Question

Find a point P on the line segment joining \(\mathrm{A}(14, -5)\) and \(\mathrm{B}(-4, 4)\), which is twice as far from A as from B. 

[Hint: \(AP = 2PB \Rightarrow AP : PB = 2 : 1\).]

Answer 1

Since $$AP$$ is twice $$PB$$, we have $$AP : PB = 2 : 1$$. So $$m = 2,\ n = 1$$.

$$x = \frac{2 \times (-4) + 1 \times 14}{2 + 1}$$

$$= \frac{-8 + 14}{3}$$

$$= \frac{6}{3}$$

$$= 2$$

$$y = \frac{2 \times 4 + 1 \times (-5)}{2 + 1}$$

$$= \frac{8 - 5}{3}$$

$$= \frac{3}{3}$$

$$= 1$$

Hence the required point is $$P(2, 1)$$.