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Question 15
The coordinates of the point which is
equivalent from the three vertices of the AOB shown in the figure, is

(A) (x,y)
(B) (y,x)
(C) (x2,y2)
(D) (y2,x2)

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Solution

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,[(h0)2+(k0)2]2
=[(h0)2+(k2y)2]2
=[(h2x)2+(k0)2]2h2+k2=h2+(k2y)2=(h2x)2+k2 ...(i)Taking first two terms of equation (i), we geth2+k2=h2+(k2y)2Required points=(h,k)=(x,y)k2=k2+4y24yk4y(yk)=0y=k [y0]Taking first and third term of equation(i), we geth2+k2=(h2x)2+k2h2=h2+4x24xh4x(xh)=0x=h [x0]
Therefore, coordinates of the point P is (x,y).

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