The x-coordinate of a point P is twice its y-coordinate. If P is equidistant from the points Q(2, -5) and R(-3, 6), then find the coordinates of P.

Let y-coordinate of point P be k.

∴ x-coordinate = 2k.

P = (2k, k)

By distance formula,

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

Given,

P is equidistant from the points Q(2, -5) and R(-3, 6).

$\Rightarrow \sqrt{(2 - 2k)^2 + (-5 - k)^2} = \sqrt{(-3 - 2k)^2 + (6 - k)^2} \\[1em] \Rightarrow \sqrt{4 + 4k^2 - 8k + 25 + k^2 + 10k} = \sqrt{9 + 4k^2 + 12k + 36 + k^2 - 12k} \\[1em] \Rightarrow \sqrt{5k^2 + 2k + 29} = \sqrt{5k^2 + 45}$

On squaring both sides,

$\Rightarrow 5k^2 + 2k + 29 = 5k^2 + 45 \\[1em] \Rightarrow 5k^2 - 5k^2 + 2k = 45 - 29 \\[1em] \Rightarrow 2k = 16 \\[1em] \Rightarrow k = \dfrac{16}{2} \\[1em] \Rightarrow k = 8.$

P = (2k, k) = (16, 8).

Hence, coordinates of P = (16, 8).