Berechne die Koordinaten des Punktes $\mathrm{P}(\mathrm{x}\mathrm{\mid }\mathrm{y})\backslash mathrm\{P(x|y)\}$ mit $\mathrm{x},\mathrm{y}\backslash mathrm\{x,\; y\}$ $\in \mathbb{Q}\backslash in\backslash mathbb\{Q\}$, wenn gilt:

$\mathrm{Q}(4\mathrm{\mid }9)\backslash \; \backslash mathrm\{Q(4|9)\}\backslash$und$\overrightarrow{\text{PQ}}=\left(\begin{array}{c}1\\ 3\end{array}\right)\backslash \; \backslash vec\{\backslash text\{PQ\}\}=\backslash begin\{pmatrix\}\; 1\; \backslash \backslash \; 3\; \backslash end\{pmatrix\}$

Um den Verbindungsvektor zwischen den Punkten $\text{P}\backslash text\{P\}$ und $\text{Q}\backslash text\{Q\}$ zu berechnen, muss man den Ortsvektor zu Punkt $\text{P}\backslash text\{P\}$ vom Ortsvektor zu Punkt $\text{Q}\backslash text\{Q\}$ subtrahieren.

Als Merkregel gilt: "Spitze minus Fuß"

Der Vektor hat also beim Minuend seine Spitze und beim Subtrahend seinen Fuß.

Bekannt ist aus der Aufgabenstellung der Punkt $\text{Q}\backslash text\{Q\}$ und der Vektor $\overrightarrow{\text{PQ}}\backslash vec\{\backslash text\{PQ\}\}$.

$\text{Q}(4\mathrm{/}9)\backslash text\{Q\}\; (4/9)$; $\overrightarrow{\text{PQ}}\left(\begin{array}{c}1\\ 3\end{array}\right)\backslash \; \backslash vec\{\backslash text\{PQ\}\}\backslash begin\{pmatrix\}1\backslash \backslash 3\backslash end\{pmatrix\}$

$\overrightarrow{\text{PQ}}=\left(\begin{array}{c}{\text{Q}}_x-{\text{P}}_x\\ {\text{Q}}_y-{\text{P}}_y\end{array}\right)\backslash vec\{\backslash text\{PQ\}\}=\backslash begin\{pmatrix\}\backslash text\{Q\}\_x-\backslash text\{P\}\_x\backslash \backslash \backslash text\{Q\}\_y-\backslash text\{P\}\_y\backslash end\{pmatrix\}$

$\left(\begin{array}{c}1\\ 3\end{array}\right)=\left(\begin{array}{c}4-\text{x}\\ 9-\text{y}\end{array}\right)\backslash begin\{pmatrix\}1\backslash \backslash 3\backslash end\{pmatrix\}=\backslash begin\{pmatrix\}4-\backslash text\{x\}\backslash \backslash 9-\backslash text\{y\}\backslash end\{pmatrix\}$

$1=4-\text{x}3=9-\text{y}1=\backslash ;\backslash ;\backslash ;4-\backslash text\{x\}\backslash \backslash 3=\backslash ;\backslash ;\backslash ;9-\backslash text\{y\}$

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

$-\text{y}=-6\text{ y}=6-\backslash text\{y\}=-6\backslash \backslash \backslash \; \backslash \; \backslash \; \backslash \; \backslash text\{y\}=\backslash \; \backslash \; 6$

Der Punkt $\text{P}\backslash text\{P\}$ hat die Koordinaten $(\text{x}=3)(\backslash text\{x\}=3)$ und $(\text{y}=6)(\backslash text\{y\}=6)$.