If Q(0, 1) is equidistant from P(5, -3) and R(x, 6), find the values of x.

By distance formula,

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

Given,

∴ Distance between Q(0, 1) and P(5, -3) = Distance between Q(0, 1) and R(x, 6).

$\Rightarrow \sqrt{(5 - 0)^2 + (-3 - 1)^2} = \sqrt{(x - 0)^2 + (6 - 1)^2} \\[1em] \Rightarrow \sqrt{5^2 + (-4)^2} = \sqrt{x^2 + 5^2} \\[1em] \Rightarrow \sqrt{25 + 16} = \sqrt{x^2 + 25}$

On squaring both sides,

$\Rightarrow 25 + 16 = x^2 + 25 \\[1em] \Rightarrow x^2 = 16 \\[1em] \Rightarrow x = \sqrt{16} \\[1em] \Rightarrow x = \pm 4.$

Hence, x = ±4.