Area of Parallelogram from Determinant

Theorem
Let $OABC$ be a parallelogram in the Cartesian plane whose vertices are located at:

The area of $OABC$ is given by:
 * $\map \Area {OABC} = \begin {vmatrix} a & b \\ c & d \end {vmatrix}$

where $\begin {vmatrix} a & b \\ c & d \end {vmatrix}$ denotes the determinant of order $2$.

Proof
Arrange for the parallelogram to be situated entirely in the first quadrant.


 * Area-of-Parallelogram-determinant.png

First need we establish that $OABC$ is actually a parallelogram in the first place.

Indeed:

Thus:
 * $OA = CB$
 * $OC = AB$

and it follows from Opposite Sides Equal implies Parallelogram that $OABC$ is indeed a parallelogram.

Now we calculate the area of $OABC$ as equal to:
 * the area occupied by the large rectangle in the diagram above

less:
 * the $4$ triangles
 * the $2$ small rectangles.

Thus: