Question
Find the relation between x and y such that the point $(x, y)$ is equidistant from the points $(7, 1)$ and $(3, 5).$
✓
Answer
Let $P(x, y)$ be equidistant from the points $A(7, 1)$ and $B(3, 5)$
$AP = BP$ (Given)
$ \Rightarrow AP^2 = BP^2$
$ \Rightarrow (x - 7)^2 + (y - 1)^2 = (x - 3)^2 + (y - 5)^2$
$ \Rightarrow x^2 + 49 - 14x + y^2 + 1 - 2y = x^2 + 9 - 6x + y^2 + 25 - 10y$
$ \Rightarrow 49 - 14x ^ + 1 - 2y = 9 - 6x + 25 - 10y$
$ \Rightarrow - 14x + 6x - 2y + 10 y = 34 - 50$
$ \Rightarrow -8x + 8y = -16$
$ \Rightarrow x - y = 2$
$AP = BP$ (Given)
$ \Rightarrow AP^2 = BP^2$
$ \Rightarrow (x - 7)^2 + (y - 1)^2 = (x - 3)^2 + (y - 5)^2$
$ \Rightarrow x^2 + 49 - 14x + y^2 + 1 - 2y = x^2 + 9 - 6x + y^2 + 25 - 10y$
$ \Rightarrow 49 - 14x ^ + 1 - 2y = 9 - 6x + 25 - 10y$
$ \Rightarrow - 14x + 6x - 2y + 10 y = 34 - 50$
$ \Rightarrow -8x + 8y = -16$
$ \Rightarrow x - y = 2$
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