# Thread: Perpendicular bisectors

1. ## Perpendicular bisectors

I can't seem to find a formula to find the perpendicular bisector of a line. Anyone know what it is?

2. ## Re: Perpendicular bisectors

Originally Posted by alexpasty2013
I can't seem to find a formula to find the perpendicular bisector of a line. Anyone know what it is?

Suppose each of $P: (x_1,y_1)~\&~Q: (x_2,y_2)$ is a point and $y_1\ne y_2$.

The perpendicular bisector of $\overline{PQ}$ is $y = \left( { - \frac{{{x_1} - {x_2}}}{{{y_1} - {y_2}}}} \right)\left( {x - \frac{{{x_1} + {x_2}}}{2}} \right) + \left( {\frac{{{y_1} + {y_2}}}{2}} \right)$

3. ## Re: Perpendicular bisectors

So we have 2 points A (-2, 2) and B (3 , 1) the the perpendicular bisector is;

AB is y = (-(-2)-3/-2+1) (x- (-2)+3/2) + (-2+1/2) =

y = -5/3(x-1/2)+ (-1/2)

and our final equation is

y= -5x-1

is that right?

4. ## Re: Perpendicular bisectors

Originally Posted by alexpasty2013
So we have 2 points A (-2, 2) and B (3 , 1) the the perpendicular bisector is;

AB is y = (-(-2)-3/-2+1) (x- (-2)+3/2) + (-2+1/2) =

y = -5/3(x-1/2)+ (-1/2)

and our final equation is

y= -5x-1

is that right?
The slope of the line through A and B is \displaystyle \begin{align*} \frac{1 - 2}{3 - (-2)} = -\frac{1}{5} \end{align*}. So the slope of your perpendicular bisector is 5, since the two slopes will multiply to give -1.

The midpoint of AB is \displaystyle \begin{align*} \left( \frac{-2 + 3}{2} , \frac{ 2 + 1}{2} \right) = \left( \frac{ 1}{2} , \frac{3}{2} \right) \end{align*}

So the perpendicular bisector is of the form \displaystyle \begin{align*} y = mx + c \end{align*} with \displaystyle \begin{align*} m = 5 \end{align*} and a point it passes through is \displaystyle \begin{align*} (x, y) = \left( \frac{1}{2} , \frac{3}{2} \right) \end{align*}. Substituting these gives

\displaystyle \begin{align*} \frac{3}{2} &= 5 \left( \frac{1}{2} \right) + c \\ \frac{3}{2} &= \frac{5}{2} + c \\ c &= -1 \end{align*}

and so the equation of your perpendicular bisector is \displaystyle \begin{align*} y = 5x - 1 \end{align*}.

5. ## Re: Perpendicular bisectors

So I got the right answer by the wrong means?

6. ## Re: Perpendicular bisectors

this is wrong - the point (1/2, 3, 2) does not lie on the line, my original point (0.5, -0.5) is on the line and apparently in the centre

7. ## Re: Perpendicular bisectors

You cannot find the bisector of a line for you cannot measure a line.