Berechnen der Koordinaten des Punktes $\mathrm{B}(\mathrm{x}\mathrm{\mid }\mathrm{y})\backslash mathrm\{B(x|y)\}$

$\overrightarrow{AB}=\left(\begin{array}{c}{\text{B}}_x-{\text{A}}_x\\ {\text{B}}_y-{\text{A}}_y\end{array}\right)\backslash overrightarrow\{AB\}=\backslash begin\{pmatrix\}\backslash text\{B\}\_x-\backslash text\{A\}\_x\backslash \backslash \backslash text\{B\}\_y-\backslash text\{A\}\_y\backslash end\{pmatrix\}$

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

$2=x-13=y+22=\backslash ;\backslash ;\backslash ;x-1\backslash \backslash 3=\backslash ;\backslash ;\backslash ;y+2$

$-\text{x}=-3\text{ x}=3-\backslash text\{x\}=-3\backslash \backslash \backslash \; \backslash \; \backslash \; \backslash \; \backslash text\{x\}=\backslash \; \backslash \; 3$

$-\text{y}=-1\text{ y}=1-\backslash text\{y\}=-1\backslash \backslash \backslash \; \backslash \; \backslash \; \backslash \; \backslash text\{y\}=\backslash \; \backslash \; 1$

Der Punkt $\text{B}\backslash text\{B\}$ hat die Koordinaten $B(3\mathrm{\mid }1)B(3|1)$ , also$(\text{x}=3)(\backslash text\{x\}=3)$ und $(\text{y}=1)(\backslash text\{y\}=1)$.