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.

Remember that the perpendicular bisector is perpendicular to the line through the two points. In other words, the slope shouldn't be $-\frac29$, but rather $\frac92=4.5$. — Rushabh Mehta, Sep 09 '19 at 17:20

score 1 · Answer 1 · answered Sep 09 '19 at 17:18

1

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

answered Sep 09 '19 at 17:18

Arthur

Oh yeah, as soon as you said that I saw that I didn’t find the perpendicular gradient! That was an easy fix. – Jamminermit Sep 09 '19 at 17:20

score 0 · Answer 2 · answered Sep 09 '19 at 18:27

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:

$18 x -4y-1=0$

2 Answers2