# Gujarat Board Solutions Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.1

Gujarat Board Solutions Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.1

## Gujarat Board Textbook Solutions Class 10 Maths 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:

(i) Let the given points be A(2, 3) and B(4, 1). We know that the distance between two points A(x_{1}, y_{1}) and B(x_{2}, y_{2}) is given bý

Question 6.

Name the type of quadrilateral formed, if any, by the following points and give reasons for your answer:

(j) (-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:

(i) Let the given points be A(-1, -2), B(1, 0), C(-1, 2) and D(-3, 0).

Then,

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:

We have, P(2, -3), Q( 10, y) and PQ = 10 units.

Now, PQ^{2} = (10)^{2} = 100

(10 – 2)^{2} + [(y – (-3)]^{2}= 100

(8)^{2} + (y + 3)^{2} = 100

64+y^{2} + 6y + 9 = 100

y^{2} + 6y – 27 = 0

y^{2} + 9y – 3y – 27 = o

y(y + 9) -3 (y + 9) = 0

(y + 9)(y – 3) = 0

y + 9 = 0

or y – 3 = 0

y = -9

or y = 3

y = -9, 3

Hence, the required value of y is -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:

We have, P(5, 3) and R(x, 6) and Q(0, 1). It is also given that,

PQ = RQ

PQ^{2} = RQ^{2}

(0 – 5)^{2} + [1 – (3)]^{2}

= (0 – x)^{2} + (1 – 6)^{2}

25 + 16 = x^{2} + 25

x^{2} = 16

x = ±4

Therefore, co-ordinates of R are R(±4, 6).

Question 10.

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

Solution:

Let P(x, y), A(3, 6) and B(-3, 4) be the given points.

It is given that,

AP = BP

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