Question

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

Answer 1

Find the coordinates of the point of trisection:

Step-1 : Coordinates of point P:

Let $A(4,-1)=\left(x_1,y_1\right)$ and $B(-2,-3)=\left(x_2,y_2\right)$

Let $P(x_p,y_p)$ and $Q(x_q,y_q)$ be the points of trisection of the line segment joining the given points, that is, $AP=PQ=QB$

Let $P$ divide $AB$ internally in the ratio $1:2=m:n$

By section formula

$P(x_p,y_p)=\left(\frac{m{x}_2+n{x}_1}{m+n},\frac{m{y}_2+n{y}_1}{m+n}\right)$

Substituting the values we get,

$P(x_p,y_p)=\left(\frac{1\times \left(-2\right)+2\times 4}{1+2},\frac{1\times \left(-3\right)+2\times \left(-1\right)}{1+2}\right)$

$P(x_p,y_p)=\left(\frac{6}{3},\frac{-5}{3}\right)$

$P(x_p,y_p)=\left(2,\frac{-5}{3}\right)$

The co-ordinates of the point $P$ are $\left(2,\frac{-5}{3}\right)$

Step-2 : Coordinates of point Q :

Now, to find point $Q$,

Point $Q$ divides $AB$ internally in the ratio of $2:1=m:n$.

$Q(x_q,y_q)=\left(\frac{m{x}_2+n{x}_1}{m+n},\frac{m{y}_2+n{y}_1}{m+n}\right)$

$Q(x_q,y_q)=\left(\frac{2\times \left(-2\right)+1\times 4}{2+1},\frac{2\times \left(-3\right)+1\times \left(-1\right)}{2+1}\right)$

$Q(x_q,y_q)=\left(\frac{0}{3},\frac{-7}{3}\right)$

$Q(x_q,y_q)=\left(0,\frac{-7}{3}\right)$

The coordinates of the point $Q$ are $\left(0,-\frac{7}{3}\right)$

Hence the coordinates of the points of trisection are $P$ $\left(2,\frac{-5}{3}\right)$ and $Q$ $\left(0,-\frac{7}{3}\right)$.