The diagram (not drawn to scale) shows a quadrilateral $ABCD$ such that $AB = BC$ and angle $BCD = 90^\circ$. The point $A$ is $(-4, 6)$ and the point $C$ is $(6, 6)$. Given that the area of triangle $ABC$ is 15 square units and the point $D$ lies on the line $3y + 5x + 42 = 0$.

1. Find the length of the base $AC$.

Since $A(-4, 6)$ and $C(6, 6)$ have the same $y$-coordinate, the line segment $AC$ is horizontal.
Length of $AC = 6 - (-4) = 10$ units.

2. Use the area of the triangle to find the perpendicular height of $B$ from $AC$.

Let the perpendicular height of $B$ from $AC$ be $h$.
$\text{Area of } \triangle ABC = \frac{1}{2} \times AC \times h$
$15 = \frac{1}{2} \times 10 \times h$
$15 = 5h \implies h = 3$

3. Determine the coordinates of $B$ using the properties of the isosceles triangle.

Since $AB = BC$, $riangle ABC$ is an isosceles triangle. The $x$-coordinate of $B$ must lie exactly halfway between $A$ and $C$.
$x$-coordinate of $B = \frac{-4 + 6}{2} = 1$
Since $B$ is in the first quadrant and forms a simple quadrilateral with $D$ (which is in the third quadrant), $B$ must be above $AC$.
$y$-coordinate of $B = 6 + 3 = 9$
Therefore, $B$ is $(1, 9)$.

1. Find the coordinates of $B$ (from part i) to calculate the gradient of $BC$.

From part (i), $B$ is $(1, 9)$.
Gradient of $BC = \frac{6 - 9}{6 - 1} = -\frac{3}{5}$

2. Find the gradient of $CD$ using the perpendicular property.

Since angle $BCD = 90^\circ$, $CD$ is perpendicular to $BC$.
Gradient of $CD = \frac{-1}{-\frac{3}{5}} = \frac{5}{3}$

3. Find the equation of the line $CD$.

Using point $C(6, 6)$ and gradient $\frac{5}{3}$:
$y - 6 = \frac{5}{3}(x - 6)$
$3y - 18 = 5x - 30$
$3y = 5x - 12$

4. Solve simultaneously with the given line equation to find $D$.

The point $D$ lies on $3y + 5x + 42 = 0$, so $3y = -5x - 42$.
Substitute $3y = 5x - 12$ into this equation:
$5x - 12 = -5x - 42$
$10x = -30 \implies x = -3$
Substitute $x = -3$ into $3y = 5x - 12$:
$3y = 5(-3) - 12 = -27 \implies y = -9$
The coordinates of $D$ are $(-3, -9)$.

1. Find the coordinates of $B$ and $D$ (from earlier parts).

Since $A(-4, 6)$ and $C(6, 6)$ have the same $y$-coordinate, the length of $AC = 6 - (-4) = 10$.
Given Area of $\triangle ABC = 15$, $\frac{1}{2}(10)(h) = 15 \implies h = 3$.
Since $AB = BC$, $B$ lies on the perpendicular bisector of $AC$ ($x = 1$). As $B$ is above the $x$-axis, $B$ is $(1, 6+3) = (1, 9)$.
Gradient of $BC = \frac{9-6}{1-6} = -\frac{3}{5}$.
Since $\angle BCD = 90^\circ$, gradient of $CD = \frac{5}{3}$.
Equation of $CD$: $y - 6 = \frac{5}{3}(x - 6) \implies 5x - 3y - 12 = 0$.
Solving simultaneously with $3y + 5x + 42 = 0$ gives $10x + 30 = 0 \implies x = -3$, and $y = -9$. Thus, $D$ is $(-3, -9)$.

2. Calculate the area of quadrilateral $ABCD$ using the shoelace formula.

Using the vertices in counter-clockwise order: $A(-4, 6)$, $D(-3, -9)$, $C(6, 6)$, $B(1, 9)$:
$\text{Area} = \frac{1}{2} \left| \begin{matrix} -4 & -3 & 6 & 1 & -4 \\ 6 & -9 & 6 & 9 & 6 \end{matrix} \right|$
$\text{Area} = \frac{1}{2} | [(-4)(-9) + (-3)(6) + (6)(9) + (1)(6)] - [(6)(-3) + (-9)(6) + (6)(1) + (9)(-4)] |$
$\text{Area} = \frac{1}{2} | (36 - 18 + 54 + 6) - (-18 - 54 + 6 - 36) |$
$\text{Area} = \frac{1}{2} | 78 - (-102) | = \frac{1}{2} (180) = 90$

1. State the coordinates of $A$ and $B$ (from part i) and set up the ratio

From part (i), $B$ is $(1, 9)$. We are given $A(-4, 6)$.
Since $X$ lies on $AB$ produced, the points are in the order $A, B, X$.
The ratio of the lengths is $AB : BX = 2 : 3$.

2. Use vectors or coordinate differences to find $X$

Let the coordinates of $X$ be $(x, y)$.
Using the ratio theorem or similar triangles, the horizontal and vertical changes from $B$ to $X$ are $\frac{3}{2}$ of the changes from $A$ to $B$.
$x - 1 = \frac{3}{2}(1 - (-4)) \implies x - 1 = \frac{3}{2}(5) = 7.5 \implies x = 8.5$
$y - 9 = \frac{3}{2}(9 - 6) \implies y - 9 = \frac{3}{2}(3) = 4.5 \implies y = 13.5$
Thus, $X = (8.5, 13.5)$.