Finding the equation of the perpendicular to the line through $(-4,3)$ and $(5,1)$

Question

I am struggling with the following question:

This seems like a question I should be able to do and here is my working:

I get an answer of $$2x+9y-17=0$$, but the books says the answer is $$18x-4y-1=0$$.

Could someone explain to me where I went wrong? thanks.

Answer 1

Hint: You've found the line that goes through $P$ and $Q$. That's not what you were asked to find.

Answer 2

The gradient of PQ is:

$m=\frac{3-1}{-4-5}=-\frac{2}{9}$

Gradient of line perpendicular on it is:

$m'=\frac{9}{2}$

The mid point M coordinates are:

$x_M=\frac{3+1}{2}=2$

$y_M=\frac{-4+5}{2}=\frac{1}{2}$

So equation of perpendicular bisector of PQ is:

$\frac{9}{2}(x-\frac{1}{2})=y-2$

Which after reduction is:

$18 x -4y-1=0$