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

**Solution:**

We know that the distance between the two points is given by the Distance Formula = √[( x₂_{ - }x₁_{ })^{2} + (y₂ - y₁)^{2}]

By substituting the values of points P (2, - 3) and Q (10, y) in the distance formula, we get

PQ = √(2 - 10)² + (- 3 - y)² = 10

PQ = √(- 8)² + (3 + y)² = 10

Squaring on both sides, we get

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

(y + 3)^{2} = 36

y + 3 = √36

y + 3 = ± 6

y + 3 = 6 or y + 3 = - 6

Therefore, y = 3 or - 9 are the possible values for y.

**☛ Check: **NCERT Solutions for Class 10 Maths Chapter 7

**Video Solution:**

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

NCERT Class 10 Maths Solutions Chapter 7 Exercise 7.1 Question 8

**Summary:**

The values of y for which the distance between the points P (2, - 3) and Q (10, y) is 10 units are y = 3 or - 9.

**☛ Related Questions:**

- Check whether (5, - 2), (6, 4) and (7, - 2) are the vertices of an isosceles triangle.
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- Find the point on the x-axis which is equidistant from (2, - 5) and (- 2, 9).

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