Question

The line segment joining the points $A(2,1)$ and $B(5,-8)$ is trisected at the point $P$ and $Q$ such that $P$ is nearer to $A$. If $P$ also lies on line given by $2x-y+k=0$, find value of $k$.

Answer 1

Using the section formula, if a point $(x,y)$ divides the line joining the points $(x_1,y_1)$ and $(x_2,y_2)$ in the ratio $m:n$, then

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

Since, $P$ and $Q$ are points of trisection and $P$ is nearer to $A$

$⟹AP:PB=1:2$

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

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

$⟹P(-3,2)$ pasees through $2x-y+k=0$

$⟹k=8$