# Area of Parallelogram from Determinant

## Theorem

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

 $\ds O$ $=$ $\ds \tuple {0, 0}$ $\ds A$ $=$ $\ds \tuple {a, c}$ $\ds B$ $=$ $\ds \tuple {a + b, c + d}$ $\ds C$ $=$ $\ds \tuple {b, d}$

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.

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

Indeed:

 $\ds \vec {AB}$ $=$ $\ds \tuple {a + b - a, c + d - c}$ $\ds$ $=$ $\ds \tuple {b, d}$ $\ds$ $=$ $\ds \vec {CB}$ $\ds \vec {OA}$ $=$ $\ds \tuple {a + b - b, c + d - d}$ $\ds$ $=$ $\ds \tuple {a, c}$ $\ds$ $=$ $\ds \vec {OA}$

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:

 $\ds \map \Area {OABC}$ $=$ $\ds \paren {a + b} \paren {c + d}$ the large rectangle $\ds$  $\, \ds - \,$ $\ds \paren {\dfrac {a c} 2} - \paren {\dfrac {\paren {a + b - b} \paren {c + d - d} } 2}$ the $2$ triangles at top and bottom $\ds$  $\, \ds - \,$ $\ds \paren {\dfrac {b d} 2} - \paren {\dfrac {\paren {a + b - a} \paren {c + d - c} } 2}$ the $2$ triangles at left and right $\ds$  $\, \ds - \,$ $\ds \paren {a + b - a} c - b \paren {c + d - c}$ the $2$ small rectangles $\ds$ $=$ $\ds a c + a d + b c + b d - \dfrac {a c} 2 - \dfrac {a c} 2 - \dfrac {b d} 2 - \dfrac {b d} 2 - 2 b c$ multiplying out and simplifying $\ds$ $=$ $\ds a c - b d$ simplifying $\ds$ $=$ $\ds \begin {vmatrix} a & b \\ c & d \end {vmatrix}$ Definition of Determinant of Order 2

$\blacksquare$