# Find the coordinates of the point of intersection of perpendicular bisectors

• May 30th 2012, 10:55 AM
Ragnarok
Find the coordinates of the point of intersection of perpendicular bisectors
Hello, in a pre-calculus book I am given this problem:

Find the coordinates of the point of intersection of the perpendicular bisectors of the sides of a triangle whose vertices are located at (-a, 0), (b, c), and (a, 0).

The only apparent tools with which we are supposed to find the answer are the Pythagorean theorem, the distance formula, the midpoint formula, and the equation of a line.

I have tried this a couple of times but end up drowned in variables. Could anyone give me a hand?
• May 30th 2012, 12:12 PM
Soroban
Re: Find the coordinates of the point of intersection of perpendicular bisectors
Hello, Ragnarok!

Did you make a sketch?

Quote:

Find the coordinates of the point of intersection of the perpendicular bisectors of the sides of a triangle
whose vertices are located at (-a, 0), (b, c), and (a, 0).

Code:

                |                 |    R                 |    * (b,c)                 | *  *               *|      *             *  |      *     P  *      |        * Q   - - * - - - - + - - - - * - -   (-a,0)      |      (a,0)                 |
The perpendicular bisector of side $PQ$ is the y-axis, $x = 0.$ .[1]

The slope of side $QR$ is $\tfrac{c}{b-a}$
. . The perpendicular slope is $\tfrac{a-b}{c}$
. . . . The midpoint of side $QR$ is $\left(\tfrac{a+b}{2},\:\tfrac{c}{2}\right)$

The equation of the perpendicular bisector of side $QR$ is:
. . $y - \tfrac{c}{2} \:=\:\tfrac{a-b}{c}\left(x - \tfrac{a+b}{2}\right) \quad\Rightarrow\quad y \:=\:\tfrac{a-b}{c}x + \tfrac{b^2+c^2-a^2}{2c}$ .[2]

The intersection of [1] and [2] is: . $\left(0,\:\frac{b^2+c^2-a^2}{2c}\right)$