Nehme $CC$ als Fußpunkt des Dreiecks.

$A_{\mathrm{\Delta }}=\frac{1}{2}\cdot \mathrm{\mid }\overrightarrow{CA}\times \overrightarrow{CB}\mathrm{\mid }A\_\{\backslash Delta\}=\backslash frac\{1\}\{2\}\backslash cdot|\backslash overrightarrow\{CA\}\backslash times\backslash overrightarrow\{CB\}|$

$\overrightarrow{CA}\backslash overrightarrow\{CA\}$ ist der Gegenvektor von $\overrightarrow{AC}\backslash overrightarrow\{AC\}$ => $\overrightarrow{CA}=\left(\begin{array}{c}-4\\ -1\end{array}\right)\backslash overrightarrow\{CA\}\; =\; \backslash begin\{pmatrix\}\; -4\; \backslash \backslash \; -1\; \backslash end\{pmatrix\}$

$\overrightarrow{CB}\backslash overrightarrow\{CB\}$ ist der Gegenvektor von $\overrightarrow{BC}=\left(\begin{array}{c}1\\ 2\end{array}\right)\backslash overrightarrow\{BC\}\; =\; \backslash begin\{pmatrix\}\; 1\; \backslash \backslash \; 2\; \backslash end\{pmatrix\}$ => $\overrightarrow{CB}=\left(\begin{array}{c}-1\\ -2\end{array}\right)\backslash overrightarrow\{CB\}\; =\; \backslash begin\{pmatrix\}\; -1\; \backslash \backslash \; -2\; \backslash end\{pmatrix\}$

$A_{\mathrm{\Delta }}=\mathrm{3,5}FEA\_\{\backslash Delta\}=\backslash \; 3\{,\}5\backslash \; FE$