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