If the distance between the points (2, -2) and (-1, x) is 5 units, then one of the value of x is

1. -2

2. 2

3. -1

4. 1

By distance formula,

d = $\sqrt{(x$2 - x1)^2 + (y2 - y1)^2}

Given,

Distance between (2, -2) and (-1, x) is 5 units.

$\therefore 5 = \sqrt{(-1 - 2)^2 + [x - (-2)]^2} \\[1em] \Rightarrow 5 = \sqrt{(-3)^2 + [x + 2]^2} \\[1em] \Rightarrow 5 = \sqrt{9 + x^2 + 4 + 4x} \\[1em] \Rightarrow 5 = \sqrt{x^2 + 4x + 13} \\[1em] \Rightarrow x^2 + 4x + 13 = 5^2 \text{ [On squaring both sides]} \\[1em] \Rightarrow x^2 + 4x + 13 = 25 \\[1em] \Rightarrow x^2 + 4x - 12 = 0 \\[1em] \Rightarrow x^2 + 6x - 2x - 12 = 0 \\[1em] \Rightarrow x(x + 6) - 2(x + 6) = 0 \\[1em] \Rightarrow (x - 2)(x + 6) = 0 \\[1em] \Rightarrow x - 2 = 0 \text{ or } x + 6 = 0 \\[1em] \Rightarrow x = 2 \text{ or } x = -6.$

As distance cannot be negative,

∴ x = 2

Hence, Option 2 is the correct option.