Key Concepts and Formulas

• Equilateral Triangle: All sides are equal, all angles are 60 degrees. The altitude from a vertex to the opposite side bisects that side and is perpendicular to it. The area is given by $A = \frac{\sqrt{3}}{4}s^2$, where $s$ is the side length. Also, $A = \frac{h^2}{\sqrt{3}}$, where $h$ is the length of the altitude.
• Symmetry with respect to the origin: If a point $(x, y)$ is symmetric to $(x', y')$ with respect to the origin, then $x' = -x$ and $y' = -y$.
• Perpendicular Lines: If two lines have slopes $m_1$ and $m_2$, they are perpendicular if and only if $m_1 m_2 = -1$.

Step-by-Step Solution

• Step 1: Analyze the symmetry and deduce the altitude's path. Since $B$ and $C$ are symmetric with respect to the origin and lie on the line $y+x=0$, the origin is the midpoint of $BC$. Because $\triangle ABC$ is equilateral, the altitude from $A$ to $BC$ bisects $BC$ and is perpendicular to it. Therefore, the altitude from $A$ passes through the origin $O(0, 0)$. This means $AO$ is the altitude.

• Step 2: Find the slope of line $BC$. The equation of line $BC$ is $y+x=0$, which can be rewritten as $y = -x$. The slope of line $BC$ is $m_{BC} = -1$.

• Step 3: Find the slope of line $AO$ (the altitude). Since $AO$ is perpendicular to $BC$, the product of their slopes is $-1$. Let the slope of $AO$ be $m_{AO}$. $m_{BC} \cdot m_{AO} = -1$ $(-1) \cdot m_{AO} = -1$ $m_{AO} = 1$

• Step 4: Find the equation of line $AO$. Line $AO$ passes through the origin $(0, 0)$ and has a slope of $1$. Using the point-slope form $y - y_1 = m(x - x_1)$: $y - 0 = 1(x - 0)$ $y = x$ Thus, the equation of line $AO$ is $y = x$.

• Step 5: Find the coordinates of point $A$. Point $A$ lies on both the line $y - 2x = 2$ and the line $y = x$. To find the coordinates of $A$, we solve the system of equations:

  $$\begin{cases} y = x \\ y - 2x = 2 \end{cases}$$

  Substitute $y = x$ into the second equation: $x - 2x = 2$ $-x = 2$ $x = -2$ Since $y = x$, we have $y = -2$. Therefore, the coordinates of point $A$ are $(-2, -2)$.

• Step 6: Calculate the length of the altitude $AO$. The altitude is the distance between $A(-2, -2)$ and $O(0, 0)$. Using the distance formula: $h = AO = \sqrt{(-2 - 0)^2 + (-2 - 0)^2}$ $h = \sqrt{(-2)^2 + (-2)^2}$ $h = \sqrt{4 + 4}$ $h = \sqrt{8} = 2\sqrt{2}$

• Step 7: Calculate the area of $\triangle ABC$. Using the formula for the area of an equilateral triangle in terms of its altitude, $A = \frac{h^2}{\sqrt{3}}$: $A = \frac{(2\sqrt{2})^2}{\sqrt{3}}$ $A = \frac{8}{\sqrt{3}}$

Common Mistakes & Tips

• Remember that the altitude of an equilateral triangle bisects the base. This is crucial for understanding why the altitude passes through the origin in this problem.
• Choosing the right formula for the area of an equilateral triangle (in terms of altitude) can save time.
• Be careful with signs when using the distance formula and calculating slopes.

Summary By exploiting the symmetry of points $B$ and $C$ with respect to the origin and the properties of equilateral triangles, we found that the altitude from $A$ passes through the origin. This allowed us to determine the coordinates of point $A$, calculate the length of the altitude, and finally find the area of the triangle. The area of $\triangle ABC$ is $\frac{8}{\sqrt{3}}$.

Final Answer The final answer is \boxed{\frac{8}{\sqrt{3}}}, which corresponds to option (D).