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NCERT Solutions for Class 10 Maths Chapter 7:Coordinate Geometry Exercise 7.1




Coordinate Geometry Exercise 7.1

In this page we have NCERT Solutions for Class 10 Maths Chapter 7:Coordinate Geometry for Exercise 7.1 on pages 161 and 162. Hope you like them and do not forget to like , social share and comment at the end of the page.
Question 1.
Find the distance between the following pairs of points :
(i) (2, 3), (4, 1)
(ii) (-5, 7), (-1, 3)
(iii) (a,b), (-a,-b)
Solution
Distance between the points AB is given by
NCERT Solutions  for Class 10 Maths Chapter 7:Coordinate Geometry Exercise 7.1

Question 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 .
Solution
Let P(0,0) and Q(36,15) be the given points.
Distance between the points PQ is given b

The position of town A and B can be represented as points P and Q respectively,So distance between town will be 39 km

Question 3.
Determine if the points (1, 5), (2, 3) and (-2, -11) are collinear.
Solution
Lets us denote point by P(1,5) , Q(2,3) and R(-2,-11)
If the points are not collinear, then we should be able to form the triangle
Lets us find the length of PQ, QR and PR by distance formula

Clearly None of these is true
PQ+QR=PR
PR+PQ=QR
PQ=QR+PR
Hence they are not collinear

Question 4.
Check whether (5, -2), (6, 4) and (7, -2) are the vertices of an isosceles triangle
Solution
Lets us denote point by P(5,-2) , Q(6,4) and R(7,-2)
Lets us find the length of PQ, QR and PR by distance formula

Now PQ =QR ,so it is an isosceles triangle

Question 5.
In a classroom, 4 friends are seated at the points A, B, C and D as shown below. Champaand Chameli walk into the class and after observing for a few minutes Champa asks Chameli,
Solution
As per the figure given,The coordinates of the points A,B,C and D are (3,4), (6,7) ,(9,4) and (6,1)
Lets us find the length of AB, BC ,CD and AD by distance formula
NCERT Solutions  for Class 10 Maths Chapter 7:Coordinate Geometry Exercise 7.1
So all the sides are equal. But we cannot still say that it is square as rhombus has all the sides equal also.
Now we know that a square has both the diagonal equal also,So lets us calculate the diagonal’s

Hence AC=BD
So, it is a square.
So, champa is correct
“Don’t you think ABCD is a square?” Chameli disagrees.

Using distance formula, find which of them is correct.

Question 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), (-3, 0)
(ii) (-3, 5), (3, 1), (0, 3), (-1, -4)
iii) (4,5) ,(7,6), (4,3) (1,2)
Solution
In these type of problem, we need to find of length of each segment, then check with the properties of different type of quadrilateral
i.e for points A,B,C and D
Line segments are AB,BC,AC,CD,AD and BD
We need to find the length of each of these
(i) (-1, -2), (1, 0), (-1, 2), (-3, 0)

Here the side AB,BC,CD and AD are equal and diagonal AC and BD are equal. So this is a square
(ii) (-3, 5), (3, 1), (0, 3), (-1, -4)

Now here AC+BC=AB
So, that means ABC are collinear points.
So, it is not a quadrilateral infact
(iii) (4,5) ,(7,6), (4,3) (1,2)

Now AB=CD and BC=DA
Now it could be rectangle or parallelogram
But diagonal AC is not equal diagonal BD
So, It is a parallelogram



Question 7.
Find the point on the x-axis which is equidistant from (2, –5) and (–2, 9).
Solution
Since the point lies on X axis, the point should be of the form (a,0)
Now (a,0) is equidistant from both the given points

Squaring both the sides
(x-2)2 + (0+5)2 =(x+2)2 + (0-9)2
Solving it we get
x=-7

Question 8.
Find the values of y for which the distance between the points P(2, – 3) and Q(10, y) is 10 units
Solution
According to the question
PQ=10

Squaring both the sides
y2 +6y-27=0
(y+9)(y-3)=0
So y=-9 or 3

Question 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.
Solution
Now
QP=QR

Squaring both the sides
25+16=x2 +25
x= -4 or 4
So point R is either (4,6) or (-4,6)


Question 10.
Find a relation between x and y such that the point (x, y) is equidistant from the point
(3, 6) and (– 3, 4).
Solution
Let the point P(x,y) is equidistant from the point Q( 3,6) and R(-3,4)
PQ=PR
NCERT Solutions  for Class 10 Maths Chapter 7:Coordinate Geometry
Squaring both the sides
x2 -6x+9 +y2-12y +36=x2 +6x+9 +y2 -8y+16
-12x-4y+20=0
Or
3x+y-5=0 ( dividing by -4)

Download Coordinate Geometry Exercise 7.1 as pdf
link to this page by copying the following text
Also Read



Reference Books for class 10

Given below are the links of some of the reference books for class 10 math.

  1. Oswaal CBSE Question Bank Class 10 Hindi B, English Communication Science, Social Science & Maths (Set of 5 Books)
  2. Mathematics for Class 10 by R D Sharma
  3. Pearson IIT Foundation Maths Class 10
  4. Secondary School Mathematics for Class 10
  5. Xam Idea Complete Course Mathematics Class 10

You can use above books for extra knowledge and practicing different questions.



Go back to Class 10 Main Page using below links
Class 10 Maths Class 10 Science

Practice Question

Question 1 What is $1 - \sqrt {3}$ ?
A) Non terminating repeating
B) Non terminating non repeating
C) Terminating
D) None of the above
Question 2 The volume of the largest right circular cone that can be cut out from a cube of edge 4.2 cm is?
A) 19.4 cm3
B) 12 cm3
C) 78.6 cm3
D) 58.2 cm3
Question 3 The sum of the first three terms of an AP is 33. If the product of the first and the third term exceeds the second term by 29, the AP is ?
A) 2 ,21,11
B) 1,10,19
C) -1 ,8,17
D) 2 ,11,20






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