Find the coordinates of the point of intersection of perpendicular bisectors
Hello, in a precalculus 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?
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 $\displaystyle PQ$ is the yaxis, $\displaystyle x = 0.$ .[1]
The slope of side $\displaystyle QR$ is $\displaystyle \tfrac{c}{ba}$
. . The perpendicular slope is $\displaystyle \tfrac{ab}{c}$
. . . . The midpoint of side $\displaystyle QR$ is $\displaystyle \left(\tfrac{a+b}{2},\:\tfrac{c}{2}\right)$
The equation of the perpendicular bisector of side $\displaystyle QR$ is:
. . $\displaystyle y  \tfrac{c}{2} \:=\:\tfrac{ab}{c}\left(x  \tfrac{a+b}{2}\right) \quad\Rightarrow\quad y \:=\:\tfrac{ab}{c}x + \tfrac{b^2+c^2a^2}{2c}$ .[2]
The intersection of [1] and [2] is: .$\displaystyle \left(0,\:\frac{b^2+c^2a^2}{2c}\right)$