A(20, 0) and B (10, -20) are two fixed points. Find the co-ordinates of the point P in AB such that : 3PB = AB. Also, find the co-ordinates of some other point Q in AB such that AB = 6AQ.

From figure,

⇒ AB = AP + PB

⇒ 3PB = AP + PB

⇒ AP = 2PB

⇒ $\dfrac{AP}{PB} = \dfrac{2}{1}$

⇒ AP : PB = 2 : 1.

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

$\therefore x = \dfrac{m$1x2 + m2x1}{m1 + m2} \\[1em] = \dfrac{2 \times 10 + 1 \times 20}{2 + 1} \\[1em] = \dfrac{20 + 20}{3} \\[1em] = \dfrac{40}{3}.

and,

$y = \dfrac{m$1y2 + m2y1}{m1 + m2} \\[1em] = \dfrac{2 \times -20 + 1 \times 0}{2 + 1} \\[1em] = \dfrac{-40 + 0}{3} \\[1em] = \dfrac{-40}{3}.

From figure,

⇒ AB = AQ + QB

⇒ 6AQ = AQ + QB

⇒ QB = 5AQ

⇒ $\dfrac{AQ}{QB} = \dfrac{1}{5}$

⇒ AQ : QB = 1 : 5.

Let the co-ordinates of Q be (p, q)

$\therefore p = \dfrac{m$1x2 + m2x1}{m1 + m2} \\[1em] = \dfrac{1 \times 10 + 5 \times 20}{1 + 5} \\[1em] = \dfrac{10 + 100}{6} \\[1em] = \dfrac{110}{6} = \dfrac{55}{3}.

and,

$q = \dfrac{m$1y2 + m2y1}{m1 + m2} \\[1em] = \dfrac{1 \times -20 + 5 \times 0}{1 + 5} \\[1em] = \dfrac{-20 + 0}{6} \\[1em] = \dfrac{-10}{3}.

Hence, P = $\Big(\dfrac{40}{3}, -\dfrac{40}{3}\Big) \text{ and Q} = \Big(\dfrac{55}{3}, -\dfrac{10}{3}\Big)$.