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(A) Let $P(x, y)$ be equidistant from the points $A(7, 1)$ and $B(3, 5)$.Given that $AP = BP$,it follows that $AP^2 = BP^2$.Using the distance formula

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$x - y = 2$

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(A) Let $P(x, y)$ be equidistant from the points $A(7, 1)$ and $B(3, 5)$.Given that $AP = BP$,it follows that $AP^2 = BP^2$.Using the distance formula,we have:$(x - 7)^2 + (y - 1)^2 = (x - 3)^2 + (y - 5)^2$Expanding both sides:$(x^2 - 14x + 49) + (y^2 - 2y + 1) = (x^2 - 6x + 9) + (y^2 - 10y + 25)$Simplifying by cancelling $x^2$ and $y^2$ from both sides:$-14x - 2y + 50 = -6x - 10y + 34$Rearranging the terms:$-14x + 6x - 2y + 10y = 34 - 50$$-8x + 8y = -16$Dividing by $-8$:$x - y = 2$Thus,the required relation is $x - y = 2$.

Find a relation between $x$ and $y$ such that the point $(x, y)$ is equidistant from the points $(7, 1)$ and $(3, 5)$.

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