NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry

NCERT
SOLUTIONS
CLASS - 10th

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Class : 10th
Subject : Maths
Chapter : 7
Chapter Name : Coordinate Geometry

Exercise 7.1

Q1 Find the distance between the following pairs of points :
(i) (2, 3), (4, 1)
(ii) (– 5, 7), (– 1, 3)
(iii) (a, b), (– a, – b)

Answer. (i) distance between the two points is given by
2 2
√(x − x ) + (y1 − y2 )
1 2

Therefore. distance between (2, 3) and (4, 1) is given by
2 2 2 2
l = √(2 − 4) + (3 − 1) = √(−2) + (2)

= √4 + 4 = √8 = 2√2

(ii) Distance between (-S,7) and (-1,3) is given by
2 2 2 2
l = √(−5 − (−1)) + (7 − 3) = √(−4) + (4)

= √16 + 16 = √32 = 4√2

(iii) Distance between (a, b) and (-a -b) is given by
2 2
l = √(a − (−a)) + (b − (−b))

= √4a = 2 √a
2 2 2 2 2 2
= √((2a) + (2b) + 4b + b

Page : 161 , Block Name : Exercise 7.1

Q2 Find the distance between the points (0, 0) and (36, 15). Can you now nd the distance
between the two towns A and B discussed in Section 7.2.

Answer. Distance between points (0.0) and (36,15)
2 2
= √36
2 2
= √(36 − 0) + (15 − 0) + 15

= √1296 + 225 = √1521 = 39

Yes, we can nd the distance between the given towns A and B Assume town A at origin point
(O, O),
Therefore, town a Rill be at point (36, 15) with respect to town A.

And hence, as calculated above, the distance between town A and B Bill be 39 km

Page : 161 , Block Name : Exercise 7.1

Q3 Determine if the points (1, 5), (2, 3) and (– 2, – 11) are collinear.

Answer. Let the points (1, 5), (2, 3), and (-2, -11) be representing the vertices A, B, and C of the
given triangle respectively.
Let A=(1,5) , B =(2,3) , C=(-2,-11)
2 2
∴ AB = √(1 − 2) + (5 − 3) = √5

2 2 √ 2 2
BC = √(2 − (−2)) + (3 − (−11)) = 4 + 14 = √16 + 196 = √212

2 2
= √3
2 2
CA = √(1 − (−2)) + (5 − (−11)) + 16 = √9 + 256 = √265

Since AB + BC ≠ CA
Therefore, the points (1, 5), (2, 3), and (-2, -11) are not collinear,

Page : 161 , Block Name : Exercise 7.1

Q4 Check whether (5, – 2), (6, 4) and (7, – 2) are the vertices of an isosceles triangle.

Answer. Let the points (5, -2), (6, 4), and (7, -2) are representing the vertices A, B, and C of the
given triangle respectively.
2 2 2 2
AB = √(5 − 6) + (−2 − 4) = √(−1) + (−6) = √1 + 36 = √37

2 2 2 2
BC = √(6 − 7) + (4 − (−2)) = √(−1) + (6) = √1 + 36 = √37

2 2 2 2
CA = √(5 − 7) + (−2 − (−2)) = √(−2) + 0 = 2

Therefore AB=BC
As two sides are equal in length, therefore, ABC is an isosceles triangle.

Page : 161 , Block Name : Exercise 7.1

Q5 In a classroom, 4 friends are seated at the points A, B, C and D as shown in Fig. 7.8.
Champa and Chameli walk into the class and after observing for a few minutes Champa asks
Chameli, “Don’t you think ABCD is a square?” Chameli disagrees. Using distance formula, nd
which of them is correct.

