# The coordinates of the point which is equidistant from the three vertices of the $\triangle \mathrm{AOB}$ as shown in the figure is(A) $(x AcademicMathematicsNCERTClass 10 #### Complete Python Prime Pack 9 Courses 2 eBooks #### Artificial Intelligence & Machine Learning Prime Pack 6 Courses 1 eBooks #### Java Prime Pack 9 Courses 2 eBooks Given: The three vertices of the $\triangle \mathrm{AOB}$ are$0(0, 0), A(0,2y)$and$B(2x, 0)$. To do: We have to find the coordinates of the point which is equidistant from the three vertices of the $\triangle \mathrm{AOB}$. Solution: Let the coordinates of the point which is equidistant from the three vertices $0(0,0)$, $A(0,2 y)$ and $B(2 x, 0)$ be $P(h, k)$. This implies,$P O=P A=P B$On squaring, we get,$(P O)^{2}=(P A)^{2}=(P B)^{2}$Using distance formula$[\sqrt{(h-0)^{2}+(k-0)^{2}}]^{2} }=[\sqrt{(h-0)^{2}+(k-2 y)^{2}}]^{2}=[\sqrt{(h-2 x)^{2}+(k-0)^{2}}]^{2}h^{2}+k^{2}=h^{2}+(k-2 y)^{2} =(h-2 x)^{2}+k^{2}$Therefore,$h^{2}+k^{2}=h^{2}+(k-2 y)^{2}k^{2} =k^{2}+4 y^{2}-4 k y4 y(y-k)=0y=kh^{2}+k^{2} =(h-2 x)^{2}+k^{2}h^{2} =h^{2}+4 x^{2}-4 x h4 x(x-h) =0x =h $The required point is$(x, y)\$.

Updated on 10-Oct-2022 13:28:26