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If a point $P(x,y)$ is equidistant from $(7,1)$ and $(3,5)$, which relation must $x$ and $y$ satisfy?
A$x+y=2$
B$x-y=2$
C$x+y=6$
D$x-y=-2$
Answer & Solution
Correct answer: B. $x-y=2$
Equidistant means the squared distances are equal: $(x-7)^2+(y-1)^2=(x-3)^2+(y-5)^2$. Expanding and simplifying gives $x-y=2$.
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