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