Question:

Step 1: Identify vertex $A$ as the intersection of lines $AB: 7x + y = 0$ and $AC: x + y = 0$. Solving $7x + y = 0$ and $x + y = 0$: subtracting gives $6x = 0 \Rightarrow x = 0, y = 0$. So $A = (0, 0)$. Step 2: Since $AB = AC$, the triangle is isosceles with $A$ at the apex. The line $BC$ is the base and must be such that $A$ is equidistant from $B$ and $C$. The perpendicular from $A$ to $BC$ bisects $BC$, meaning $BC$ is perpendicular to the angle bisector of $\angle BAC$. Step 3: Find the angle bisectors of lines $7x + y = 0$ and $x + y = 0$. $$\frac{7x + y}{\sqrt{50}} = \pm \frac{x + y}{\sqrt{2}}$$ Using $+$: $\frac{7x+y}{5\sqrt{2}} = \frac{x+y}{\sqrt{2}} \Rightarrow 7x + y = 5(x+y) \Rightarrow 2x - 4y = 0 \Rightarrow x = 2y$. Using $-$: $7x + y = -5(x+y) \Rightarrow 12x + 6y = 0 \Rightarrow 2x + y = 0$. Step 4: Since $AB = AC$, $BC$ must be perpendicular to one of the angle bisectors (the one that bisects $\angle BAC$). The angle bisector through $A$ that lies inside the triangle is $x = 2y$, i.e., slope $= \frac{1}{2}$. So $BC$ has slope $-2$ (perpendicular to $x = 2y$). Step 5: Line $BC$ passes through $(2, 3)$ with slope $-2$: $$y - 3 = -2(x - 2) \Rightarrow y - 3 = -2x + 4 \Rightarrow 2x + y = 7$$ This matches option (a): $BC: 2x + y = 7$. Step 6: Find $B$ and $C$ by intersecting $BC: 2x + y = 7$ with $AB: 7x + y = 0$ and $AC: x + y = 0$. Intersection with $AB: 7x + y = 0$: $2x + y = 7$ and $7x + y = 0 \Rightarrow 5x = -7 \Rightarrow x = -\frac{7}{5}, y = \frac{49}{5}$. So $B = \left(-\frac{7}{5}, \frac{49}{5}\right)$. Intersection with $AC: x + y = 0$: $2x + y = 7$ and $x + y = 0 \Rightarrow x = 7, y = -7$. So $C = (7, -7)$. Step 7: Compute area of $\triangle ABC$ with $A=(0,0)$, $B=\left(-\frac{7}{5}, \frac{49}{5}\right)$, $C=(7,-7)$. $$\text{Area} = \frac{1}{2}|x_A(y_B - y_C) + x_B(y_C - y_A) + x_C(y_A - y_B)|$$ $$= \frac{1}{2}\left|0\left(\frac{49}{5}+7\right) + \left(-\frac{7}{5}\right)(-7-0) + 7\left(0 - \frac{49}{5}\right)\right|$$ $$= \frac{1}{2}\left|0 + \frac{49}{5} - \frac{343}{5}\right| = \frac{1}{2} \cdot \frac{294}{5} = \frac{147}{5}$$ This matches option (c).