) 2 = 4 + 4
= 8 = 2 2 units
DA = (- 3 + 1) 2 + (0 + 2) 2
= (- 2) 2 + (2) 2 = 4 + 4
= 8 = 2 2 units
\ AB = BC = CD = DA
Now, AC = (- 1 + 1) 2 + (2 + 2) 2
= 0 + 42 = 4 units
and BD = (- 3 - 1) 2 + (0 – 0) 2
= (- 4) 2 + 0 = 4 units
\ AC = BD
All four sides AB, BC, CD and DA are equal and diagonals AC and BD are also equal. Hence,
quadrilateral ABCD is a square.
(ii) Let, the given points are A(–3, 5), B(3, 1), C(0, 3) and D(–1, – 4).
AB = (3 + 3) + (1 – 5)
2 2

= (6) + (–4) 2 2

= 36 + 16
= 52 = 2 13 units
BC = (0 – 3) + (3 – 1)
2 2

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 = (–3) + (2) 2 2

= 9 + 4 = 13 units
CD = (–1 – 0) + (–4 – 3)
2 2

= (–1) + (–7) 2 2

= (1 + 49)
= 50 = 5 2 units
AD = (–1 + 3) + (–4 – 5)
2 2

= (2) + (–9) 2 2

= 4 + 81 = 85 units
AC = (0 + 3) + (3 – 5)
2 2

= (3) + (–2) 2 2

= 9 + 4 = 13 unit
BD = (–1 – 3) + (– 4 – 1)
2 2

= (–4) + (–5) 2 2

= 16 + 25 = 41 units
Here, BC + AC = AB
Point A, B and C are collinear i.e., all the points lie on same line.
Hence, ABCD is not a quadrilateral.
(iii) Let, the given points are P(4, 5), Q(7, 6), R(4, 3) and S(1, 2).
\ PQ = (7 - 4) 2 + (6 - 5) 2
= 32 + 12 = 9 + 1 = 10 units
QR = (4 - 7) 2 + (3 - 6) 2
= (- 3) 2 + (- 3) 2 = 9 + 9
= 18 = 3 2 units
RS = (1 - 4) 2 + (2 - 3) 2
= (- 3) 2 + (- 1) 2 = 9 + 1
= 10 units
SP = (4 - 1) 2 + (5 - 2) 2
= 32 + 32 = 9 + 9
= 18 = 3 2 units
PR = (4 - 4) 2 + (3 - 5) 2
= 0 + (- 2) 2 = 2 units
QS = (1 - 7) 2 + (2 - 6) 2
= (- 6) 2 + (- 4) 2 = 36 + 16
= 52 = 2 13 units
\ PQ = RS and QR = SP
and PR ≠ QS
Hence, PQRS is a parallelogram.

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 7. Find the point on the x-axis which is equidistant from (2, –5) and (–2, 9).
Sol. : ∵ The point lies on the X-axis
Hence, ordinate or y-coordinate = 0. Hence, point A (x, 0) lies on X-axis.
∵ Point A(x, 0) is equidistant from points B(2, – 5) and C(– 2, 9)
\ AB = AC
( 2 – x ) + ( – 5 – 0 ) = ( – 2 – x ) + ( 9 – 0)
2 2 2 2

⇒(2 – x)2 + (– 5 – 0)2 = (– 2 – x)2 + (9 – 0)2

⇒4 + x2 – 4x + 25= 4 + x2 + 4x + 81
⇒ x2 – 4x + 29 = x2 + 4x + 85
⇒ – 4x – 4x = 85 – 29
⇒ – 8x = 56
56
\ x = - 8 = –7
Hence, the point lies on the X-axis is (–7, 0) which is equidistant from (2, – 5) and (– 2, 9).
8. Find the values of y for which the distance between the points P(2, –3) and Q(10, y) is 10
units. 
Sol. : According to the question,
PQ = 10 units
(10 - 2) + [y - (- 3)] 2 = 10
2

