### Straight Lines Question 6

#### Question 6 - 29 January - Shift 1

Let $B$ and $C$ be the two points on the line $y+x=0$ such that $B$ and $C$ are symmetric with respect to the origin. Suppose $A$ is a point on $y-2 x=2$ such that $\triangle ABC$ is an equilateral triangle. Then, the area of the $\triangle ABC$ is

(1) $3 \sqrt{3}$

(2) $2 \sqrt{3}$

(3) $\frac{8}{\sqrt{3}}$

(4) $\frac{10}{\sqrt{3}}$

## Show Answer

#### Answer: (3)

#### Solution:

#### Formula: Distance between point and line

$y-2 x=2$

At A $x=y$

$Y-2 x=2$

$(-2,-2)$

Height from line $x+y=0$

$h=\frac{4}{\sqrt{2}}$

Area of $\Delta=\frac{\sqrt{3}}{4} \frac{h^{2}}{\sin ^{2} 60}=\frac{8}{\sqrt{3}}$