Question

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

Options

• (x, y)

• (y, x)

• `(x/2, y/2)`

• `(y/2, x/2)`

Answer 1

The coordinates of the point which is equidistant from the three vertices of the ∆AOB as shown in the figure is (x, y).

Explanation:

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

Then, PO = PA = PB

⇒ (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`

⇒ 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

⇒ 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)