Die Gerade $gg$ mit $y=\frac{1}{4}x-1y=\backslash frac\{1\}\{4\}\; x\; -1$ soll mit dem Winkel $\alpha =50\mathrm{°}\backslash alpha=50°$ um das Zentrum $Z(1\mathrm{\mid }2)Z\; (1|2)$ gedreht werden.

Gesucht ist jetzt also die Geradengleichung von $g^{\mathrm{\prime }}g\text{'}$.

Man wählt einen allgemeinen Punkt $P_n(x\mathrm{\mid }\frac{1}{4}x-1)P\_n(x|\backslash frac\{1\}\{4\}x-1)$ auf der Geraden und dreht diesen um $50\mathrm{°}50°$ um $ZZ$.

$\overrightarrow{ZP}=\left(\begin{array}{c}x-1\\ \frac{1}{4}x-1-2\end{array}\right)=\left(\begin{array}{c}x-1\\ \frac{1}{4}x-3\end{array}\right)\backslash overrightarrow\{ZP\}\; =\; \backslash begin\{pmatrix\}x-1\backslash \backslash \; \backslash frac\{1\}\{4\}x-1-2\backslash end\{pmatrix\}=\; \backslash begin\{pmatrix\}\; x-1\backslash \backslash \; \backslash frac\{1\}\{4\}x\; -\; 3\backslash end\{pmatrix\}$

Jetzt muss nur noch die Parallelverschiebung durchgeführt werden:

$\Rightarrow P_n^{\mathrm{\prime }}(\mathrm{0,83}x-\mathrm{1,95}\mathrm{\mid }-\mathrm{0,61}x+\mathrm{0,85})\backslash Rightarrow\; P\_n\text{'}\backslash left(0\{,\}83x-1\{,\}95|-0\{,\}61x+0\{,\}85\backslash right)$

$\Rightarrow P^{\mathrm{\prime }}(\frac{(\sqrt{3}-2)x}{4}+\frac{9-3\sqrt{3}}{2}\mid \frac{x(2\sqrt{3}+1)}{4}-\frac{3\sqrt{3}+5}{2})\backslash Rightarrow\; P\text{'}\backslash left(\backslash left.\backslash frac\{(\backslash sqrt\{3\}-2)x\}\{4\}\; +\; \backslash frac\{9-3\backslash sqrt\{3\}\}\{2\}\backslash right|\backslash frac\{x(2\backslash sqrt\{3\}\; +1)\}\{4\}\; -\; \backslash frac\{3\backslash sqrt\{3\}+5\}\{2\}\backslash right)$

Löse $(1)(1)$ nach $xx$ auf.

Setze $(1^{\mathrm{\prime }})(1\text{'})$ in $(2)(2)$ ein.

$y^{\mathrm{\prime }}=\frac{2\sqrt{3}+1}{\sqrt{3}-2}{x}^{\mathrm{\prime }}+\frac{19(-\sqrt{3}+1)}{2(\sqrt{3}-2)}y\text{'}\; =\; \backslash frac\{2\backslash sqrt3+1\}\{\backslash sqrt3-2\}x\text{'}+\backslash frac\{19(-\backslash sqrt3+1)\}\{2(\backslash sqrt3-2)\}$

Das kannst du nun noch runden und erhähst die Geradengleichung für $g^{\mathrm{\prime }}g\text{'}$.

$y^{\mathrm{\prime }}=\mathrm{16,66}{x}^{\mathrm{\prime }}+\mathrm{25,95}y\text{'}=16\{,\}66x\text{'}+25\{,\}95$

$\overrightarrow{Z{P}_n}=\left(\begin{array}{c}x-(-3)\\ -3x+2-\mathrm{0,5}\end{array}\right)\backslash overrightarrow\{ZP\_n\}\; =\; \backslash begin\{pmatrix\}x-(-3)\backslash \backslash -3x\; +\; 2-0\{,\}5\backslash end\{pmatrix\}$ $=\left(\begin{array}{c}x+3\\ -3x+\mathrm{1,5}\end{array}\right)=\; \backslash begin\{pmatrix\}\; x+3\backslash \backslash -3x\; +\; 1\{,\}5\backslash end\{pmatrix\}$

$\Rightarrow P^{\mathrm{\prime }}(\frac{4}{\sqrt{2}}x+\frac{3-6\sqrt{2}}{2\sqrt{2}}\mid -\frac{2}{\sqrt{2}}x+\frac{9+\sqrt{2}}{2\sqrt{2}})\backslash Rightarrow\; P\text{'}\backslash left(\backslash left.\backslash frac\{4\}\{\backslash sqrt2\}x\; +\; \backslash frac\{3-6\backslash sqrt\{2\}\}\{2\backslash sqrt2\}\backslash right|-\backslash frac\{2\}\{\backslash sqrt2\}x\; +\; \backslash frac\{9+\backslash sqrt\{2\}\}\{2\backslash sqrt2\}\backslash right)$

$\Rightarrow P^{\mathrm{\prime }}(\frac{4}{\sqrt{2}}x+\frac{3-6\sqrt{2}}{2\sqrt{2}}\mid -\frac{2}{\sqrt{2}}x+\frac{9+\sqrt{2}}{2\sqrt{2}})\backslash Rightarrow\; P\text{'}\backslash left(\backslash left.\backslash frac\{4\}\{\backslash sqrt2\}x\; +\; \backslash frac\{3-6\backslash sqrt\{2\}\}\{2\backslash sqrt2\}\backslash right|-\backslash frac\{2\}\{\backslash sqrt2\}x\; +\; \backslash frac\{9+\backslash sqrt\{2\}\}\{2\backslash sqrt2\}\backslash right)$

Löse $(1)(1)$ nach $xx$ auf.

Setze $(1^{\mathrm{\prime }})(1\text{'})$ in $(2)(2)$ ein.

$y^{\mathrm{\prime }}=-\frac{1}{2}{x}^{\mathrm{\prime }}+\frac{21\sqrt{2}-8}{8}y\text{'}=-\backslash frac12x\text{'}+\backslash frac\{21\backslash sqrt2-8\}\{8\}$

Das kannst du nun noch runden und erhälst die Geradengleichung für $g^{\mathrm{\prime }}g\text{'}$.

$g^{\mathrm{\prime }}:y^{\mathrm{\prime }}=-\mathrm{0,5}{x}^{\mathrm{\prime }}+\mathrm{2,71}g\text{'}:\; y\text{'}=-0\{,\}5x\text{'}+2\{,\}71$