Question

Find the co-ordinates of the points of trisection of the line segment joining the points $(-3,3)$ and $(3,-3)$.

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 $(x_1,y_1)$ and Q $(x_2,y_2)$ are the points of trisection of the line segment joining the given points i.e., AP = PQ = QB
Therefore, point P divides AB internally in the ratio $1:2$.
$x_1$ = $\frac{1\left(3\right)+2(-3)}{1+2}$
$y_1$ = $\frac{1(-3)+2\left(3\right)}{1+2}$
$x_1$ = $\frac{3-6}{3}=-1$
$y_1$ = $\frac{-3+6}{3}=1$
Therefore, $p(x_1,y_1)$ =$-1,1$

Point Q divides internally by $2:1$

$x_2$ = $\frac{2\left(3\right)+1(-3)}{1+2}$
$y_2$ = $\frac{2(-3)+1\left(3\right)}{1+2}$
$x_2$ = $\frac{6-3}{3}=1$
$y_2$ = $\frac{-6+3}{3}=-1$
Therefore, $p(x_2,y_2)$ =$(1,-1)$