distance Geometry Problems

Find the relation between x and y, if the point (x, y) is equidistant from (7, -6),(-3, 4)

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How to find the relation of x and y.

Problem: How to Find the relation between x and y, if the point (x, y)is equidistant from (7, -6),(-3, 4).

(7,-6),(-3,4)\,

Solutions:

Find the distance between (x,y) and the first point is
\sqrt{(x-7)^2+(y+6)^2}\,
\sqrt{x^2+49-14x+y^2+36+12y}\,
\sqrt{x^2+y^2-14x+12y+85}\,
Let this expression be 1.

Find the distance between (x,y) and the second point is
\sqrt{(x+3)^2+(y-4)^2}\,
\sqrt{x^2+9+6x+y^2+16-8y}\,
\sqrt{x^2+y^2+6x-8y+25}\,
Let this expression be 2.
As per the given condition,equating both the expressions,
\sqrt{x^2+y^2-14x-12y+85}=\sqrt{x^2+y^2+6x-8y+25}\,
Squaring on both sides,we get
x^2+y^2-14x+12y+22=x^2+y^2+6x-8y+25\,
cancelling the terms which are common on both sides,
-14x-6x+12y+8y=-60\,
-20x+20y=-60\,

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