Question

Show that the line segment joining the points $(-5,8)$ and $(10,-4)$ is trisected by the co-ordinate axes.

Answer 1

Using the section formula, if a point $(x,y)$ divides the line joining the points $(x_1,y_1)$ and $(x_2,y_2)$ in the ratio $m:n$, then

$(x,y)=(\frac{m{x}_2+n{x}_1}{m+n},\frac{m{y}_2+n{y}_1}{m+n})$

Let $P$ and $Q$ be the point of trisection of the line segment joining the points $A(-5,8)$ and $B(10,-4)$.

So, $AP=PQ=QB$. That is, $AP:PQ:QB=1:1:1$

$\Rightarrow AP:PB=1:2$

Co-ordinates of the point $P$ are $(\frac{1\cdot 10+2\cdot -5}{1+2},\frac{1\cdot -4+2\cdot 8}{1+2})=(\frac{10-10}{3},\frac{12}{3})=(0,4)$

Therefore point $P$ lie on $y-axis$

Since $AP:PQ:QB=1:1:1$

$\Rightarrow AQ:QB=2:1$

Co-ordinates of the point $Q$ are $(\frac{2\cdot 10+1\cdot -5}{2+1},\frac{2\cdot -4+1\cdot 8}{2+1})=(\frac{20-5}{3},\frac{-8+8}{3})=(5,0)$

Therefore point $Q$ lie on $x-axis$

Hence, the line segment joining the given points $A$ and $B$ is trisected by the co-ordinate axes.