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The coordinates of the point whichis equidistant from the three verticesof the ΔAOB as shown in the figure is

(A) (x, y)

(B) (y, x)

(C) \frac{x}{2},\frac{y}{2}



Answers (1)

In the given figure, it is clear that DAOB is a right-angle triangle.

And in a right-angle triangle, the mid-point of the hypotenuse is equidistant from the three vertices. Thus, co-ordinates must be mid-point of AB

\\ A (0, 2y), B(2x, 0)\\ (x\textsubscript{1}, y\textsubscript{1}) = (0, 2y) ,(x\textsubscript{2}, y\textsubscript{2}) = (2x, 0)\\

Now find mid-point of AB using mid-point formula 

\left( \frac{{{x}_{1}}+{{x}_{2}}}{2},\frac{{{y}_{1}}+{{y}_{2}}}{2} \right) \\

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

Hence, option A is correct.

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