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

## 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

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

$\blacksquare$