Question

In the equilateral triangle $ABC$ , the equation of the side $BC$ is $x+y-2=0$ and the centroid of $\Delta ABC$ is $\left(0 , 0\right)$ . If the points $A,B,C$ are in anticlockwise order, then the midpoint of the line segment joining $A$ and $C$ is

• A. $\left(\frac{- \sqrt{3} + 1}{2} , \frac{\sqrt{3} + 1}{2}\right)$
• B. $\left(\frac{- \sqrt{3} - 1}{2} , \frac{\sqrt{3} - 1}{2}\right)$
• C. $\left(\frac{- \sqrt{3} - 1}{2} , \frac{\sqrt{3} + 1}{2}\right)$
• D. $\left(\frac{- \sqrt{3} + 1}{2} , \frac{\sqrt{3} - 1}{2}\right)$

Answer 1

Answer: (B)

Foot of (0,0) on $B C$ is $D(1,1) \Rightarrow O D=\sqrt{2}$
$\Rightarrow $ Slope of $O D$ is 1
$\Rightarrow O E$ makes an angle $\left(45^{\circ}+120^{\circ}\right)=165^{\circ}$ with the positive $x$ -axis
$\Rightarrow E=\left(\sqrt{2} \cos 165^{\circ}, \sqrt{2} \sin 165^{\circ}\right)=\left(\frac{-\sqrt{3}-1}{2}, \frac{\sqrt{3}-1}{2}\right)$