NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry

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Written by Team Trustudies
Updated at 2021-02-22


NCERT solutions for class 10 Maths Chapter 7 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)



NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry


Answer :

Distance formula =(x2?x1)2+(y2?y1)2
(i) Applying Distance Formula to find distance between points (2, 3) and (4,1), we get
d = (4?2)2+(1?3)2
=22+(?2)2=4+4
=8
=22

(ii) Applying Distance Formula to find distance between points (–5, 7) and (–1, 3), we get
d = (?5+1)2+(7?3)2
=((?4)2+(4)2=16+16
=32
=42

(iii) Applying Distance Formula to find distance between points (a, b) and (–a, –b), we get
d = (?a?a)2+(?b?b)2
=((?2a)2+(?2b)2
=4a2+4b2
=2a2+b2

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



NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry


Answer :

The coordinates of point A are (0, 0) and coordinates of point B are (36, 15).
To find the distance between them, we use Distance formula:
d = (36?0)2+(15?0)2
=362?152
=1296+225
=1521
=39km

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



NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry


Answer :

Let A = (1, 5), B = (2, 3) and C = (–2, –11)
Using Distance Formula to find distance AB, BC and CA.
AB = (2?1)2+(3?5)2
=12?(?2)2
=1+5
=5
BC = (?2?2)2+(?11?3)2
=(?4)2?(?14)2
=16+196
=212
CA = (?2?1)2+(?11?5)2
=(?3)2?(?15)2
=9+256
=265
Since AB + AC ? BC, BC + AC ? AB and AC ? BC.
Therefore, the points A, B and C are not collinear.

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



NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry


Answer :

Let A = (5, –2), B = (6, 4) and C = (7, –2)
Using Distance Formula to find distances AB, BC and CA.

AB = (6?5)2+(4+2)2
=12?62
=1+36
=37

BC = (7?6)2+(?2?4)2
=12?(?6)2
=1+36
=37

CA = (7?5)2+(?2+2)2
=22?0
=4
=2
Since AB = BC.

Therefore, A, B and C are vertices of an isosceles triangle.

Q5 ) In a classroom, 4 friends are seated at the points A (3, 4), B (6, 7), C (9, 4) and D (6, 1). 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, find which of them is correct.



NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry


Answer :

We have A = (3, 4), B = (6, 7), C = (9, 4) and D = (6, 1)
Using Distance Formula to find distances AB, BC, CD and DA, we get

AB = (6?3)2+(7?4)2
=32+32
=9+9
=18
=32

BC = (9?6)2+(4?7)2
=32+32
=9+9
=18
=32

CD = (6?9)2+(1?4)2
=32+(?3)2=9+9
=18
=32

DA = (6?3)2+(1?4)2
=32+(?3)2
=9+9
=18
=32

Therefore, All the sides of ABCD are equal here. … (1)

Now, we will check the length of its diagonals.

AC = (9?3)2+(4?4)2
=62?02
=36
=6

BD = (6?6)2+(1?7)2
=02?62
=36
=6

So, Diagonals of ABCD are also equal. … (2)

From (1) and (2), we can definitely say that ABCD is a square.

Therefore, Champa is correct.

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)



NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry


Answer :

i) Let A = (–1, –2), B = (1, 0), C= (–1, 2) and D = (–3, 0)
Using Distance Formula to find distances AB, BC, CD and DA, we get

AB = (1+1)2+(0+2)2
=4+4
=22

BC = (?1?1)2+(2?0)2
=4+4
=22

CD = (?3+1)2+(0?2)2
=4+4
=22

DA = (?3+1)2+(0?2)2
=4+4
=22

Therefore, all four sides of quadrilateral are equal. … (1)
Now, we will check the length of diagonals.

AC = (?1+1)2+(2+2)2
=0+16
=4

BD = (?3?1)2+(0+0)2
=16+0
=4

Therefore, diagonals of quadrilateral ABCD are also equal. … (2)

From (1) and (2), we can say that ABCD is a square.

