Question

Find the co-ordinates of the points of trisection of the line segment joining the points \(\mathrm{A}(5, -3)\) and \(\mathrm{B}(2, -9)\).

Answer 1

Let $$P$$ and $$Q$$ be the two points of trisection, so $$AP = PQ = QB = r$$.

Then $$\dfrac{AP}{PB} = \dfrac{r}{2r} = \dfrac{1}{2}$$, so P divides AB in the ratio $$1 : 2$$:

$$x = \frac{1 \times 2 + 2 \times 5}{1 + 2}$$

$$= \frac{12}{3}$$

$$= 4$$

$$y = \frac{1 \times (-9) + 2 \times (-3)}{1 + 2}$$

$$= \frac{-15}{3}$$

$$= -5$$

So $$P(4, -5)$$.

Also $$\dfrac{AQ}{QB} = \dfrac{2r}{r} = 2 : 1$$, so Q divides AB in the ratio $$2 : 1$$:

$$x = \frac{2 \times 2 + 1 \times 5}{2 + 1}$$

$$= \frac{9}{3}$$

$$= 3$$

$$y = \frac{2 \times (-9) + 1 \times (-3)}{2 + 1}$$

$$= \frac{-21}{3}$$

$$= -7$$

So $$Q(3, -7)$$

Hence the points of trisection are $$P(4, -5)$$ and $$Q(3, -7)$$.