Let the coordinate of the point is A(3,0).

equidistant from the three vertices

0(0, 0), A(0, 2y) and B(2x, 0) is P(h, k)

Then, PO = PA = PB

$\Rightarrow$  PO² = PA² =  PB²            (i)

By distance formula,

${\left[\sqrt{{(h–0)}^2+{(k–0)}^2}\right]}^2$

  = ${\left[\sqrt{{(h –0)}^2+{(k–2y)}^2}\right]}^2$

   =${\left[\sqrt{{{(h–2x)}^2+k–0)}^2}\right]}^2$

$\Rightarrow$ h² + k² = h² + (k – 2y)²

      = (h – 2x)² + k²                       (ii)

Taking first two equations, we get

 = h² + k² = h² + (k – 2y)²

k² = k² + 4y² – 4yk $\Rightarrow$ 4y(y –k) = 0

    y = k

Taking First and third equations, we get

h² + k² = (h – 2x)² + k²

$\Rightarrow$  h² = h² + 4x² –4xh

$\Rightarrow$ 4x(x – h) = 0

    x = h

$\therefore$ Required points = (h, k) = (x, y)