Thread:

1. 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!

2. 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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