question_answer

The centroid of an equilateral triangle is (0, 0). If two vertices of the triangle lie on \[x+y=2\sqrt{2},\]then one of them will have its coordinates

A)  \[(\sqrt{2}+\sqrt{6},\sqrt{2}-\sqrt{6})\]

B)  \[(\sqrt{2}+\sqrt{3},\sqrt{2}-\sqrt{3})\]

C)  \[(\sqrt{2}+\sqrt{5},\sqrt{2}-\sqrt{5})\]

D)  None of these

Correct Answer: A

Solution :

Let the vertices B and C lie on the given line. Then, \[OD=\frac{2\sqrt{2}}{\sqrt{2}}=2,\] Equation of OD is \[y=x\]\[\Rightarrow \]\[y=\sqrt{2}\](for point D) Also, \[BD=OD\times \tan 60{}^\circ =2\sqrt{3},\] for the coordinate of B and C. Using parametric equation of line, we get \[\frac{x-\sqrt{2}}{-\frac{1}{\sqrt{2}}}=\frac{y-\sqrt{2}}{\frac{1}{\sqrt{2}}}=\pm \,\,2\sqrt{3}\] \[\Rightarrow \]\[\frac{-x+\sqrt{2}}{\frac{1}{\sqrt{2}}}=\frac{y-\sqrt{2}}{\frac{1}{\sqrt{2}}}=\pm \,\,2\sqrt{3}\] \[\Rightarrow \]\[-x+\sqrt{2}=y-\sqrt{2}=\pm \,\,\sqrt{6}\] \[\Rightarrow \]\[x=\sqrt{2}+\sqrt{6},\]\[\sqrt{2}-\sqrt{6}\]and\[y=\sqrt{2}+\sqrt{6},\]\[\sqrt{2}-\sqrt{6}\] Hence, coordinates are \[(\sqrt{2}+\sqrt{6},\sqrt{2}-\sqrt{6})\].