Question

Find the coordinates of the points of trisection of the line segment joining the points $A(2,-2)$ and $B(-7,4)$.

Answer 1

Let the given points be $A(2,-2)\&B(-7,4)$

$P$ & $Q$ are two points on $AB$ such that

$AP=PQ=QB$

Let $k=AP=PQ=QB$

Hence comparing $AP$ & $PB$

$AP=k$

$PB=PQ+QB$

$=k+k=2k$

Hence, ratio of $AP$ & $PB$ $=\frac{m}{2m}$

$=\frac{1}{2}$

Thus $P$ divides $AB$ in the ratio $1:2$

Now, we have to find $P$

Let $P$ be $(x,y)$

Hence,

$m_1=1,m_2=2$

And for AB

$x_1=2,x_2=-2$

$y_1=-7,y_2=4$

$x=\frac{m_1{x}_2+m_2{x}_1}{m_1+m_2}$

$=\frac{1\times (-7)+2\times 2}{1+2}$

$=\frac{-7+4}{3}$

$=-1$

$y=\frac{m_1{x}_2+m_2{x}_1}{m_1+m_2}$

$=\frac{1\times 4+2\times (-2)}{1+2}$

$=\frac{4-4}{3}$

$=0$

Hence, point $P$ is $P(-1,0)$

Similarly, Point $A$ divides $AB$ in the ratio $AQ$ & $QB$

$=\frac{AQ}{QB}$

$=\frac{AP+PQ}{QB}$

$=\frac{k+k}{k}$

$=\frac{2}{1}$

$=2:1$

Now, we have to find $Q$

Let $Q$ be $(x,y)$

Hence,

$m_1=2,m_2=1$

$x_1=2,x_2=-2$

$y_1=-7,y_2=4$

$x=\frac{m_1{x}_2+m_2{x}_1}{m_1+m_2}$

$=\frac{2\times (-7)+1\times 2}{1+2}$

$=\frac{-14+2}{3}$

$=-4$

$y=\frac{m_1{x}_2+m_2{x}_1}{m_1+m_2}$

$=\frac{2\times 4+1\times (-2)}{1+2}$

$=\frac{8-2}{3}$

$=2$

Hence, point $Q$ is $Q(-4,2)$