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 A P^2=B P^2 $
$ \Rightarrow(x-7)^2+(y-1)^2=(x-3)^2+(y-5)^2 $
$ \Rightarrow x^2+49-14 x+y^2+1-2 y=x^2+9-6 x+y^2+25-10 y $
$ \Rightarrow 49-14 x+1-2 y=9-6 x+25-10 y $
$ \Rightarrow-14 x+6 x-2 y+10 y=34-50 $
$ \Rightarrow-8 x+8 y=-16 $
$ \Rightarrow x-y=2$
$AP = BP$ (Given)
$ \Rightarrow A P^2=B P^2 $
$ \Rightarrow(x-7)^2+(y-1)^2=(x-3)^2+(y-5)^2 $
$ \Rightarrow x^2+49-14 x+y^2+1-2 y=x^2+9-6 x+y^2+25-10 y $
$ \Rightarrow 49-14 x+1-2 y=9-6 x+25-10 y $
$ \Rightarrow-14 x+6 x-2 y+10 y=34-50 $
$ \Rightarrow-8 x+8 y=-16 $
$ \Rightarrow x-y=2$
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