Answer. AB = √(3 − 6) 2
+ (4 − 7)
2
= √(−3)
2
+ (−3)
2
= √9 + 9 = √18 = 3√2

2 2 2 2
BC = √(6 − 9) + (7 − 4) = √(−3) + (3) = √9 + 9 = √18 = 3√2

2 2 2 2
CB = √(9 − 6) + (4 − 1) = √(3) + (3) = √9 + 9 = √18 = 3√2

2 2 2 2
AD = √(3 − 6) + (4 − 1) = √(−3) + (3) = √9 + 9 = √18 = 3√2

2 2 2 2
AC = √(3 − 9) + (4 − 4) = √(−6) + 0 = 6

2 2 2 2
BD = √(6 − 6) + (7 − 1) = √0 + (6) = 6

It can be observed that all sides of this quadrilateral ABCD are of the same length
and also the diagonals are Of the sarne length,
Therefore, ABCD is a square and hence, Champa correct

Page : 161 , Block Name : Exercise 7.1

Q6 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), (– 3, 0)
(ii) (–3, 5), (3, 1), (0, 3), (–1, – 4)
(iii) (4, 5), (7, 6), (4, 3), (1, 2)

Answer. (i) Let one points (-1, -2), (1, 0), (-1, 2), and (-3, O) be representing the vertices
A, B, C, and D Of the given quadrilateral respectively,
2 2 2 2
∴ AB = √(−1 − 1) + (−2 − 0) = √(−2) + (−2) = √4 + 4 = √8 = 2 √2

2 2 2 2
BC = √(1 − (−1)) + (0 − 2) = √(2) + (−2) = √4 + 4 = √8 = 2 √2

2 2 2 2
CD = √(−1 − (−3)) + (2 − 0) = √(2) + (2) = √4 + 4 = √8 = 2√2

2 2 2 2
AD = √(−1 − (−3)) + (−2 − 0) = √(2) + (−2) = √4 + 4 = √8 = 2 √2

Diagonal AC = √(−1 − (−1)) 2
+ (−2 − 2)
2
= √0
2
+ (−4)
2
= √16 = 4

Diagonal = √(1 − (−3)) 2 2 2 2
+ (0 − 0) = √(4) + 0 = √16 = 4

It can be observed that all sides of this quadrilateral are of the same length end also,
the diagonals are of the same length. Therefore, the given points are the vertices of
a square.

(ii)Let the points(- 3, 5), (3, 1), (O, 3), and (-1,-4) be representing the vertices A, B, C, and D Of
the given quadrilateral respectively.

2 2 2 2
AB = √(−3 − 3) + (5 − 1) = √(−6) + (4) = √36 + 16 = √52 = 2√13

2 2 2 2
BC = √(3 − 0) + (1 − 3) = √(3) + (−2) = √9 + 4 = √13

2 2 2 2
CD = √(0 − (−1)) + (3 − (−4)) = √(1) + (7) = √1 + 49 = √50 = 5√2

2 2 2 2
AD = √(−3 − (−1)) + (5 − (−4)) = √(−2) + (9) = √4 + 81 = √85

It can be observed that all sides of this quadrilateral are of different lengths
Therefore, it can be said that it is only a general quadrilateral, and not speci c such
as square, rectangle, etc.

(iii)Let Y-,e points (4, 5), (7, 6), (4, 3), and (1, 2) be representing the vertices A, a,
C, and D of the given quadrilateral respectively.
2 2 2 2
AB = √(4 − 7) + (5 − 6) = √(−3) + (−1) = √9 + 1 = √10

2 2 2 2
BC = √(7 − 4) + (6 − 3) = √(3) + (3) = √9 + 9 = √18

2 2 2 2
CD = √(4 − 1) + (3 − 2) = √(3) + (1) = √9 + 1 = √10

2 2 2 2
AD = √(4 − 1) + (5 − 2) = √(3) + (3) = √9 + 9 = √18

2 2 2 2
AC = √(4 − 4) + (5 − 3) = √(0) + (2) = √0 + 4 = 2

2 2 2 2
CD = √(7 − 1) + (6 − 2) = √(6) + (4) = √36 + 16 = √52 = 13√2

It can be observed that opposite sides of this quadrilateral are of the same length.
However, the diagonals are Of different lengths, Therefore, the given points are the
vertices Of a parallelogram

Page : 161 , Block Name : Exercise 7.1

Q7 Find the point on the x-axis which is equidistant from (2, –5) and (–2, 9).

