# Finding the coordinates of a perpendicular bisector

Okay, so in the question, I think I understood part $a$ and $b$ (although I'm not sure) and for part $c$ I'm getting a bit confused.

Question:

The line 1(1) passes through the points $P (−1, 5)$ and $Q(11, 11)$.

(a) Find an equation for $11$ in the form $y = mx + b$ where $m$ and $c$ are constants.

The line L(2) passes through the point $R(9, 0)$ and is perpendicular to $11$ . The lines 1(1) and 1(2) intersect at the point $S$.

(b) Calculate the coordinates of S.

(c) Show that the length of $RS$ is $\sqrt{80}$.

Okay, so for part a I got: $2y=x+5$ part b: the gradient will be $-2$ then $y-0=-2(x-9)$, giving me $y=−2x+18$, giving me the coordinates $(9, 18)$ part c: for this part, would I just get the point $R$ and point $S$ and use the equation of the distance of a line?

I think I know the theory but for some reason I can't get it right. Can you explain please? It would be greatly appreciated. Thanks!

• Perpendicular to 11? Commented Feb 27, 2014 at 18:34
• the 1(1) is a way of typing L1 or L2 Commented Feb 27, 2014 at 18:38
• it says 11 instead of your intended 1(1)? I have worked the first three, where are you stuck? Commented Feb 27, 2014 at 18:54

@Debby

$y=\frac{x+11}{2}$

for the equation that passes through the points. Can you get m and b?

Use the slope from the last question and get the negative reciprocal of it. The equation that passes through (9,0) and is perpendicular to the first one is:

$y=-2 x+18$

The point of intersection of those two equations is done by solving them simultaneously you will get (5,8). Please work through that on your own.

For c)

You use the distance formula, you will get:

$4 \sqrt{5} = \sqrt{80}$