P and Q are the points on the line segment joining the points A(3, -1) and B(-6, 5) such that AP = PQ = QB. Find the coordinates of P and Q.

Given, AP = PQ = QB

∴ P divides AB in the ratio of 1 : 2 and Q divides it in 2 : 1.

Let coordinates of P be (a, b) and of Q be (c, d)

Applying section formula for x coordinate of P we get,

$a = \dfrac{m$1x2 + m2x1}{m1 + m2} \\[1em] = \dfrac{1 \times (-6) + 2 \times 3}{1 + 2} \\[1em] = \dfrac{-6 + 6}{3} \\[1em] = 0.

Similarly, applying section formula for y coordinate of P we get,

$b = \dfrac{m$1y2 + m2y1}{m1 + m2} \\[1em] = \dfrac{1 \times 5 + 2 \times (-1)}{1 + 2} \\[1em] = \dfrac{5 - 2}{3} \\[1em] = \dfrac{3}{3} \\[1em] = 1.

∴ Coordinates of P = (a, b) = (0, 1).

Applying section formula for x coordinate of Q we get,

$c = \dfrac{m$1x2 + m2x1}{m1 + m2} \\[1em] = \dfrac{2 \times (-6) + 1 \times 3}{2 + 1} \\[1em] = \dfrac{-12 + 3}{3} \\[1em] = -3.

Similarly, applying section formula for y coordinate of Q we get,

$d = \dfrac{m$1y2 + m2y1}{m1 + m2} \\[1em] = \dfrac{2 \times 5 + 1 \times (-1)}{2 + 1} \\[1em] = \dfrac{10 + (-1)}{3} \\[1em] = \dfrac{9}{3} \\[1em] = 3.

∴ Coordinates of Q = (c, d) = (-3, 3).

Hence, coordinates of P = (0, 1) and Q = (-3, 3).