Question

The line joining $A(2,1)$ and $B(5,-8)$ is trisected at the points $P$ and $Q$. If $P$ lies on the line $2x-y+K=0$, the value of $K$ is ?

• A. $-5$
• B. $-4$
• C. $-3$
• D. $-8$

Answer 1

The correct option is D $-8$

Here $\frac{AP}{PB}=\frac{1}{2}$
$P=[\frac{\left(1\right)\left(5\right)+2\left(2\right)}{1+2},\frac{\left(1\right)(-8)+2\left(1\right)}{1+2}]$
$\Rightarrow P=(\frac{9}{3},\frac{-6}{3})=(3,-2)$

$P(3,-2$ ) lies on $2x-y+K=0$
$\Rightarrow 2\left(3\right)-(-2)+K=0$
$\Rightarrow 6+2+K=0\Rightarrow K=-8$