Area of a parallelogram, vertices $(-1,-1), (4,1), (5,3), (10,5)$. I need to find the area of a parallelogram with vertices $(-1,-1), (4,1), (5,3), (10,5)$.
If I denote $A=(-1,-1)$, $B=(4,1)$, $C=(5,3)$, $D=(10,5)$, then I see that $\overrightarrow{AB}=(5,2)=\overrightarrow{CD}$. Similarly $\overrightarrow{AC}=\overrightarrow{BD}$. So I see that these points indeed form a parallelogram.
It is assignment from linear algebra class. I wasn't sure if I had to like use a matrix or something. 
 A: Hints:


*

*The area of a parallelogram with side vectors $\bf a$ and $\bf b$ is $\det(\bf a\ \bf b)$.

*For a parallelogram $A,B,C,D$ its side vectors are e.g. $B-A$ and $C-A$.

A: Hint: the area of a parallelogram (see left-most image) is equal to the determinant of the $2\times 2$ matrix formed by the column vectors representing component vectors determined by the given points. $$A = \text{det}\,\left(\vec u \;\; \vec v\right)$$
The area of a parallelogram is also equal to the magnitutude of the cross product of the component vectors $\vec u, \vec v$ = $$|\vec u\times \vec v| = |\vec u|\,|\vec v|\sin \theta$$
where $\theta$ is the measure of the angle formed by the component vectors $\vec u, \vec v$.
Use your points to determine the component vectors $\vec u, \vec v$. 
For parallelogram formed by $p_1, p_2, p_3, p_4$, put $\vec u = p_2 - p_1$, $\vec v = p_3 - p_1$ (where $p_4$ is the point opposite $p_1$, $p_2$ the point opposite $p_3$):
$$\vec u = \langle 4 -(- 1), 1-(-1)\rangle = \langle 5, 2\rangle$$
$$\vec v = \langle 5 - (-1), 3 - (-1)\rangle = \langle 6, 4\rangle$$
So compute $$A = \det \begin{pmatrix} 5 & 6 \\ 2 & 4 \end{pmatrix},\;\;\text{or}\;\; A = \vert \vec u\times \vec v\vert$$
See the parallelogram whose area can be determined by component vectors $\vec u, \vec v\;\;$ (left of the image):

A: As the diagonal of a parallelogram $A(-1,-1),B(4,1),C(5,3),D(10,5)$ divides into two congruent triangles, which implies the triangles have same area.
So, the area of the parallelogram will be $2\cdot$  area of any one of  $\triangle ABC,\triangle ADC, \triangle ABD, \triangle BCD$
For example, the area of $$\triangle ABC=\frac12\left|\det\begin{pmatrix} -1&-1&1 \\ 4&1&1 \\ 5&3&1\end{pmatrix}\right|$$
$$=\frac12\left|\det\begin{pmatrix} -1&-1&1 \\ 5&2&0 \\ 6&4&0\end{pmatrix}\right|\text{ applying} R_2'=R_2-R_1,R_3'=R_3-R_1$$
$$=\frac12\left|5\cdot4-6\cdot2\right|=4$$
