Question

Find the coordinates of the points which divide the line segment joining $A(2,-3)$ and $B(-4,-6)$into three equal parts.

Answer 1

Let P & Q be the points of trisection. Then AP$:$PB$=1:2$ & AQ$:$QB$=2:1$

(i) Let P divides AB in ratio $1:2$

Hence $m_1=1;m_2=2;x_1=2;y_1=-3;x_2=-4,y_2=-6$

$P(x,y)=P(\frac{m_1{x}_2+m_2{x}_1}{m_1+m_2},\frac{m_1{y}_2+m_2{y}_1}{m_1+m_2})=P(\frac{1(-4)+2\left(2\right)}{1+2},\frac{1(-6)+2(-3)}{1+2})$

$=P(\frac{-4+4}{3},\frac{-6-6}{3})=P(0,0)$

(ii) Let Q divides AB in ratio $2:1$

Here $m_1=2,m_2=1,x_1=2,y_1=-3,x_2=-4,y_2=-6$

$=Q(\frac{2(-4)+1\left(2\right)}{2+1},\frac{2(-6)+1(-3)}{2+1})$

$=Q(\frac{-8+2}{3},\frac{-12-3}{3})$

$=Q(-2,-5)$

$\therefore$ Hence the coordinates of points of trisection are $(0,0)$ & $(-2,-5)$.