A) \[2x-a=0\]
B) \[x+a=0\]
C) \[2x+a=0\]
D) \[3x-2a=0\]
Correct Answer: C
Solution :
[c] Since the equilateral triangle is inscribed in the circle with centre at the origin, centroid lies on the origin. |
So, \[\frac{AO}{OD}=\frac{2}{1}\] |
and \[OD=\frac{1}{2}AO=\frac{a}{2}\] |
So, other vertices of triangle have coordinates, |
\[\left( -\frac{a}{2},\frac{\sqrt{3a}}{2} \right)\] and \[\left[ -\frac{a}{2},-\frac{\sqrt{3}}{2}a \right]\] |
\[\left( -\frac{a}{2}\frac{\sqrt{3a}}{2} \right)y\] |
\[\left( -\frac{a}{2},\frac{-\sqrt{3a}}{2} \right)\] |
\[\therefore \] Equation of line BC is: |
\[x=-\frac{a}{2}\] |
\[\Rightarrow \,\,\,\,\,2x+a=0\] |
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