How do you find the area of a parallelogram with the vertices? How do you find the area of a parallelogram with the following vertices; $A(4,2)$, $B(8,4)$, $C(9,6)$ and $D(13,8)$.
The absolute value of the cross product of two vectors $\vec{a}, \vec{b} \in \mathbb{R}^3$ spanning the parallelogram is its area:

$$A_\text{parallelogram}= \left|\vec{a}\times\vec{b}\right|$$



So in your case we have to write the points in $\mathbb{R}^2$ as vectors in $\mathbb{R}^3$ and apply the formula:

$\vec{AB} = \begin{pmatrix}8\\4\\0\end{pmatrix}  -\begin{pmatrix}4\\2\\0\end{pmatrix} =\begin{pmatrix}4\\2\\0\end{pmatrix}$

$\vec{AD} = \begin{pmatrix}13\\8\\0\end{pmatrix}  -\begin{pmatrix}4\\2\\0\end{pmatrix} =\begin{pmatrix}9\\6\\0\end{pmatrix}$

$A_\text{parallelogram}= \left|\vec{AB}\times\vec{AD}\right| = \left| \begin{pmatrix}4\\2\\0\end{pmatrix} \times \begin{pmatrix}9\\6\\0\end{pmatrix} \right| = \left|\begin{pmatrix}0\\0\\6\end{pmatrix}  \right| = 6$



You might have noticed that this simplifies to 

$$A_\text{parallelogram}= (b_1 - a_1)(d_2-a_2)-(b_2-a_2)(d_1-a_1)$$
$$= (8 - 4)(8-2)-(4-2)(13-4)=-24-(-18)=6$$
I am just providing you with the simplest shortcut to doing this

Pick the first three points A(4,2), B(8, 4) and C(9, 6)

Negate point A to get (-4, -2) and add to the other two points B and C. Add x's and y's so you have a new point

(4, 2)
(5, 4)   Now use determinant to find the area. 

16-10 = 6sq unit

Neglect any negative sign that arises