The line $y + 2x = 12$ intersects the curve $y = 2x^2 - 6x - 4$ at two distinct points $A$ and $B$.

1. Express $y$ in terms of $x$ from the linear equation and substitute into the quadratic equation

From $y + 2x = 12$, we have $y = 12 - 2x$.
Substitute this into $y = 2x^2 - 6x - 4$:
$12 - 2x = 2x^2 - 6x - 4$

2. Simplify and solve the resulting quadratic equation for $x$

$2x^2 - 4x - 16 = 0$
$x^2 - 2x - 8 = 0$
$(x - 4)(x + 2) = 0$
$x = 4 \text{ or } x = -2$

3. Find the corresponding $y$-coordinates and state the points

When $x = 4$, $y = 12 - 2(4) = 4$.
When $x = -2$, $y = 12 - 2(-2) = 16$.
The coordinates are $(-2, 16)$ and $(4, 4)$.

1. Find the coordinates of $A$ and $B$ from part (i)

Substitute $y = 12 - 2x$ into the curve equation:
$12 - 2x = 2x^2 - 6x - 4$
$2x^2 - 4x - 16 = 0$
$x^2 - 2x - 8 = 0$
$(x - 4)(x + 2) = 0$
$x = 4$ or $x = -2$
When $x = 4$, $y = 12 - 2(4) = 4$.
When $x = -2$, $y = 12 - 2(-2) = 16$.
The coordinates are $(4, 4)$ and $(-2, 16)$.

2. Find the midpoint of $AB$

Midpoint $= \left( \frac{4 + (-2)}{2}, \frac{4 + 16}{2} \right)$
$= (1, 10)$

3. Find the gradient of the perpendicular bisector

The line $AB$ has equation $y = -2x + 12$, so its gradient is $m_1 = -2$.
Let the gradient of the perpendicular bisector be $m_2$.
$m_1 m_2 = -1$
$-2 m_2 = -1 \implies m_2 = \frac{1}{2}$

4. Form the equation of the perpendicular bisector

Using the point-slope form with midpoint $(1, 10)$ and gradient $\frac{1}{2}$:
$y - 10 = \frac{1}{2}(x - 1)$
$2y - 20 = x - 1$
$2y = x + 19$