Question

Find the coordinates of points which trisect the line segment joining $(1,-2)and(-3,4).$

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 A $(1,-2)$ and $B(-3,4)$ be the given points.

Let the points of trisection be $P(a,b)$ and $Q(c,d)$.

Then, $AP=PQ=QB=\lambda \left(say\right)$

.$\therefore PB=PQ+QB=2\lambda$

$AQ=AP+PQ=2\lambda$

$\Rightarrow AP:PB=\lambda :2\lambda$

$=1:2$

$andAQ:QB=2\lambda :\lambda$

$b=2:1$

So, $P$ divides $AB$ internally in the ratio $1:2$ while $Q$ divides internally in the ratio $2:1$ . Thus, the coordinates of $P(a,b)$ and $Q(c,d)$ are

$P(a,b)=(\frac{1\times -3+2\times 1}{1+2},\frac{1\times 4+2\times -2}{1+2})$

$\Rightarrow P(a,b)=(\frac{-1}{3},0)$

And,

$Q(c,d)=(\frac{2\times -3+1\times 1}{2+1},\frac{2\times 4+1\times (-2)}{2+1})$

$\Rightarrow Q(c,d)=(\frac{-5}{3},2)$

Hence, the two points of trisection are $(\frac{-1}{3},0)$ and $(\frac{-5}{3},2)$

REMARK: As $Q$ is the midpoint of $BP$. So, the coordinates of $Q$ can also be obtained by using the midpoint formula.