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,

=

=

= h^{2} + k^{2} = h^{2} + (k – 2y)^{2} …..(ii)

= (h – 2x)^{2} + k^{2} ….. (iii)

Taking first two equations, we get

h^{2} + k^{2} = h^{2} + (k – 2y)^{2}

→ k^{2} = k^{2} + 4y^{2} – 4yk → 4y(y – k) = 0

→ y = k [∵ y ≠ 0]

Taking first and third equations, we get

h^{2} + k^{2} = (h – 2x)^{2} + k^{2}

→ h^{2} + h^{2} + 4x^{2} – 4xh

→ 4x (x – h) = 0

→ x = h [∵ x ≠ 0]

∴ Required points = (h, k) = (x, y)

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