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. Represent the coordinates of A and identify the relevant parallel sides

Since $A$ lies on the $y$-axis, let
$A=(0,y).$
In parallelogram $ABCD$, opposite sides are parallel, so
$AD \parallel BC.$
Hence
$\text{gradient of }AD=\text{gradient of }BC=-\frac{4}{3}.$

2. Use the gradient formula to find the y-coordinate of A

Using $A=(0,y)$ and $D=(6,-1)$,
$\text{gradient of }AD=\frac{-1-y}{6-0}.$
So
$\frac{-1-y}{6}=-\frac{4}{3}.$
Multiply by $6$:
$-1-y=-8.$
Hence
$y=7.$
Therefore,
$A=(0,7).$

1. Identify the equal opposite sides in the parallelogram

Since $ABCD$ is a parallelogram, opposite sides are equal.

Hence,
$BC=AD.$

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

Using $A(0,7)$ and $D(6,-1)$,
$AD=\sqrt{(6-0)^2+(-1-7)^2}$
$=\sqrt{6^2+(-8)^2}$
$=\sqrt{36+64}$
$=\sqrt{100}=10.$

Therefore,
$BC=10.$

1. Use the fact that $AC$ is horizontal to get the y-coordinate of $C$.

Since $AC$ is horizontal and $A(0,7)$,

$C$ has the same $y$-coordinate as $A$.

Let $C=(x,7)$.

2. Use the gradient of the side parallel to $AB$ to find the x-coordinate of $C$.

In a parallelogram, $AB \parallel DC$, so

$\text{gradient of }DC=\text{gradient of }AB=0.8.$

Using $D(6,-1)$ and $C=(x,7)$,

$\frac{7-(-1)}{x-6}=0.8$

$\frac{8}{x-6}=0.8$

$8=0.8(x-6)$

$x-6=10$

$x=16$

Therefore, $C=(16,7)$.

1. Find the base and perpendicular height of the parallelogram using earlier coordinates.

From earlier parts, $A(0,7)$ and $C(16,7)$, so
$AC=16-0=16\text{ units}.$
Since $AC$ is horizontal with equation $y=7$ and $D(6,-1)$, the perpendicular height from $D$ to $AC$ is
$7-(-1)=8\text{ units}.$

2. Calculate the area of the parallelogram.

Area of parallelogram $=$ base $\times$ height
$=16\times 8=128.$