Answer. We have to nd a point on x-axis. Therefore y-coordinate will be 0.
Let the point on x-axis be (x,0)
Distance between (x,0) , (2,-5)=√(x − 2) + (0 − (−5)) = √(x − 2) + (5)
2 2 2 2

Distance between (x,0) , (-2,-9)= √(x − (−2)) 2
+ (0 − (−9))
2
= √(x + 2)
2
+ (9)
2

By the given condition, these distances are equal in measure.
√(x − 2)2 + (5)2 = √(x + 2)2 + (9)2

2 2
(x − 2) + 25 = (x + 2) + 81
2 2
x + 4 − 4x + 25 = x + 4 + 4x + 81

8x=25-81
8x=-56
x=-7
Therefore the point is (-7,0)

Page : 161 , Block Name : Exercise 7.1

Q8 Find the values of y for which the distance between the points P(2, – 3) and Q(10, y) is 10
units.

Answer. It is given that the distance between (2 -3) and (10, y) is 10.
Therefore √(2 − 10) 2
+ (−3 − y)
2
= 10

√(−8)2 + (3 + y)2 = 10

2
64 + (y + 3) = 100
2
(y + 3) = 36

y + 3 = ±6

y + 3 = 6 or y + 3 = −6

Therefore ,y=3 or -9

Page : 161 , Block Name : Exercise 7.1

Q9 If Q(0, 1) is equidistant from P(5, –3) and R(x, 6), nd the values of x. Also nd the
distances QR and PR.

Answer. PQ= QR
√(5 − 0)2 + (−3 − 1)2 = √(0 − x)2 + (1 − 6)2

√(5)2 + (−4)2 = √(−x)2 + (−5)2

2
√25 + 16 = √x + 25
2
41 = x + 25
2
16 = x

x = ±4

Therefore , point R is (4,6) or (-4,6).
When the point is ( 4,6 )
PQ=√(5 − 4) 2
+ (−3 − 6)
2
= √1
2
+ (−9)
2
= √1 + 81 = √82

qR = √(0 − 4) 2
+ (1 − 6)
2
= √(−4)
2
+ (−5)
2
= √16 + 25 = √41

When the point R is (-4,6)
PR=√(5 − (−4)) + (−3 − 6) 2 2
= √(9)
2
+ (−9)
2
= √81 + 81 = 9√2

qR= √(0 − (−4)) 2
+ (1 − 6)
2
= √(4)
2
+ (−5)
2
= √16 + 25 = √41

Page : 162 , Block Name : Exercise 7.1

Q10 Find a relation between x and y such that the point (x, y) is equidistant from the point (3,
6) and (– 3, 4).

Answer. Point (x, y) is equidistant from (3, 6) and (-3, 4).
2 2 2 2
∴ √(x − 3) + (y − 6) = √(x − (−3)) + (y − 4)

√(x − 3)2 + (y − 6)2 = √(x + 3)2 + (y − 4)2

2 2 2 2
(x − 3) + (y − 6) = (x + 3) + (y − 4)
2 2 2 2
x + 9 − 6x + y + 36 − 12y = x + 9 + 6x + y + 16 − 8y

36 − 16 = 6x + 6x + 12y − 8y

20 = 12x + 4y

3x + y = 5

3x + y − 5 = 0

Page : 162 , Block Name : Exercise 7.1

Exercise 7.2

Q1 Find the coordinates of the point which divides the join of (–1, 7) and (4, –3) in the ratio 2 :
3.

Answer. Let P(x, y) be the required point. using the section formula, we obtain
2×4+3×(−1) 8−3 5
x = = = = 1
2+3 5 5

2×(−3)+3×7 −6+21 15
y = = = = 3
2+3 5 5

Therefore, the point is (1, 3).

Page : 167 , Block Name : Exercise 7.2

Q2 Find the coordinates of the points of trisection of the line segment joining (4, –1) and
(–2, –3).

Answer.