⇒ 8 2 + (y + 3) 2 = 10
⇒ 64 + y 2 + 9 + 6y = 10
Squaring both the sides,
73 + y2 + 6y = 100
⇒y2 + 6y + 73 – 100 = 0
⇒ y2 + 6y – 27 = 0
⇒y2 + 9y – 3y – 27 = 0
⇒y(y + 9) – 3(y + 9) = 0
⇒ (y + 9)(y – 3) = 0
\ y = – 9 or y = 3
9. If Q(0, 1) is equidistant from P(5, –3) and R(x, 6), find the values of x. Also find the distances QR
and PR.
Sol. : ∵ Point Q(0, 1) is equidistant from P(5, – 3) and R(x, 6).
So, PQ = QR
(5 – 0) + (– 3 – 1) = (x – 0) + (6 – 1)
2 2 2 2

⇒(5 – 0)2 + (– 3 – 1)2 = (x – 0)2 + (6 – 1)2

⇒ (5)2 + (– 4)2 = x2 + 52
⇒ 25 + 16 = x2 + 25
⇒ x2 = 16 ⇒ x = ± 4
⇒ x = – 4, 4
\ Point R is either (4, 6) or (– 4, 6)
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 Condition-I : When point R is (4, 6), then
QR = (4 – 0) + (6 – 1)
2 2

= (4) + (5) 2 2

= 16 + 25 = 41 units
PR = (4 – 5) + (6 + 3)
2 2

= (–1) + (9) 2 2

= 1 + 81 = 82 units
Condition-II : When point R is (– 4, 6) , then
QR = (–4 – 0) + (6 – 1)2 2

= (–4) + (5) 2 2

= 16 + 25 = 41 units
PR = (– 4 – 5) + (6 + 3) 2 2

= (–9) + (9) 2 2

= 81 + 81
= 162 = 9 2 units
10. Find a relation between x and y such that the point (x, y) is equidistant from the point (3,
6) and (–3, 4).
Sol. : Let, point A(x, y), is equidistant from B(3, 6) and C(–3, 4).
\ AB = AC
⇒ (3–x) + (6–y) = (–3–x) + (4–y)
2 2 2 2

⇒ 9 + x – 6x + 36 + y2 – 12y
2

= 9 + x2 + 6x + 16 + y2 – 8y
⇒x2 + y2 – 6x – 12y + 45 = x2 + y2 + 6x – 8y + 25
⇒ x2 + y2 – 6x – 12y + 45 – x2 – y2 – 6x + 8y – 25 = 0
⇒ –12x – 4y + 20 = 0
⇒ – 4(3x + y – 5) = 0
⇒ 3x + y – 5 = 0

Exercise – 7.2
1. Find the coordinates of the point which divides the join of (–1, 7) and (4, –3) in the ratio 2 :
3.
Sol. : Let, the given points are A(–1, 7) and B(4, –3)
Let, the coordinates of required point is (x, y)
Given, m1 = 2 and m2 = 3
By section formula,
m1 x2 + m2 x1
x= m1 + m2
2 (4) + 3 (- 1) 8-3
= 2+3
= 5

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 5
= 5
=1
m1 y2 + m2 y1
and y= m1 + m2
2 (- 3) + 3 (7) - 6 + 21
= 2+3
= 5
15
= 5
=3
Thus, the coordinates of required point is (1, 3).
2. Find the coordinates of the points of trisection of the line segment joining (4, –1) and
(–2, –3).
Sol. : Let the points P and Q trisect the line segment joining the points A and B.
x x x
A P Q B
(4, –1) (–2, –3)
AP = PQ = QB = x
PB = PQ + QB = x + x = 2x
AQ = AP + PQ = x + x = 2x
\ AP : PB = x : 2x = 1 : 2
andAQ : QB = 2x : x = 2 : 1
Since, point P divides AB in the ratio 1 : 2.
e o
m1 x2 + m2 x1 m1 y2 + m2 y1
\Coordinates of P
= m1 + m2
,
m1 + m2

d n
1 (- 2) + 2 (4) 1 (- 3) + 2 (- 1)
= 1+2
,
1+2
d n
-2 + 8 -3 - 2
= 3
,
3
d , n = c 2, m
6 -5 –5
= 3 3 3
and Q divides AB in the ratio 2 : 1.
m1 x2 + m2 x1 m1 y2 + m2 y1
\Coordinates of Q
= m1 + m2
,
m1 + m2

