A(-8, 0), B(0, 16) and C(0, 0) are the vertices of a triangle ABC. Point P lies on AB and Q lies on AC such that AP : PB = 3 : 5 and AQ : QC = 3 : 5.

Show that : PQ = $\dfrac{3}{8}$BC.

The triangle is shown in the figure below:

Let the co-ordinates of P be (x, y)

$\therefore x = \dfrac{m$1x2 + m2x1}{m1 + m2} \\[1em] = \dfrac{3 \times 0 + 5 \times -8}{3 + 5} \\[1em] = \dfrac{0 - 40}{8} \\[1em] = \dfrac{-40}{8} = -5.

and,

$y = \dfrac{m$1y2 + m2y1}{m1 + m2} \\[1em] = \dfrac{3 \times 16 + 5 \times 0}{3 + 5} \\[1em] = \dfrac{48 + 0}{8} \\[1em] = 6.

P = (x, y) = (-5, 6).

Let the co-ordinates of Q be (m, n)

$\therefore m = \dfrac{m$1x2 + m2x1}{m1 + m2} \\[1em] = \dfrac{3 \times 0 + 5 \times -8}{3 + 5} \\[1em] = \dfrac{0 - 40}{8} \\[1em] = \dfrac{-40}{8} = -5.

and,

$n = \dfrac{m$1y2 + m2y1}{m1 + m2} \\[1em] = \dfrac{3 \times 0 + 5 \times 0}{3 + 5} \\[1em] = \dfrac{0 + 0}{8} \\[1em] = 0.

Q = (m, n) = (-5, 0).

Distance between two points = $\sqrt{(x$2 - x1)^2 + (y2 - y1)^2}

$PQ = \sqrt{(-5 - (-5))^2 + (0 - 6)^2} \\[1em] = \sqrt{0^2 + (-6)^2} \\[1em] = \sqrt{36} \\[1em] = 6 \text{ units}. \\[1em] BC = \sqrt{(0 - 0)^2 + (0 - 16)^2} \\[1em] = \sqrt{0^2 + (-16)^2} \\[1em] = \sqrt{256} \\[1em] = 16 \text{ units}.$

⇒ BC = 16 units

⇒ $\dfrac{3}{8}BC = 16 \times \dfrac{3}{8}$ = 6 units = PQ.

Hence, proved that PQ = $\dfrac{3}{8}$BC.