The diagram above shows a triangle $ABC$ with vertices $B(7,2)$ and $C(-6,-2).$ $AB$ is perpendicular to $AC$ and is parallel to the line $y = -2x$. $M$ and $N$ are the mid-points of $AB$ and $BC$ respectively.

1. Determine the gradient of line $AB$

Since $AB$ is parallel to $y = -2x$, the gradient of $AB$, $m_{AB} = -2$.

2. Find the equation of line $AB$

Using point $B(7, 2)$ and $m_{AB} = -2$:
$y - 2 = -2(x - 7)$
$y = -2x + 14 + 2$
$y = -2x + 16$

3. Determine the gradient of line $AC$

Since $AB$ is perpendicular to $AC$, $m_{AB} \times m_{AC} = -1$.
$-2 \times m_{AC} = -1 \implies m_{AC} = \frac{1}{2}$.

4. Find the equation of line $AC$

Using point $C(-6, -2)$ and $m_{AC} = \frac{1}{2}$:
$y - (-2) = \frac{1}{2}(x - (-6))$
$y + 2 = \frac{1}{2}x + 3$
$y = \frac{1}{2}x + 1$

5. Solve the equations simultaneously to find the coordinates of $A$

Equating the two expressions for $y$:
$-2x + 16 = \frac{1}{2}x + 1$
$15 = \frac{5}{2}x$
$x = 15 \times \frac{2}{5} = 6$
Substitute $x = 6$ into $y = -2x + 16$:
$y = -2(6) + 16 = 4$
The coordinates of $A$ are $(6, 4)$.

1. Use the Midpoint Theorem to establish parallel lines

In $\triangle ABC$, $M$ and $N$ are given as the mid-points of sides $AB$ and $BC$ respectively. By the Midpoint Theorem, the line segment connecting these mid-points is parallel to the third side, so $MN \parallel AC$.

2. Conclude the shape of the quadrilateral

A quadrilateral with at least one pair of parallel opposite sides is a trapezium. Since $MN \parallel AC$, quadrilateral $ACNM$ is a trapezium.

1. State the coordinates of $A$ (from part a) and find the coordinates of $M$ and $N$.

From part (a), $A$ is $(6, 4)$.
$M$ is the mid-point of $AB$:
$M = \left(\frac{6+7}{2}, \frac{4+2}{2}\right) = (6.5, 3)$
$N$ is the mid-point of $BC$:
$N = \left(\frac{7+(-6)}{2}, \frac{2+(-2)}{2}\right) = (0.5, 0)$

2. Apply the shoelace formula to find the area of quadrilateral $ACNM$.

Using the vertices $A(6, 4)$, $C(-6, -2)$, $N(0.5, 0)$, and $M(6.5, 3)$ in counter-clockwise order:
$\text{Area} = \frac{1}{2} \left| \begin{matrix} 6 & -6 & 0.5 & 6.5 & 6 \\ 4 & -2 & 0 & 3 & 4 \end{matrix} \right|$
$\text{Area} = \frac{1}{2} \left| (6)(-2) + (-6)(0) + (0.5)(3) + (6.5)(4) - [ (4)(-6) + (-2)(0.5) + (0)(6.5) + (3)(6) ] \right|$
$\text{Area} = \frac{1}{2} \left| (-12 + 0 + 1.5 + 26) - (-24 - 1 + 0 + 18) \right|$
$\text{Area} = \frac{1}{2} \left| 15.5 - (-7) \right|$
$\text{Area} = \frac{1}{2} (22.5) = 11.25$