question_answer

One vertex of the equilateral triangle with centroid at the origin and one side as \[x+y-2=0\]is

A)            \[(-1,-1)\]                                  

B)            \[(2,2)\]

C)            \[(-2,-2)\]                                  

D)            None of these

Correct Answer: C

Solution :

           Let the co-ordinate of vertex A be\[(h,k)\]. Then \[AD\]is perpendicular to \[BC\], therefore \[OA\,\bot \,BC\]                    Þ \[\frac{k-0}{h-0}\times \frac{-1}{1}=-1\Rightarrow k=h\]     .....(i)                               Let the coordinates of D be\[(\alpha ,\beta )\]. Then the co-ordinates of O are \[\left( \frac{2\alpha +h}{2+1},\frac{2\beta +k}{2+1} \right)\]. Therefore \[\frac{2\alpha +h}{3}=0\] and \[\frac{2\beta +k}{3}=0\] \[\Rightarrow \alpha =-\frac{h}{2},\beta =\frac{-k}{2}\].                    Since \[(\alpha ,\beta )\]lies on \[x+y-2=0\]Þ\[\alpha +\beta -2=0\]                    Þ  \[-h/2-k/2-2=0\]Þ\[h+k+4=0\]                    Þ \[2h+4=0\Rightarrow h=k=-2\],               [from (i)]                    Hence the coordinates of vertex A are\[(-2,-2)\].