Let B and C be two points on the line y+x=0 s | vedclass

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(C) Let the coordinates of $B$ be $(-t, t)$ and $C$ be $(t, -t)$ since they lie on $y+x=0$ and are symmetric about the origin.The length of the side $

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

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(C) Let the coordinates of $B$ be $(-t, t)$ and $C$ be $(t, -t)$ since they lie on $y+x=0$ and are symmetric about the origin.The length of the side $BC$ is $a = \sqrt{(t - (-t))^2 + (-t - t)^2} = \sqrt{(2t)^2 + (-2t)^2} = \sqrt{8t^2} = 2\sqrt{2}|t|$.The midpoint of $BC$ is the origin $(0, 0)$. The altitude from $A$ to $BC$ lies on the line perpendicular to $y+x=0$ passing through the origin,which is $y=x$.Point $A$ is the intersection of $y=x$ and $y-2x=2$. Substituting $y=x$ into $y-2x=2$ gives $x-2x=2$,so $x=-2$ and $y=-2$. Thus,$A = (-2, -2)$.The height $h$ of the equilateral triangle is the distance from $A(-2, -2)$ to the line $x+y=0$,which is $h = \frac{|-2 + (-2)|}{\sqrt{1^2 + 1^2}} = \frac{4}{\sqrt{2}} = 2\sqrt{2}$.For an equilateral triangle,the height $h = \frac{\sqrt{3}}{2}a$,so $a = \frac{2h}{\sqrt{3}} = \frac{2(2\sqrt{2})}{\sqrt{3}} = \frac{4\sqrt{2}}{\sqrt{3}}$.The area of the equilateral triangle is $\frac{\sqrt{3}}{4}a^2 = \frac{\sqrt{3}}{4} \left(\frac{4\sqrt{2}}{\sqrt{3}}\right)^2 = \frac{\sqrt{3}}{4} \cdot \frac{16 \cdot 2}{3} = \frac{\sqrt{3}}{4} \cdot \frac{32}{3} = \frac{8\sqrt{3}}{3} = \frac{8}{\sqrt{3}}$.

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 the line $y-2x=2$ such that $\triangle ABC$ is an equilateral triangle. Then,the area of the $\triangle ABC$ is

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