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.
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:
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 |
$\blacksquare$
Sources
- 1998: Richard Kaye and Robert Wilson: Linear Algebra ... (previous) ... (next): Part $\text I$: Matrices and vector spaces: $1$ Matrices: $1.6$ Determinant and trace