Let P(x , y ) and Q(x , y ) are the points of trisection of the line segment joining
1 1 2 2

the given points i.e., AP PQ Q3
Therefore, point p divides AB internally in the ratio 1:2.
1×(−2)+2×4
x1 =
1+2

2×(−3)+1×(−1)
y2 =
2+1
−4+4
x2 = = 0
3
−6−1 −7
y2 = =
3 3

7
Q (x2 , y2 ) = (0, − )
3

Page : 167 , Block Name : Exercise 7.2

Q3 To conduct Sports Day activities, in your rectangular shaped school ground ABCD, lines
have been drawn with chalk powder at a distance of 1m each. 100 ower pots have been placed

at a distance of 1m from each other along AD, as shown in Fig. 7.12. Niharika runs 1 4 th the
distance AD on the 2nd line and posts a green ag. Preet runs 1 5 th the distance AD on the
eighth line and posts a red ag. What is the distance between both the ags? If Rashmi has to
post a blue ag exactly halfway between the line segment joining the two ags, where should
she post her ag?

Answer. It can be observed that Niharika posted the green ag at 1

4
of the distance AD i.e.,
1
( × 100) m = 25
4

Similarly, Preet posted red ag at of the distance i.e. ( metre from the
1 1
× 100) m = 20
5 5

starting point of 8th line. Therefore , the coordinate of this point R are (8,20).
Distance between these ags by using distance formula = GR
2 2
= √(8 − 2) + (25 − 20) = √36 + 25 = √61m

The point at which Rashmi should post her bue ag is the mid-point of the line
joining these points. Let this point be A (x, y).
,y =
2+8 25+20
x =
2 2
2

x =
10

2
= 5 ,y = 45

2
= 22.5

Hence A(x, y) = (5, 22.5)
Therefore , rashmi should post her blue ag at 22.5m on 5th line.

Page : 167 , Block Name : Exercise 7.2

Q4 Find the ratio in which the line segment joining the points (– 3, 10) and (6, – 8) is divided
by (– 1 6).

Answer. Let the ratio in which the line segment joining (-3, 10) and (6, -8) is divided by
point (-1, 6) be k : 1.
Therefore −1 =
6k−3

k+1

−k − 1 = 6k − 3

7k=2
2
k =
7

Therefore , the required ratio is 2:7.

Page : 167 , Block Name : Exercise 7.2

Q5 Find the ratio in which the line segment joining A(1, – 5) and B(– 4, 5) is divided by the x-
axis. Also nd the coordinates of the point of division.

Answer. Let the ratio in which the line segment joining A (1, -5) and B (-4, 5) is divided by X-
axis be k:1.
Therefore , the coordinates of the point of division is (
−4k+1 5k−5
, )
k+1 k+1

When we know that y -coordinates of any point on x-axis is 0.
5k−5
∴ = 0
k+1

k=1
Therefore x-axis divides it in the ratio 1:1.
−4(1)+1 5(1)−5
Division point = (
−4+1 5−5 −3
, ) = ( , ) = ( , 0)
1+1 1+1 2 2 2

Page : 167 , Block Name : Exercise 7.2

Q6 If (1, 2), (4, y), (x, 6) and (3, 5) are the vertices of a parallelogram taken in order, nd x and
y.

Answer.

Let (1, 2), (4, y), (x, 6), and (3, 5) are the coordinates of A, B, C, D vertices of a
parallelogram ABCD. Intersection point O of diagonal AC and BD also divides these
diagonals.
Therefore, O is the mid-point of AC and BD.
If O is the mid-point of AC, then the coordinates of O are
1+x 2+6 x+1
( , ) ⇒ ( , 4)
2 2 2

If O is the mid-point of BD, then the coordinates of O are
4+3 5+y 7 5+y
( , ) ⇒ ( , )
2 2 2 2

Since both the coordinates are of the sane point O,
x+1 7
∴ =
2 2

And 94 =
5+y

2

x+1=7 and 5+y=8
x=6 and y=3

Page : 167 , Block Name : Exercise 7.2

Q7 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. Let the coordinates of point A be (x, y).
Mid-point of AS is (2, -3), which is the center of the circle,
x+1 y+4
∴ (2, −3) = ( , )
2 2

y+4
and
x+1
⇒ = 2 = −3
2 2

x+1=4 and y+4=-6
x=3 and y=-10
Therefore , the coordinates are (3 , -10)

Page : 167 , Block Name : Exercise 7.2