$\backslash$$\overrightarrow{AB}=\left(\begin{array}{c}{x}_B-x_A\\ y_B-y_A\end{array}\right)\backslash overrightarrow\{\{AB\}\}=\backslash begin\{pmatrix\}x\_\{B\}-x\_\{A\}\backslash \backslash y\_\{B\}-y\_\{A\}\backslash end\{pmatrix\}$ setze die entsprechenden Werte ein:

$\left(\begin{array}{c}2\\ 3\end{array}\right)=\left(\begin{array}{c}{x}_B-(-1)\\ y_B-2\end{array}\right)\backslash begin\{pmatrix\}\; 2\; \backslash \backslash \; 3\; \backslash end\{pmatrix\}\backslash \; =\backslash \; \backslash begin\{pmatrix\}\backslash \; x\_\{B\}-(-1)\backslash \backslash y\_\{B\}-\backslash \; \backslash \; \backslash \; \backslash \; \backslash \; 2\backslash end\{pmatrix\}$

$2=x_B+1\Rightarrow x_B=12\backslash \; =\backslash \; x\_\{B\}\backslash \; +\backslash \; 1\backslash \; \backslash Rightarrow\backslash \; x\_\{B\}\backslash \; =\backslash \; 1$

$3=y_B-2\Rightarrow y_B=53\backslash \; =\backslash \; y\_\{B\}\backslash \; -\backslash \; 2\backslash \; \backslash Rightarrow\backslash \; y\_\{B\}\backslash \; =\backslash \; 5$

Hiermit erhältst Du: $B(1\mathrm{\mid }5)B(1|5\; )$