Question

Show that the line segment joining the points \(\mathrm{A}(-5, 8)\) and \(\mathrm{B}(10, -4)\) is trisected by the co-ordinate axes. Also, find the points of trisection of AB. 

[Hint: Let x-axis divide AB in the ratio k : 1 at the point P(x, 0). Let y-axis divide AB in the ratio p : 1 at the point Q(0, y).]

Answer 1

x-axis cuts AB at $P(x,0)$ in the ratio $$k : 1$$:

$$\frac{k \times (-4) + 1 \times 8}{k + 1} = 0$$

$$\Rightarrow -4k + 8 = 0$$

$$\Rightarrow k = 2$$

So the x-axis divides AB in the ratio $$2 : 1$$ and

$$x = \frac{2 \times 10 + 1 \times (-5)}{2 + 1}$$

$$= \frac{20 - 5}{3}$$

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

$$= 5$$ 

Hence $$P(5,0)$$.

y-axis cuts AB at $Q(0,y)$ in the ratio $p:1$:

$$\frac{p \times 10 + 1 \times (-5)}{p + 1} = 0$$

$$\Rightarrow 10p - 5 = 0$$

$$\Rightarrow p = \frac{1}{2}$$

So the y-axis divides AB in the ratio $$\dfrac{1}{2} : 1 = 1 : 2$$ and

$$y = \frac{1 \times (-4) + 2 \times 8}{1 + 2}$$

$$= \frac{-4 + 16}{3}$$

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

$$= 4$$

Hence $$Q(0,4)$$.

The x-axis cuts AB in the ratio $$2 : 1$$ (two parts from A) while the y-axis cuts it in the ratio $$1 : 2$$ (one part from A). So the two axes cut the segment into three equal parts – AB is trisected by the co-ordinate axes.

The points of trisection are $$Q(0, 4)$$ and $$P(5, 0)$$.