(ii) Let A = (–3, 5), B= (3, 1), C= (0, 3) and D= (–1, –4)
Using Distance Formula to find distances AB, BC, CD and DA, we get
AB = (?3?3)2+(1?5)2
=36+16
=213

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

CD = (?1?0)2+(?4?3)2
=1+49
=52

DA = (?1+3)2+(?4?5)2
=4+81
=85

We cannot find any relation between the lengths of different sides.

Therefore, we cannot give any name to the figure ABCD.

(iii) Let A = (4, 5), B= (7, 6), C= (4, 3) and D= (1, 2)
Using Distance Formula to find distances AB, BC, CD and DA, we get

AB = (7?4)2+(6?5)2
=9+1
=10

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

CD = (1?4)2+(2?3)2
=9+1
=10

DA = (1?4)2+(2?5)2
=9+9
=18

Here opposite sides of quadrilateral ABCD are equal. … (1)

We can now find out the lengths of diagonals.

AC = (4?4)2+(3?5)2
=0+4
=2

BD = (1?7)2+(2?6)2
=36+16=
213

Here diagonals of ABCD are not equal. … (2)

From (1) and (2), we can say that ABCD is not a rectangle therefore it is a parallelogram.

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



NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry


Answer :

Let the point be (x, 0) on x–axis which is equidistant from (2, –5) and (–2, 9).

Using Distance Formula and according to given conditions we have:

(x?2)2+(0?(?5))2
=(x?(?2))2+(0?9)2

=>x2+4?4x+25
=x2+4+4x+81

Squaring both sides, we get
?x2+4?4x+25=x2+4+4x+81
??4x+29=4x+85
?8x=?56
?x=?7

Therefore, point on the x–axis which is equidistant from (2, –5) and (–2, 9) is (–7, 0)

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



NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry


Answer :

Using Distance formula, we have
?10=(2?10)2+(?3?y)2
?10=(?8)2+9+y2+6y
?10=64+9+y2+6y
Squaring both sides, we get
?100=73+y2+6y
?y2+6y+27=0

Solving this Quadratic equation by factorization, we can write
?y2+9y?3y?27=0
?y(y+9)3(y+9)=0
?(y+9)(y?3)=0
?y=3,?9

Q9 ) 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.



NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry


Answer :

It is given that Q is equidistant from P and R. Using Distance Formula, we get
PQ = RQ

PQ=(5?0)2+(?3?1)2
=(?5)2+(?4)2
=25+16
=41
QR= (0?x)2+(1?6)2
=(?x)2+(?5)2
=x2+25
?41=x2+25
On squaring both sides,
?41=x2+25
?x2=16
?x=±4
Coordinates of point R will be (4,6) or (-4,6),
If R is (4,6),
QR = (0?4)2+(1?6)2
=42+52
=16+25
=41
PR = (5?4)2+(?3?6)2
=12+92
=1+81
=82
If R is (-4,6),
QR = (0+4)2+(1?6)2
=42+52
=16+25
=41
PR = (5+4)2+(?3?6)2
=92+92
=81+81
=92

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



NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry


Answer :

It is given that (x, y) is equidistant from (3, 6) and (–3, 4).
Using Distance formula, we can write

?(x?3)2+(y?6)2=(x?(?3))2+(y?4)2
?x2+9?6x+y2+36?12y=x2+9+6x+y2+16?8y
Squaring both sides, we get
?x2+9?6x+y2+36?12y=x2+9+6x+y2+16?8y
??6x?12y+45=6x?8y+25
?12x+4y=20
?3x+y=5

NCERT solutions for class 10 Maths Chapter 7 Coordinate Geometry 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.



NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry


Answer :

Let the points be P(x,y), Using the section formula

P(x,y)=[mx2+nx1m+n,my2+ny1m+n]

x=2(4)+3(?1)2+3=8?35=1
y=2(?3)+3(7)2+3=?6+215=3
Therefore the point is(1,3)

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



NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry


Answer :




Let P (x1, y1) and Q (x2, y2) are the points of trisection of the line segment joining the given points i.e., AP = PQ = QB

Therefore, point P divides AB internally in the ratio 1:2.

x1 = (1×(?2)+2×4)1+2=(?2+8)3=63=2

y1 = (1×(?3)+2×?1)1+2=(?3?2)3=?53

?P = (x1, y1) = P(2, ?53)

Point Q divides AB internally in the ratio 2:1.

x2 = (2×(?2)+1×4)2+1=(?4+4)3=0

y2 = (2×(?3)+1×?1)2+1=(?6?1)3=?73

The coordinates of the point Q(x2 ,y2 ) = (0, ?73 )

Q3 ) 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, as shown in the following figure. Niharika runs 14 th the distance AD on the 2nd line and posts a green flag. Preet runs 15th 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?



NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry


Answer :

The green flag is 14th of total distance =
14×100=25 m in second line

? the coordinate of green flag
= (2 , 25 )
Similarly coordinate of red flag
= (8 , 20)

Distance between red and green flag
= (8?2)2+(20?25)2
=62+(?5)2
=36+25
=61m

Now blue flag is posted at the mid point of two flag
Then let, the coordinate of blue flag = (x , y )

?(x,y)=((8+2)2,20+252)
=(5,452)=(5,22.5)

? the blue flag is posted in fifth line at a distance of 22.5 m

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



NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry


Answer :

Let the required ratio be k : 1
By section formula ,

x=m1x2+m2x1m1+m2
??1=k×6+1×?3)k+1
??k?1=6k?3
?7k=2
?k=27

The required ratio = k : 1
= 27 : 1 = 27×7:1×7 = 2 : 7

Q5 ) 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.



NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry


Answer :



Let p(x , 0) be the point which devides the line segment joining A(1, -5 ) and B(-4,5 ) in the ratio m : 1

Then the using section formula

(x , 0 ) = (m×(?4)+1×1m+1,m×5+1×(?5)m+1)
?0=m×5+1×(?5)m+1
?0=5m?5
?m=5/5=1

Hence the required ratio is 1 : 1

Since the ratio is 1 : 1 , so P is the mid point

? x = m×(?4)+1×1m+1
x = 1×(?4)+1×11+1=?4+12=?32

?(?32,0) is the required point

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



NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry


Answer :



Mid point of AC = Mid point BD
?x+12,6+22=4+32,y+52
?x+12=72
?x+1=7
?x=7?1=6
?y+52=82
?y+5=8
?y=8?5=3

Q7 ) Find the coordinates of a point A, where AB is the diameter of circle whose centre is (2, – 3) and B is (1,4).



NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry


Answer :

Let the coordinates of point A be (x, y).

Mid-point of AB is (2, – 3), which is the centre of the circle.

Coordinate of B = (1, 4)
(2, -3) =(x+12,y+42)
x+12=2
?x+1=4
?x=4?1=3
andy+42=?3
?y+4=?6
?y=?6?4=?10
? x = 3 and y = -10
The coordinates of A(3,-10).

Q8 ) If A and B are (-2, -2) and (2, -4), respectively, find the coordinates of P such that AP = 37 AB and P lies on the line segment AB.



NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry


Answer :

AP = 37 AB
7 AP = 3 AB
AB : AP = 7: 3
Let AB = 7x
? AP = 3x
AB = AP + BP
7x = 3x + BP
BP = 7x - 3x = 4x

?APBP=3x4x
? AP : BP = 3:4
by section formula ,

(x, y ) = (m1x2+m2x1m1+m2,m1y2+m2y1m1+m2)
(x, y) = (3×2+4×(?2)3+4,3×(?4)+4(?2)3+4)
(x ,y ) =