Ans: (a) $P’=(5,3)$, $Q’=(-19,-7)$
- $P'(5,3)$, $Q'(-19,-7)$.
- The slope of $PQ$
$\begin{array}{rcl}
m_{PQ} & = & \dfrac{5-(-7)}{(-3) – 2} \\
& = & \dfrac{-12}{5}
\end{array}$The slope of $P’Q’$
$\begin{array}{rcl}
m_{P’Q’} & = & \dfrac{3-(-7)}{5-(-19)} \\
& = & \dfrac{5}{12}
\end{array}$Since $m_{PQ} \times m_{P’Q’} = -1$, then $PQ$ is perpendicular to $P’Q’$.
