Question

The slope of a line joining $P(6,k)$ and $Q(1-3k,3)$ is $\frac{1}{2}$. Find midpoint of $PQ$

Answer 1

Let $P(6,k)=(x_1,y_1)$ and $Q(1-3k,3)=(x_2,y_2)$
$\therefore$ Slope of $PQ=\frac{y_2-y_1}{x_2-x_1}$
$\Rightarrow \frac{3-k}{1-3k-6}=\frac{1}{2}$
$6-2k=-5-3k$
$11=-k$
$k=-11$

Hence, $P\equiv (6,-11)$ and $Q\equiv (1-(3\times -11),3)\equiv (34,3)$

Midpoint of $PQ\equiv (\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})\equiv (\frac{6+34}{2},\frac{-11+3}{2})\equiv (20,-4)$