The coordinates of the point which is equidis | vedclass

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(B) Let the coordinates of the point $P(h, k)$ be equidistant from the three vertices $O(0, 0)$,$A(0, 2y)$,and $B(2x, 0)$.Then,$PO = PA = PB$.$\Righta

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$(x, y)$

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(B) Let the coordinates of the point $P(h, k)$ be equidistant from the three vertices $O(0, 0)$,$A(0, 2y)$,and $B(2x, 0)$.Then,$PO = PA = PB$.$\Rightarrow (PO)^2 = (PA)^2 = (PB)^2$ ..........$(i)$By the distance formula,we have:$[\sqrt{(h-0)^2 + (k-0)^2}]^2 = [\sqrt{(h-0)^2 + (k-2y)^2}]^2 = [\sqrt{(h-2x)^2 + (k-0)^2}]^2$$\Rightarrow h^2 + k^2 = h^2 + (k-2y)^2 = (h-2x)^2 + k^2$Taking the first two parts:$h^2 + k^2 = h^2 + (k-2y)^2$$k^2 = k^2 + 4y^2 - 4yk$$4y(y - k) = 0 \Rightarrow k = y$ (since $y 
eq 0$).Taking the first and third parts:$h^2 + k^2 = (h-2x)^2 + k^2$$h^2 = h^2 + 4x^2 - 4xh$$4x(x - h) = 0 \Rightarrow h = x$ (since $x 
eq 0$).Therefore,the required point is $(h, k) = (x, y)$.

The coordinates of the point which is equidistant from the three vertices of the $\triangle AOB$ as shown in the figure are

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