Find the equation of the line passing through the point (−3, 5) and perpendicular to the line joining (2, 5) and (−3, 6).

The given points are $A \left(2, 5\right) \mathrm{and} B \left(-3, 6\right)$.

$\therefore$ Slope of AB $=\frac{6-5}{-3-2}=-\frac{1}{5}$

Let m be the slope of the required line. Then,

$$\begin{gathered} m\times \mathrm{Slope} \mathrm{of} AB=-1 \\ \Rightarrow m\times \frac{-1}{5}=-1 \\ \Rightarrow m=5 \end{gathered}$$


So, the equation of the line that passes through (−3, 5) and has slope 5 is

$$\begin{gathered} y-5=5\left(x+3\right) \\ \Rightarrow 5x-y+20=0 \end{gathered}$$

Hence, the equation of the required line is $5x-y+20=0$