Ans: (a) $A’=(-4-3)$, $B’=(9,9)$
- $A’ = (-4,-3)$, $B’ = (9,9)$.
- Consider the slope of $AB$, we have
$\begin{array}{rcl}
m_{AB} & = & \dfrac{-9 -4}{9 – (-3)} \\
& = & \dfrac{-13}{12}
\end{array}$Consider the slope of $A’B’$, we have
$\begin{array}{rcl}
m_{A’B’} & = & \dfrac{9 -(-3)}{9-(-4)} \\
& = & \dfrac{12}{13}
\end{array}$Hence, we have
$\begin{array}{rcl}
m_{AB} \times m_{A’B’} & = & \dfrac{-13}{12} \times \dfrac{12}{13} \\
& = & -1
\end{array}$Therefore, $AB$ is perpendicular to $A’B’$.
