A(2, 5), B(-1, 2) and C(5, 8) are the vertices of △ABC. P and Q are points on AB and AC respectively such that AP : PB = AQ : QC = 1 : 2.

(a) Find the coordinates of points P and Q.

(b) Show that BC = 3 × PQ.

(a) Let coordinates of P be (x, y)

By section formula,

P(x, y) = $\Big(\dfrac{m$1x2 + m2x1}{m1 + m2}, \dfrac{m1y2 + m2y1}{m1 + m2}\Big)

Substituting values we get,

$\Rightarrow P = \Big(\dfrac{1 \times -1 + 2 \times 2}{1 + 2}, \dfrac{1 \times 2 + 2 \times 5}{1 + 2}\Big) \\[1em] \Rightarrow P = \Big(\dfrac{-1 + 4}{3}, \dfrac{2 + 10}{3}\Big) \\[1em] \Rightarrow P = \Big(\dfrac{3}{3}, \dfrac{12}{3}\Big) \\[1em] \Rightarrow P = (1, 4).$

Let coordinates of Q be (a, b)

By section formula,

Q(a, b) = $\Big(\dfrac{m$1x2 + m2x1}{m1 + m2}, \dfrac{m1y2 + m2y1}{m1 + m2}\Big)

Substituting values we get,

$\Rightarrow Q = \Big(\dfrac{1 \times 5 + 2 \times 2}{1 + 2}, \dfrac{1 \times 8 + 2 \times 5}{1 + 2}\Big) \\[1em] \Rightarrow Q = \Big(\dfrac{5 + 4}{3}, \dfrac{8 + 10}{3}\Big) \\[1em] \Rightarrow Q = \Big(\dfrac{9}{3}, \dfrac{18}{3}\Big) \\[1em] \Rightarrow Q = (3, 6).$

Hence, coordinates of P = (1, 4) and Q = (3, 6).

(b) By distance formula,

$BC = \sqrt{[5 - (-1)]^2 + (8 - 2)^2} \\[1em] = \sqrt{[5 + 1]^2 + 6^2} \\[1em] = \sqrt{6^2 + 6^2} \\[1em] = \sqrt{36 + 36} \\[1em] = \sqrt{72} = 6\sqrt{2} \text{ units}. \\[1em] PQ = \sqrt{(3 - 1)^2 + (6 - 4)^2} \\[1em] = \sqrt{2^2 + 2^2} \\[1em] = \sqrt{4 + 4} \\[1em] = \sqrt{8} \\[1em] = 2\sqrt{2}.$

3PQ = 3 × $2\sqrt{2} = 6\sqrt{2}$ = BC.

Hence, proved that BC = 3PQ.