Question:

Step 1: Write down the given equation of the line. The equation of the line is given as: $$(k + 1)2x + ky - 2k^2 - 2 = 0$$ This equation can be expanded and rearranged as: $$2kx + 2x + ky - 2k^2 - 2 = 0$$ Step 2: Generate two distinct lines by choosing specific values for $k$. To find a point through which the line passes regardless of the value $k$, we can consider the intersection of two distinct lines from this family. Such an intersection point is often the desired fixed point. Let's choose $k=1$ and $k=-1$ for convenience. For $k=1$: Substitute $k=1$ into the given equation: $$(1 + 1)2x + (1)y - 2(1)^2 - 2 = 0$$ $$2(2x) + y - 2 - 2 = 0$$ $$4x + y - 4 = 0 \quad (\text{Equation 1})$$ For $k=-1$: Substitute $k=-1$ into the given equation: $$(-1 + 1)2x + (-1)y - 2(-1)^2 - 2 = 0$$ $$0 \cdot 2x - y - 2(1) - 2 = 0$$ $$-y - 2 - 2 = 0$$ $$-y - 4 = 0$$ $$y = -4 \quad (\text{Equation 2})$$ Step 3: Find the point of intersection of these two lines. Now we solve the system of Equation 1 and Equation 2 to find the coordinates $(x,y)$ of the fixed point. Substitute $y=-4$ from Equation 2 into Equation 1: $$4x + (-4) - 4 = 0$$ $$4x - 8 = 0$$ $$4x = 8$$ $$x = 2$$ Thus, the point through which the line passes regardless of the value $k$ is $(2, -4)$. Step 4: Find the equation of the line with slope 2 passing through the fixed point. We need to find the equation of a line with slope $m=2$ that passes through the point $(x_1, y_1) = (2, -4)$. Using the point-slope form of a linear equation, $y - y_1 = m(x - x_1)$: $$y - (-4) = 2(x - 2)$$ $$y + 4 = 2x - 4$$ $$y = 2x - 4 - 4$$ $$y = 2x - 8$$ The equation of the line with slope 2 passing through the point $(2, -4)$ is $y = 2x - 8$. Comparing this with the given options, we find that it matches Option 1. The final answer is $\boxed{\text{y = 2x - 8}}$.