If the distance between the points (2, –2) and (–1, x) is 5, one of the values of x is
a. -2
b. 2
c. -1
d. 1

Solution:

Let us consider the points as

A = (2, -2)

B = (-1, x)

AB = 5 units

Using the distance formula

AB2 = (x₂ - x₁)2 + (y₂ - y₁)2

Substituting the values

52 = (-1 - 2)2 + (x + 2)2

25 = (-3)2 + (x + 2)2

Using the algebraic identity

(a + b)2 = a2 + b2 + 2ab

25 = 9 + x2 + 4 + 4x

By further calculation

25 = x2 + 4x + 13

x2 + 4x + 13 - 25 = 0

x2 + 4x - 12 = 0

By splitting the middle term

x2 + 6x - 2x - 12 = 0

Taking out the common terms

x(x + 6) - 2(x + 6) = 0

(x + 6)(x - 2) = 0

So we get

x + 6 = 0

x = -6

And

x - 2 = 0

x = 2

Therefore, one of the values of x is 2.

✦ Try This: If the distance between the points (3, –3) and (–2, x) is 4, one of the values of x is

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

NCERT Exemplar Class 10 Maths Exercise 7.1 Sample Problem 1

If the distance between the points (2, –2) and (–1, x) is 5, one of the values of x is a. -2, b. 2, c. -1, d. 1

Summary:

If the distance between the points (2, –2) and (–1, x) is 5, one of the values of x is 2

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