Area of Quadrilateral in Determinant Form/Examples/Vertices at (2, -1), (4, 3), (-1, 2), (-3, -2)

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Example of Area of Quadrilateral in Determinant Form

Let $Q$ be a quadrilateral embedded in the cartesian plane with vertices at $\tuple {2, -1}$, $\tuple {4, 3}$, $\tuple {-1, 2}$ and $\tuple {-3, -2}$.

The area of $Q$ is given by:

$\map \Area Q = 18$


Proof

From Area of Quadrilateral in Determinant Form:

\(\ds \map \Area Q\) \(=\) \(\ds \dfrac 1 2 \paren {\size {\paren {\begin{vmatrix} 2 & -1 & 1 \\ 4 & 3 & 1 \\ -1 & 2 & 1 \\ \end{vmatrix} } } + \size {\paren {\begin{vmatrix} 2 & -1 & 1 \\ -3 & -2 & 1 \\ -1 & 2 & 1 \\ \end{vmatrix} } } }\)
\(\ds \) \(=\) \(\ds \dfrac 1 2 \size {\paren {2 \times 3 - 4 \times \paren {-1} } - \paren {2 \times 2 - \paren {-1} \times \paren {-1} } + \paren {4 \times 2 - \paren {-1} \times 3} }\) Definition of Determinant
\(\ds \) \(\) \(\, \ds + \, \) \(\ds \dfrac 1 2 \size {\paren {2 \times \paren {-2} - \paren {-3} \times \paren {-1} } - \paren {2 \times 2 - \paren {-1} \times \paren {-1} } + \paren {\paren {-3} \times 2 - \paren {-1} \times \paren {-2} } }\) Definition of Determinant
\(\ds \) \(=\) \(\ds \dfrac 1 2 \size {\paren {6 - \paren {-4} } - \paren {4 - 1} + \paren {8 - \paren {-3} } }\)
\(\ds \) \(\) \(\, \ds + \, \) \(\ds \dfrac 1 2 \size {\paren {\paren {-4} - 3} - \paren {4 - 1} + \paren {\paren {-6} - 2} }\)
\(\ds \) \(=\) \(\ds \dfrac 1 2 \paren {\size {10 - 3 + 11} + \size {-7 - 3 + \paren {-8} } }\)
\(\ds \) \(=\) \(\ds \dfrac 1 2 \paren {\size {18} + \size {-18} }\)
\(\ds \) \(=\) \(\ds \dfrac 1 2 \paren {36}\)
\(\ds \) \(=\) \(\ds 18\)

$\blacksquare$


Sources