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

Options

A. `(x,y)`
B. `(y,x)`
C. `((x)/(2),(y)/(2))`
D. `((y)/(2),(x)/(2))`

Answer

A

Explanation

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
`rArr (PO)^(2)=(PA)^(2)=(PB)^(2)` ...(i)
By distance formula,
`[sqrt((h-0^(2))+(k-0)^(2))]^(2)=[sqrt((h-0^(2))+(k-2y)^(2))]^(2)=[sqrt((h-2x)^(2)+(k-0)^(2))]^(2)`
`rArr h^(2)+k^(2)=h^(2)+(k-2y)^(2)=(h-2x)^(2)+k^(2)` ...(ii)
Taking first two equations , we get
`h^(2)+k^(2)=h^(2)+(k-2y)^(2)`
`rArr k^(2)=k^(2)+4y^(2)-4ykrArr4y(y-k)=0`
`[:.yne0]`
`rArr y=k`
Taking first and equations , we get
`h^(2)+k^(2)=(h-2x)^(2)+k^(2)`
`rArr h^(2)=h^(2)+4x^(2)-4xh`
`rArr 4x(x-h)=0`
`rArr x=h " "[:.xne0]`
`:.` Required points `=(h,k)=(x,y)`