Question

Find the co-ordinates of the points of trisection (i.e point dividing the three equal parts) of the segment joining the points $A(2,-2)$ and $B(-7,4)$.

Answer 1

Given:- A line segment joining the points $A(2,-2)$ and $B(-7,4)$.

Let $P$ and $Q$ be the points on $AB$ such that,

$AP=PQ=QB$

Therefore,

$P$ and $Q$ divides $AB$ internally in the ratio $1:2$ and $2:1$ respectively.

As we know that if a point $(h,k)$ divides a line joining the point $(x_1,y_1)$ and $(x_2,y_2)$ in the ration $m:n$, then coordinates of the point is given as-

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

Therefore,

Coordinates of $P=(\frac{1\times (-7)+2\times 2}{1+2},\frac{1\times 4+2\times (-2)}{1+2})=(-1,0)$

Coordinates of $Q=(\frac{2\times (-7)+1\times 2}{1+2},\frac{2\times 4+1\times (-2)}{1+2})=(-4,2)$

Therefore, the coordinates of the points of trisection of the line segment joining $A$ and $B$ are $(-1,0)$ and $(-4,2)$.