Question

# Find the area of the parallelogram with vertices A(-3, 0), B(-1 , 3), C(5, 2), and D(3, -1).

Analytic geometry
Find the area of the parallelogram with vertices A(-3, 0), B(-1 , 3), C(5, 2), and D(3, -1).

2021-05-29

Calculations:
From the given points, we may choose any three of then such and dind a two vector that express the sides of the parallelogram "or one side and a diagonal of the parallelogram", we may chose for example the points B, C, and D.
More over, we can choose point B to be the common point for the two sides "the initial point of the two vector", thus we need to find vector BD and vector BC, using equation as follows
$$BC \leq 5+1,2-3>$$
$$\leq 6,-1>$$
And, the other vector side BD is
$$BD \leq 3+1,-1-3>$$
$$\leq 4,-4>$$
And the cross product of both vectors is
$$BC\times BD \leq 6,-1>\times<4,-4>$$
$$=\begin{vmatrix}i&j&k\\6&-1&0\\4&-4&0\end{vmatrix}$$
$$=i\begin{vmatrix}-1&0\\4&0\end{vmatrix}-j\begin{vmatrix}6&0\\4&0\end{vmatrix}+k\begin{vmatrix}6&-1\\4&-4\end{vmatrix}$$
$$=0i-0j+k((6)(-4)-(1)(-4))$$
$$=-20k$$
Knowing, the cross product of the two vectors of the parallelogram we can use equation to find the area
$$\text{Area}=|-20k|$$
$$=20$$
Thus, the area of the parallelogram is 20 units squared.