d n
2 (- 2) + 1 (4) 2 (- 3) + 1 (- 1)
= 2+1
,
2+1

d n
-4 + 4 -6 - 1
= 3
,
3

d 0, n
-7
= 3
Hence, the coordinates of the points of trisection of the line-segment are Pd 2, n Q d 0, n.
-5 -7
3
and 3
3. To conduct sport Day activities, in your rectangular shaped school ground ABCD, lines have
been drawn with chalk powder at a distance of 1m each. 100 flower pots have been placed
at a distance of 1m from each other along AD, as shown in figure below. Niharika runs 4
1

the the distance AD on the 2nd line and posts a green flag. Preet runs 5 th the distance AD on
1

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 ?
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 D C

2
1
A 1 2 3 4 5 6 7 8 9 10 B

Sol. : From the given figure the position of green flag which is post by Niharika is M c 2, 100 × 4 m i.e.,
1

M(2, 25) and the position of red flag which is post by Preet is N c8, 5 × 100 m i.e., N(8, 20).
1

Hence, distance between flags,

MN = (x2 - x1) 2 + (y2 - y1) 2
= (8 - 2) 2 + (20 - 25) 2

= 6 2 + (- 5) 2

= 36 + 25

= 61 m
Let, the position of the blue flag which is posted by Rashmi is P(x, y) and it is at half the
distance between M and N i.e., P is the mid-point of MN.
M P (x, y) N
(2, 25) (8, 20)

d n
x1 + x2 y1 + y2
Coordinates of P(x, y) = 2
,
2

d n
2 + 8 25 + 20
= 2
,
2

c , m
10 45
= 2 2
= (5, 22.5)
Thus, the blue flag should be post on the 5th line and 22.5m above.
4. Find the ratio in which the line segment joining the points (–3, 10) and (6, –8) is divided by
(–1, 6)
Sol. : Let the point R(–1, 6) divides the line segment joining the points P(–3, 10) and Q(6, –8) in the
ratio k : 1.
Coordinates of R
c m1 x2 + m2 x1 , m
m1 y2 + m2 y1
(–1, 6) = m1 + m2 m1 + m2
k × 6 + 1 × (–3) k × (– 8) + 1 × 10
(–1, 6) = k +1 , k +1
6k – 3 –8k + 10
(–1, 6) = k +1 , k +1
6k – 3
then –1= k +1

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 6k – 3 = – k – 1
7k = 2
2
k= 7
–8k + 10
and 6= k +1

6k + 6 = – 8k + 10
14k = 4
4 2
k= 14 = 7
Hence, required ratio = 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.
Sol. : The points is on the X-axis. So, the ordinate point or the y-coordinate of the point is 0.
Hence, the coordinates of the point is (x, 0)
Now, let us consider the point P(x, 0) divides the line segment joining the points A(1, – 5) and
B(– 4, 5) in the ratio k : 1.
By section formula, coordinates of P
c m1 x2 + m2 x1 , m
m1 y2 + m2 y1
(x, 0) = m1 + m2 m1 + m2

d n
- 4k + 1 5k - 5
(x, 0) = k+1 k+1
,

Comparing the coordinates

5k - 5
0= k+1
⇒ 0 = 5k – 5
⇒ 5k =5⇒k=1
Hence, required ratio 1 : 1.
then, coordinates of the point of division
d n
1 × (– 4) + 1 × 1 1× 5 + 1 × (– 5)
= 1+1
,
1+1
d n = c , 0m
-4 + 1 5 - 5 –3
= 2
,
2 2
6. If (1, 2), (4, y), (x, 6) and (3, 5) are the vertices of a parallelogram taken in order, find x and
y.
Sol. : Let, A(1, 2), B(4, y), C(x, 6) and D(3, 5) are the vertices of a parallelogram.
D (3, 5) C (x, 6)

A (1, 2) B (4, y)

Since, ABCD is a parallelogram. So, its diagonals AC and BD bisect each other.
Let, Mid point of AC and BD is H.

