Let the coordinate of the point which is equidistant from the three vertices O(0,0), A(0,2y) and B(2x,0) is P(h,k)
Then, PO = PA = PB
⇒(PO)2=(PA)2=(PB)2By distance formula,[√(h−0)2+(k−0)2]2
=[√(h−0)2+(k−2y)2]2
=[√(h−2x)2+(k−0)2]2⇒h2+k2=h2+(k−2y)2=(h−2x)2+k2 ...(i)Taking first two terms of equation (i), we geth2+k2=h2+(k−2y)2∴Required points=(h,k)=(x,y)⇒k2=k2+4y2−4yk⇒4y(y−k)=0⇒y=k [∵y≠0]Taking first and third term of equation(i), we geth2+k2=(h−2x)2+k2⇒h2=h2+4x2−4xh⇒4x(x−h)=0⇒x=h [∵x≠0]
Therefore, coordinates of the point P is (x,y).