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

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}

(D)\frac{y}{2},\frac{x}{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.

Posted by

infoexpert21

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