Find the coordinates of the point which is equidistant from the three vertices of the $\vartriangle AOB$ as shown in the figure."

Given: A point is equidistant from the vertices of $\vartriangle AOB$. 

To do: To find the co-ordinates of the point.

Solution:

In figure, Vertices of the triangle are: 

$A( 0,\ 2y),\ B( 2x,\ 0)$ and $O( 0,\ 0)$.

It's a right angled triangle.

A point which is equidistant from vertices of triangle is known as circumcentre of that triangle. 

For a right triangle the circumcentre is the mid point of hypotenuse.

Thus here coordinates of circumcentre must be mid point of $AB=( \frac{x_1+x_2}{2},\ \frac{y_1+y_2}{2})$

$=( \frac{0+2x}{2},\ \frac{2y+0}{2})$

$=( \frac{2x}{2},\ \frac{2y}{2})$

$=( x,\ y)$

Thus, $( x,\ y)$, is equidistant from the vertices of $\vartriangle AOB$. 

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Updated on: 10-Oct-2022

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