The distance between the points (3, 1) and (0, x) is 5. Find x.

Let (3, 1) = (x₁, y₁) and (0, x) = (x₂, y₂)

⇒ Distance between the given points =

$\sqrt{(x$2 - x1)^2 + (y2 - y1)^2}\\[1em] ⇒ 5 = \sqrt{(0 - 3)^2 + (x - 1)^2}\\[1em] ⇒ 5 = \sqrt{(-3)^2 + (x - 1)^2}\\[1em] ⇒ 5^2 = (-3)^2 + (x - 1)^2\\[1em] ⇒ 25 = 9 + (x - 1)^2\\[1em] ⇒ 25 - 9 = (x - 1)^2\\[1em] ⇒ 16 = (x - 1)^2\\[1em] ⇒ \sqrt{16} = x - 1\\[1em] ⇒ 4 = x - 1 \text{ and} -4 = x -1\\[1em] ⇒ x = 4 + 1 \text{ and } x = -4 + 1\\[1em] ⇒ x = 5 \text{ and} -3\\[1em]

Hence, the value of x = 5 and -3.