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(A) Let the coordinates of point $P$ be $(x, y)$.Given that $PA = PB$,we have $(PA)^2 = (PB)^2$.Using the distance formula: $(x - (a + b))^2 + (y - (a

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$x - y = 0$

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(A) Let the coordinates of point $P$ be $(x, y)$.Given that $PA = PB$,we have $(PA)^2 = (PB)^2$.Using the distance formula: $(x - (a + b))^2 + (y - (a - b))^2 = (x - (a - b))^2 + (y - (a + b))^2$.Expanding both sides:$(x^2 - 2x(a + b) + (a + b)^2) + (y^2 - 2y(a - b) + (a - b)^2) = (x^2 - 2x(a - b) + (a - b)^2) + (y^2 - 2y(a + b) + (a + b)^2)$.Canceling $x^2, y^2, (a + b)^2,$ and $(a - b)^2$ from both sides:$-2x(a + b) - 2y(a - b) = -2x(a - b) - 2y(a + b)$.Dividing by $-2$:$x(a + b) + y(a - b) = x(a - b) + y(a + b)$.$ax + bx + ay - by = ax - bx + ay + by$.$bx - by = -bx + by$.$2bx - 2by = 0$.$x - y = 0$.

If the distance of any point $P(x, y)$ from the points $A(a + b, a - b)$ and $B(a - b, a + b)$ are equal,then the locus of $P$ is:

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