# PSEB Solutions for Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.1

PSEB Solutions for Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.1

## PSEB 10th Class Maths Solutions Chapter 7 Coordinate Geometry Ex 7.1

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:

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

discussed in section 7.2.

Solution:

Question 5.

In a classroom, 4 friends are seated at the points A, B, C and D as shown in fig. Champa and Charnel 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, and why?

Solution:

In the given diagram, the vertices of given points are : A (3, 4); B (6, 7); C (9, 4) and D (6, 1).

Now,

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:

From above discussion, it is clear that

AB = BC = CD = DA = √8 and AC = BD = 4.

Hence, given quadrilateral ABCD is a square.

From above discussion, it is clear that AB = CD and BC = DA. and AC ≠ BD.

i.e., opposite sides are equal but their diagonals are not equal.

Hence, given quadrilateral ABCD is a parallelogram.

Question 7.

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

Solution:

Let required point be P (x, 0) and given points be A (2, – 5) and B (- 2, 9).

According to question,

PA = PB

(PA)^{2} = (PB)^{2}

or (2 – x)^{2} + (- 5- 0)^{2} = (- 2 – x)^{2} + (9 – 0)^{2}

or 4 + x^{2} – 4x + 25 = 4 + x^{2}+ 4x + 81

-8x = 56

x = 4/4 = – 7

Hence, required point be (- 7, 0).

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:

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:

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:

**Follow on Facebook page –** **Click Here**

**Google News join in – Click Here**

**Read More Asia News – Click Here**

**Read More Sports News – Click Here**

**Read More Crypto News – Click Here**