In the given diagram, ΔAOB is a right-angled triangle and C is the mid-point of AB. The coordinates of the point which is equidistant from the three vertices of ΔAOB is :

1. (x, y)

2. (y, x)

3. $\Big(\dfrac{x}{2}, \dfrac{y}{2}\Big)$

4. $\Big(\dfrac{2x}{3}, \dfrac{2y}{3}\Big)$

Given,

C is midpoint of AB.

We know that,

In a right-angled triangle, the point equidistant from all three vertices is the midpoint of the hypotenuse.

Using midpoint formula :

M = $\Big(\dfrac{x$1 + x2}{2}, \dfrac{y1 + y2}{2}\Big)

Substituting values, we get :

$\Rightarrow C = \Big(\dfrac{0 + 2x}{2}, \dfrac{2y + 0}{2} \Big) \\[1em] \Rightarrow C = \Big(\dfrac{2x}{2}, \dfrac{2y}{2} \Big) \\[1em] \Rightarrow C = \text{(x, y)}.$

Hence, option 1 is the correct option.