Question

Point $P$ divides the line segment joining the points $A(2,1)$ and $B(5,-8)$ such that $\frac{AP}{AB}=\frac{1}{3}$. If $P$ lies on the line $2x-y+k=0$. Find the value of $k$.

Answer 1

We know that by section formula, the co-ordinates of the points which divide internally the line segment joining the points $(x_1,y_1)$ and $(x_2,y_2)$ in the ratio $m:n$ is

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

Since, $\frac{AP}{AB}=\frac{1}{3}\Rightarrow \frac{AB}{AP}=3\Rightarrow \frac{AP+PB}{AP}=3\Rightarrow \frac{PB}{AP}+1=3$

$\Rightarrow \frac{AP}{PB}=\frac{1}{2}$

By section formula, coordinates of $P$.

$\left(\frac{5\times 1+2\times 2}{2+1}\right),\left(\frac{-8\times 1+2\times 1}{2+1}\right)$

$=(\frac{9}{3},\frac{-6}{3})=(3,-2)$

Putting these in $2x-y+R=0$, we get $6+2+R=0$

$\Rightarrow R=-8$