Let B and C be two points on the line y+x=0 such that B and C are symmetric with respect t…

Question

Let $B$ and $C$ be 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 $y-2x=2$ such that $	riangle ABC$ is an equilateral triangle. Then, the area of the $	riangle ABC$ is:

Accepted Answer

Since $A$ has equal $x$- and $y$- coordinates related by symmetry constraints (the perpendicular from $A$ to $BC$ must pass through the origin, the midpoint of $BC$), $A$ lies on the line $y=x$ direction perpendicular consideration... Using the source's method: solve $y=x$ with $y-2x=2$: $x=y \implies -x=2 \implies x=-2,\ y=-2$, so $A=(-2,-2)$.

Height from $A$ to line $x+y=0$:
$$h = \frac{|-2-2|}{\sqrt2} = \frac4{\sqrt2} = 2\sqrt2$$

For an equilateral triangle, height $h = \dfrac{\sqrt3}2 a$ (side $a$), so $a^2 = \dfrac{4h^2}3 = \dfrac{4(8)}3=\dfrac{32}3$.

$$\text{Area} = \frac{\sqrt3}4a^2 = \frac{\sqrt3}4\cdot\frac{32}3 = \frac{8\sqrt3}3 = \frac8{\sqrt3}$$

Solutions · 1