Question:

Step 1: Factor the given curve equation using the algebraic identity. We start with the equation $x^3 + y^3 + 3xy = 1$, which can be rewritten as: $$x^3 + y^3 + 3xy - 1 = 0$$ Using the identity $a^3 + b^3 + c^3 - 3abc = (a+b+c)(a^2+b^2+c^2-ab-bc-ca)$, we set $a = x$, $b = y$, and $c = -1$: $$x^3 + y^3 + (-1)^3 - 3(x)(y)(-1) = x^3 + y^3 - 1 + 3xy$$ Therefore: $$x^3 + y^3 + 3xy - 1 = (x + y - 1)(x^2 + y^2 + 1 - xy + x + y) = 0$$ Step 2: Identify the real components of the factored equation. The equation factors into two parts: - **First factor:** $x + y - 1 = 0$, which is the line $x + y = 1$ - **Second factor:** $x^2 + y^2 + 1 - xy + x + y = 0$ For the second factor, we treat it as a quadratic in $x$: $$x^2 + x(1-y) + (y^2+y+1) = 0$$ The discriminant is: $$\Delta = (1-y)^2 - 4(y^2+y+1) = 1 - 2y + y^2 - 4y^2 - 4y - 4 = -3y^2 - 6y - 3 = -3(y+1)^2$$ Since $\Delta \leq 0$ for all real $y$, the second factor has real solutions only when $\Delta = 0$, which occurs at $y = -1$. When $y = -1$: $x = \frac{-(1-y)}{2} = \frac{-2}{2} = -1$ So the second factor represents the single point $(-1, -1)$. Step 3: Identify vertex $A$ and side $BC$. From the factorization: - Side $BC$ is the line: $x + y = 1$ - Vertex $A$ is the point: $A = (-1, -1)$ Step 4: Find the altitude from $A$ perpendicular to $BC$. Since $BC$ has equation $x + y = 1$, its slope is $-1$. The altitude from $A$ perpendicular to $BC$ has slope $1$ (negative reciprocal). The altitude from $A(-1, -1)$ is: $$y - (-1) = 1(x - (-1))$$ $$y = x$$ Step 5: Find the orthocentre $B$. Since $B$ is the orthocentre and lies on side $BC$, it is the intersection of the altitude $y = x$ and the line $BC: x + y = 1$: $$x + x = 1 \implies x = \frac{1}{2}, \quad y = \frac{1}{2}$$ Therefore: $B = \left(\frac{1}{2}, \frac{1}{2}\right)$ Step 6: Find the equation of line $AB$. The slope of line $AB$ passing through $A(-1, -1)$ and $B\left(\frac{1}{2}, \frac{1}{2}\right)$ is: $$m = \frac{\frac{1}{2} - (-1)}{\frac{1}{2} - (-1)} = \frac{\frac{3}{2}}{\frac{3}{2}} = 1$$ The equation of line $AB$ is: $$y - (-1) = 1(x - (-1))$$ $$y = x$$ In the form $y = mx + c$: $m = 1$ and $c = 0$ Step 7: Calculate the final answer. $$|4 - m - c| = |4 - 1 - 0| = |3| = 3$$ **Final Answer: The absolute value of $(4 - m - c)$ is $\boxed{3}$, which corresponds to Option 2.**