Question

The coordinates of the point P which is equidistant from the three vertices of the △AOB as shown in the figure is

• A. \((x,y)\)
• B. \((y,x)\)
• C. \((\frac x2,\frac y2)\)
• D. \((\frac y2,\frac x2)\)

Answer 1

The Correct Option is A

The point P is equidistant from the three vertices of triangle ΔOAB. This means that P is the circumcenter of ΔOAB. Since O is at the origin (0, 0), triangle ΔOAB has a right angle at O. The circumcenter of a right triangle is the midpoint of the hypotenuse. The coordinates of A are (0, 2y) and the coordinates of B are (2x, 0). The midpoint of AB is given by: 

\[\left(\frac{0 + 2x}{2}, \frac{2y + 0}{2}\right) = (x, y)\]

Thus, the coordinates of P are (x, y). Final Answer: 

The final answer is [1]