If the point (x, y) is equidistant from the p | vedclass

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(D) Let the given points be $P(a + b, b - a)$ and $Q(a - b, a + b)$. The point $(x, y)$ is equidistant from $P$ and $Q$,so $dist(P, (x, y))^2 = dist(Q

Vedclass · Accepted Answer

$bx - ay = 0$

Vedclass · Answer

(D) Let the given points be $P(a + b, b - a)$ and $Q(a - b, a + b)$. The point $(x, y)$ is equidistant from $P$ and $Q$,so $dist(P, (x, y))^2 = dist(Q, (x, y))^2$.$(x - (a + b))^2 + (y - (b - a))^2 = (x - (a - b))^2 + (y - (a + b))^2$Expanding both sides:$x^2 - 2x(a + b) + (a + b)^2 + y^2 - 2y(b - a) + (b - a)^2 = x^2 - 2x(a - b) + (a - b)^2 + y^2 - 2y(a + b) + (a + b)^2$Canceling $x^2, y^2, (a + b)^2$ from both sides:$-2ax - 2bx + (b - a)^2 - 2by + 2ay = -2ax + 2bx + (a - b)^2 - 2ay - 2by$Since $(b - a)^2 = (a - b)^2$,these terms also cancel:$-2bx + 2ay = 2bx - 2ay$$4ay = 4bx$$bx - ay = 0$Thus,the correct option is $D$.

If the point $(x, y)$ is equidistant from the points $(a + b, b - a)$ and $(a - b, a + b)$,then:

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