Area of Parallelogram from Determinant

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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.


Area-of-Parallelogram-determinant.png


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
\(\ds \) \(\) \(\, \ds - \, \) \(\ds \paren {a + b - a} c - b \paren {c + d - d}\) 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
\(\ds \) \(=\) \(\ds a d - b c\) simplifying
\(\ds \) \(=\) \(\ds \begin {vmatrix} a & b \\ c & d \end {vmatrix}\) Determinant of Order $2$

$\blacksquare$


Sources

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