The line joining the points $(2, 1)$ and $(5, -8)$ is trisected at the points P and Q. If point P lies on the line $2x – y + k = 0$. Find the value of $k$.

Given:

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

To do:

We have to find the value of $k$.

Solution:

The line segment $AB$ is trisected by the points $P$ and $Q$.

This implies, $AP: PB = 1:2$

Using section formula, we have,

$( x,\ y)=( \frac{nx_{1}+mx_{2}}{m+n}, \frac{ny_{1}+my_{2}}{m+n})$

Then, coordinates of $P$ are,

$P=( \frac{1\times5+2\times2}{1+2}, \frac{1\times(-8)+1\times2}{1+2})$

$\Rightarrow P=( \frac{5+4}{3}, \frac{-8+2}{3})$

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

$\Rightarrow P=( 3, -2)$

The point $P( 3,\ -2)$ lies on the line $2x-y+k=0$. 

This implies, point $P( 3,\ -2)$ satisfies the above equation.

$\Rightarrow 2(3)-(-2)+k=0$

$\Rightarrow 6+2+k=0$

$\Rightarrow 8+k=0$

$\Rightarrow k=-8$

Therefore, the value of $k$ is $-8$.