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i.e., Mid point of AC = Mid point of BD
d n d n
1+x 2+6 4+3 y+5
2
,
2
= 2
,
2

d , 4n d , n
1+x 7 y+5
or 2
= 2 2
Comparing the coordinates
1+x 7
2
= 2
⇒1 + x =7 ⇒x=6
5+y
and 4= 2
⇒ 8= 5 + y ⇒ y = 8 – 5 = 3
Hence, x = 6 and 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).
Sol. : Let AB is the diameter of circle whose centre is C(2, –3) and the coordinates of B are (1,
4).
Let the coordinates of A are (x, y).
∵ AB is the diameter of the circle. So, C is the mid point of AB.

A B
(x, y) C (2, –3) (1, 4)

d n
x+1 y+4 x1 + x2 y1 + y2
\Coordinates of mid point C =
2
,
2
[∵ Coordinates of mid point = 2
,
2
]

(2 – 3) = c x + 1 , m
y+4
2 2
Comparing the coordinates :
x+1 y+4
Hence, 2 = 2
and– 3 = 2
or 4 = x + 1 and– 6 = y + 4
or x = 4 – 1 = 3 and y = – 6 – 4 = – 10
Hence, the coordinates of point 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.
Sol. : According to the question,
3 AP 3
AP = 7
AB ⇒ =
AB 7
AP
AP + PB = 3
7

AP + PB 7
AP
=
3
PB
1+
AP
= 37

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 PB 4 AP 3
\
AP
= 3
⇒ =
PB 4
\
AP : PB = 3 : 4
3 4
A (–2, –2) P (x, y) B (2, –4)

Now point P(x, y) divided the line segment AB in the ratio 3 : 4.

e o
m1 x2 + m2 x1 m1 y2 + m2 y1
Coordinates of P = m1 + m2
,
m1 + m2

d n
3×2 + 4× (- 2) 3× (- 4) + 4× (- 2)
= 3+4
,
3+4

= d 6 7 8 , 127 8 n = d 72 , 720 n
- - - - -

Hence, Coordinates of point P is d 72 , 720 n .

- -

9. Find the coordinates of the points, which divide the line segment joining A(–2, 2) and B(2, 8)
into four equal parts.
Sol. : Let the points P, Q and R lie on the line segment AB such that, AP = PQ = QR = RB
Let AP = PQ = QR = RB = x
x x x x
A (–2, 2) P Q R B (2, 8)
AP x 1
Now, PB
= 3x
= 3
Now P divides AB in the ratio 1 : 3.
e o
m1 x2 + m2 x1 m1 y2 + m2 y1
\ Coordinates of P
= m1 + m2
,
m1 + m2

d n
1 (2) + 3 (- 2) 1 (8) + 3 (2)
= 1+3
,
1+3

d n = d , n
2-6 8+6 - 4 14
= 4
,
4 4 4

c- 1, m
7
= 2
Point Q is the mid point of AB.
d n
x1 + x2 y1 + y2
\Coordinates of Q
= 2
,
2

d n = c 0, m
-2 + 2 2 + 8 10
= 2
,
2 2
= (0, 5)
AR 3x 3
Now, RB
= =
x 1
Now R divides AB in the ratio 3 : 1.
\Coordinates of Point R

d n
3 × 2 + 1 × (- 2) 3 × 8 + 1 × 2
= 3+1
,
3+1

d n
6 - 2 24 + 2
= 4
,
4

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 c , m c1, m
4 26 13
= 4 4
= 2

c- 1, m , c1, m.
7 13
Here, required points are 2
(0, 5) and 2
10. Find the area of a rhombus if its vertices are (3, 0), (4, 5), (–1, 4) and (–2, –1) taken in order.
Sol. : Let, A(3, 0), B(4, 5), C(–1, 4) and D(–2, –1)
The vertices of a rhombus
\Diagonal AC =
(- 1 - 3) 2 + (4 - 0) 2

= (- 4) 2 + 4 2

= 16 + 16 = 32 = 4 2 units
andDiagonal BD = ( - 2 - 4) 2 + ( - 1 - 5) 2

= (- 6) 2 + (- 6) 2
= 36 + 36 = 72 = 6 2 units
1
\ Area of rhombus =
2
× product of diagonals
1
= 2
× AC × BD
1
= 2
×4 2 ×6 2

= 24 sq. unit

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