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Math Help - Pythagorean problem

  1. #1
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    Pythagorean problem

    Hi, I'm studying geometry and I'm stuck with this problem.

    The problem says

    "Two points, A and B, are given in the plane (no figure). Describe the set of points for which {AX}^2-{BX}^2 is constant.

    I am thinking that this triangle is isoceles and AX^2 - BX^2 = 0
    but it sounds too easy of an answer maybe someone can explain this to me in terms of a geometric theory or whatever. Please explain with some reference to something..

    Thanks!
    Last edited by mr fantastic; January 15th 2010 at 04:11 PM. Reason: Fixed latex tags
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  2. #2
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    Hello, stonedcarli!

    Two points, A and B, are given in the plane (no figure).
    Describe the set of points for which {AX}^2-{BX}^2 is constant.

    I assume the following:

    . . We want the locus of a point X(x,y)

    . . The given points are: . A(a_1,a_2) and B(b_1,b_2)

    . . AX and BX are distances.


    We have: . \begin{array}{ccc}AX^2 &=& (x-a_1)^2 + (y-a_2)^2 \\ \\[-4mm]<br />
BX^2 &=& (x-b_1)^2 + (y-b_2)^2 \end{array}


    Then: . \bigg[(x-a_1)^2 + (y-a^2)^2\bigg] - \bigg[(x-b_1)^2 + (y - b_2)^2\bigg] \;=\;k

    . . \bigg[x^2 - 2a_1x + a_1^2 + y^2 - 2a_2y + a_2^2\bigg] - \bigg[x^2 - 2b_1x + b_1^2 + y^2 - 2b_2y + b_2^2\bigg] \;=\;k

    . . x^2 - 2a_1x + a_1^2 + y^2 - 2a_2y + a_2^2 - x^2 + 2b_1x - b_1^2 - y^2 + 2b_2y - b_2^2 \;=\;k

    . . -2a_1x + 2b_1x - 2a_2y - 2b_2y +a_1^2 - b_1^2 - a_2^2 + b_2^2 \;=\;k

    . . 2(b_1-a_1)x + 2(b_2-a_2)y \;=\;k -(a_1^2 - a_2^2) + (b_1^2 - b_2^2)


    The equation is of the form: . Ax + By \:=\:C\quad\hdots\;\text{ a straight line}

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