下图显示了一个顶点为 $A(-4, -3)$、$B(6, 1)$、$C(3, 5)$ 和 $D(-2, 3)$ 的四边形 $ABCD$。$M$ 位于 x 轴上，是边 $AD$ 的中点。

1. Calculate the gradient of side AB.

Gradient of $AB = \frac{1 - (-3)}{6 - (-4)} = \frac{4}{10} = \frac{2}{5}$

2. Calculate the gradient of side DC.

Gradient of $DC = \frac{5 - 3}{3 - (-2)} = \frac{2}{5}$

3. Conclude that ABCD is a trapezium based on parallel sides.

Since the gradient of $AB$ equals the gradient of $DC$, the lines $AB$ and $DC$ are parallel. A quadrilateral with exactly one pair of parallel sides is a trapezium.

1. Apply the midpoint formula to points $A$ and $D$

Coordinates of $M = \left( \frac{-4 + (-2)}{2}, \frac{-3 + 3}{2} \right)$

2. Evaluate the coordinates

$M = \left( \frac{-6}{2}, \frac{0}{2} \right) = (-3, 0)$

1. Find the coordinates of $M$ (from part b).

$M$ is the midpoint of $A(-4, -3)$ and $D(-2, 3)$.
$M = \left(\frac{-4 + (-2)}{2}, \frac{-3 + 3}{2}\right) = (-3, 0)$

2. Find the equation of the line $BC$.

Gradient of $BC = \frac{5 - 1}{3 - 6} = \frac{4}{-3} = -\frac{4}{3}$
Equation of $BC$:
$y - 1 = -\frac{4}{3}(x - 6)$
$y = -\frac{4}{3}x + 8 + 1$
$y = -\frac{4}{3}x + 9$

3. Express the coordinates of $N$ and set up the area equation for quadrilateral $MNCD$.

Let $N$ be $(x, -\frac{4}{3}x + 9)$.
Using the shoelace formula for the area of $MNCD$ with vertices $M(-3, 0)$, $N(x, -\frac{4}{3}x + 9)$, $C(3, 5)$, and $D(-2, 3)$:
$\text{Area} = \frac{1}{2} \left| \begin{matrix} -3 & x & 3 & -2 & -3 \\ 0 & -\frac{4}{3}x + 9 & 5 & 3 & 0 \end{matrix} \right| = \frac{65}{4}$
$\frac{1}{2} \left| \left( -3\left(-\frac{4}{3}x + 9\right) + 5x + 9 + 0 \right) - \left( 0 + 3\left(-\frac{4}{3}x + 9\right) - 10 - 9 \right) \right| = \frac{65}{4}$

4. Simplify the area equation.

$\frac{1}{2} \left| (4x - 27 + 5x + 9) - (-4x + 27 - 19) \right| = \frac{65}{4}$
$\frac{1}{2} \left| (9x - 18) - (-4x + 8) \right| = \frac{65}{4}$
$\frac{1}{2} \left| 13x - 26 \right| = \frac{65}{4}$
$|13x - 26| = \frac{65}{2} = 32.5$

5. Solve for $x$ and find the corresponding $y$-coordinate.

$13x - 26 = 32.5 \quad \text{or} \quad 13x - 26 = -32.5$
$13x = 58.5 \quad \text{or} \quad 13x = -6.5$
$x = 4.5 \quad \text{or} \quad x = -0.5$
Since $N$ lies on the line segment $BC$, its $x$-coordinate must be between $x_C=3$ and $x_B=6$. Thus, $x = 4.5$.
Substitute $x = 4.5$ into the equation of $BC$:
$y = -\frac{4}{3}(4.5) + 9 = -6 + 9 = 3$
The coordinates of $N$ are $(4.5, 3)$.