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# The coordinates of the point which is equidistant from the three vertices of the $ \triangle \mathrm{AOB} $ as shown in the figure is

**(A)** $ (x

Given:

The three vertices of the \( \triangle \mathrm{AOB} \) are $0(0, 0), A(0,2y)$ and $B(2x, 0)$.

To do:

We have to find the coordinates of the point which is equidistant from the three vertices of the \( \triangle \mathrm{AOB} \).

Solution:

Let the coordinates of the point which is equidistant from the three vertices \( 0(0,0) \), \( A(0,2 y) \) and \( B(2 x, 0) \) be \( P(h, k) \).

This implies,

$P O=P A=P B$

On squaring, we get,

$(P O)^{2}=(P A)^{2}=(P B)^{2}$

Using distance formula

$[\sqrt{(h-0)^{2}+(k-0)^{2}}]^{2} }=[\sqrt{(h-0)^{2}+(k-2 y)^{2}}]^{2}=[\sqrt{(h-2 x)^{2}+(k-0)^{2}}]^{2}$

$h^{2}+k^{2}=h^{2}+(k-2 y)^{2} =(h-2 x)^{2}+k^{2}$

Therefore,

$h^{2}+k^{2}=h^{2}+(k-2 y)^{2}$

$k^{2} =k^{2}+4 y^{2}-4 k y$

$4 y(y-k)=0$

$y=k$

$h^{2}+k^{2} =(h-2 x)^{2}+k^{2}$

$h^{2} =h^{2}+4 x^{2}-4 x h$

$4 x(x-h) =0$

$x =h $

The required point is $(x, y)$.

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