$ABCD$ is a parallelogram.
Point $A$ lies on the $y$-axis and coordinates of $D$ is $(6, -1)$.
$AC$ is a horizontal line and gradient of $BC = -\frac{4}{3}$.

1. Identify the gradient of $AD$ using the properties of a parallelogram

Since $ABCD$ is a parallelogram, $AD$ is parallel to $BC$.
Therefore, gradient of $AD$ = gradient of $BC = -\frac{4}{3}$.

2. Form an equation using the coordinates of $A$ and $D$ to find $A$

Since $A$ lies on the $y$-axis, let the coordinates of $A$ be $(0, y)$.
Gradient of $AD = \frac{-1 - y}{6 - 0} = -\frac{4}{3}$
$\frac{-1 - y}{6} = -\frac{4}{3}$
$-1 - y = -8$
$y = 7$
Coordinates of $A$ are $(0, 7)$.

1. Identify the relationship between the lengths of $BC$ and $AD$

Since $ABCD$ is a parallelogram, the opposite sides are equal in length. Therefore, $BC = AD$.

2. Calculate the length of $AD$ using the coordinates of $A$ and $D$

From part (a), $A$ is $(0, 7)$ and $D$ is $(6, -1)$.
$\text{Length of } AD = \sqrt{(6 - 0)^2 + (-1 - 7)^2}$
$AD = \sqrt{6^2 + (-8)^2}$
$AD = \sqrt{36 + 64} = \sqrt{100} = 10$
Thus, the length of $BC$ is $10$ units.

1. Identify the coordinates of the vertices from earlier parts

From earlier parts, $A$ is $(0, 7)$, $C$ is $(16, 7)$, and $D$ is $(6, -1)$.

2. Calculate the length of the horizontal base $AC$

Since $AC$ is a horizontal line, its length is the difference in the $x$-coordinates:
$\text{Base } AC = 16 - 0 = 16 \text{ units}$

3. Calculate the perpendicular height of the parallelogram and find the area

The perpendicular height from $D$ to the line $AC$ is the difference in their $y$-coordinates:
$\text{Height} = 7 - (-1) = 8 \text{ units}$

$\text{Area of parallelogram} = \text{base} \times \text{height}$
$\text{Area} = 16 \times 8 = 128 \text{ units}^2$