Find the equation of the line passing through (–3, 5)

The slope of the line joining the points $(2,5)$ and $(-3,6)$ is $m=\frac{6-5}{-3-2}=\frac{1}{-5}$

We know that two non-vertical lines are perpendicular to each other if and only if their slopes are negative reciprocals of each other.

Therefore, slope of the line perpendicular to the line through the points $(2,5)$ and $(-3,6)=-\frac{1}{m}=-\frac{1}{\left(\frac{-1}{5}\right)}=5$

Now, the equation of the line passing through point (–3, 5), whose slope is 5, is

$(y-5)=5(x+3)$

$y-5=5 x+15$

i.e., $5 x-y+20=0$