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Find a relation between x and y such that the point (x, y) is equidistant from the point (3, 6) and (–3, 4).

It is given that (x, y) is equidistant from (3, 6) and (–3, 4).
Using Distance formula, we can write

$$\Rightarrow \sqrt{(x-3)^2 + (y-6)^2} = \sqrt{(x-(-3))^2 + (y-4)^2}$$
$$\Rightarrow \sqrt{x^2 + 9 - 6x + y^2 + 36 -12y } = \sqrt{x^2 + 9 + 6x + y^2 + 16 -8y }$$
Squaring both sides, we get
$$\Rightarrow x^2 + 9 - 6x + y^2 + 36 -12y = x^2 + 9 + 6x + y^2 + 16 -8y$$
$$\Rightarrow -6x - 12y + 45 = 6x - 8y + 25$$
$$\Rightarrow 12x + 4y = 20$$
$$\Rightarrow 3x + y = 5$$