Question

Find the co-ordinates of the points of trisection of the line segment joining the points $A(6,-2)$ and $B(-8,10)$.

Answer 1

Let $P$ and $Q$ be the points of trisection so that $AP=PQ=QB$.
For P:
$m_1:m_2=AP:PB=1:2;(x_1,y_1)=(6,-2)$ and $(x_2,y_2)=(-8,10)$
$\therefore x=\frac{m_1{x}_2+m_2{x}_1}{m_1+m_2}=\frac{1\times -8+2\times 6}{1+2}=\frac{4}{3}$
$\therefore y=\frac{m_1{y}_2+m_2{y}_1}{m_1+m_2}=\frac{1\times 10+2\times -2}{1+2}=2$
$\therefore$ Point $P=(\frac{4}{3},2)$
For $Q$ :
$m_1:m_2=AQ:QB=2:1;(x_1,y_1)=(6,-2)$ and $(x_2,y_2)=(-8,10)\therefore Q=(\frac{2\times -8+1\times 6}{2+1},\frac{2\times 10+1\times -2}{2+1})=(-\frac{10}{3},6)$