The coordinates of the point which is equidistant from the three vertices of the ΔAOB as shown in the figure is


Let the coordinate of the point P is (h, k).

Given,


Point P is equidistant from the three vertices O, A and B


Where;


O (0, 0),


A (0, 2y) and


B (2x, 0)


Then,


PO = PA = PB


(PO)2 = (PA)2 = (PB)2 ….(i)


By using distance formula,



=


=


= h2 + k2 = h2 + (k – 2y)2 …..(ii)


= (h – 2x)2 + k2 ….. (iii)


Taking first two equations, we get


h2 + k2 = h2 + (k – 2y)2


k2 = k2 + 4y2 – 4yk 4y(y k) = 0


y = k [ y ≠ 0]


Taking first and third equations, we get


h2 + k2 = (h – 2x)2 + k2


h2 + h2 + 4x2 – 4xh


4x (x – h) = 0


x = h [ x ≠ 0]


Required points = (h, k) = (x, y)

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