The coordinates of the point which is equidistant from the three vertices of the ΔAOB as shown in the figure is

Let the coordinate of the point P is (h, k).

Given,

Point P is equidistant from the three vertices O, A and B

Where;

O (0, 0),

A (0, 2y) and

B (2x, 0)

Then,

PO = PA = PB

→ (PO)² = (PA)² = (PB)² ….(i)

By using distance formula,

=

=

= h² + k² = h² + (k – 2y)² …..(ii)

= (h – 2x)² + k² ….. (iii)

Taking first two equations, we get

h² + k² = h² + (k – 2y)²

→ k² = k² + 4y² – 4yk → 4y(y – k) = 0

→ y = k [∵ y ≠ 0]

Taking first and third equations, we get

h² + k² = (h – 2x)² + k²

→ h² + h² + 4x² – 4xh

→ 4x (x – h) = 0

→ x = h [∵ x ≠ 0]

∴ Required points = (h, k) = (x, y)