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Math Help - Find the coordinates of the point of intersection of perpendicular bisectors

  1. #1
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    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?
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  2. #2
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    Re: Find the coordinates of the point of intersection of perpendicular bisectors

    Hello, Ragnarok!

    Did you make a sketch?


    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)
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