Berechne die Koordinaten des Schnittpunkts $NN$ der Geraden $g_{1}g\_1$ mit der

x-Achse.

Setze$g_1=0\backslash \; g\_1=0\backslash$ und berechne$x\backslash \; x$.

$-\mathrm{0,5}x+3=0-0\{,\}5x+3=0$

Die Koordinaten des Schnittpunktes der Geraden $g_1\backslash \; g\_1\backslash$mit der $x-x-$Achse sind dann:

$N(6\mathrm{\mid }0)\backslash \; N(6|0)$

Berechne $xx$ und $yy$ indem Du die entsprechenden Werte in die Funktionsgleichung einsetzt.

Berechne y:

$-\mathrm{0,5}x+3=y\Rightarrow -\mathrm{0,5}\cdot 5+3=y\Rightarrow y=-\mathrm{2,5}+3\Rightarrow y=\mathrm{0,5}\backslash \; -0\{,\}5x+3=y\backslash quad\backslash Rightarrow\backslash quad-0\{,\}5\backslash cdot5+3=y\backslash quad\backslash Rightarrow\backslash quad\; y=-2\{,\}5+3\backslash hspace\{1.8mm\}\backslash Rightarrow\backslash hspace\{1mm\}\; y=0\{,\}5$

Berechne x:

$-\mathrm{0,5}x+3=y\Rightarrow -\mathrm{0,5}x+3=21\Rightarrow -\mathrm{0,5}x=18\Rightarrow x=-36\backslash \; -0\{,\}5x+3=y\backslash quad\backslash Rightarrow\backslash quad-0\{,\}5x+3=21\backslash quad\backslash Rightarrow\backslash quad\; -0\{,\}5x=18\backslash quad\backslash Rightarrow\backslash quad\; x=-36$

Übertrage die Werte in die Wertetabelle.

$\text{Wertetabelle:}\backslash text\{\backslash Large\{Wertetabelle:\}\}$

$g_1:y=-\mathrm{0,5}x+3\Rightarrow m_1=-\mathrm{0,5}g\_\{1\}:y=-0\{,\}5x+3\backslash qquad\backslash Rightarrow\backslash qquad\; m\_\{1\}=-0\{,\}5$

da die Gerade $g_{2}g\_\{2\}\backslash$senkrecht auf der Geraden$g_1\backslash \; g\_\{1\}\backslash$steht, gilt: $m_1\cdot m_2=-1\Rightarrow m_2={\displaystyle \frac{-1}{m_1}}={\displaystyle \frac{-1}{-\mathrm{0,5}}}\Rightarrow m_2=2\backslash \backslash m\_\{1\}\backslash cdot\backslash \; m\_\{2\}\backslash \; =-1\backslash quad\backslash Rightarrow\backslash quad\backslash \; m\_\{2\}\backslash \; =\backslash \; \backslash dfrac\{-1\}\{m\_\{1\}\}=\backslash dfrac\{-1\}\{-0\{,\}5\}\backslash quad\backslash Rightarrow\backslash quad\; m\_\{2\}=2$

Die allgemeine Form der Geradengleichung lautet:

$y=mx+t\backslash hspace\{51mm\}y\backslash \; =\backslash \; mx\backslash \; +\backslash \; t$

Setze $m_2\backslash \; m\_\{2\}\backslash$und die Koordinaten des Punktes $BB$ in die Gleichung ein.

$0=2\cdot (-\mathrm{2,5})+t_2\Rightarrow t_2=5\backslash \; 0\backslash \; =\backslash \; 2\backslash cdot\; (-2\{,\}5)\backslash \; +\backslash \; t\_2\backslash quad\backslash Rightarrow\backslash quad\; t\_2\backslash \; =\backslash \; 5$

Die Funktionsgleichung der Geraden $g_{2}g\_\{2\}\backslash$lautet:

$g_2:y=2x+5\backslash \; g\_\{2\}:y\backslash \; =\backslash \; 2x\backslash \; +\backslash \; 5$

Die Skizze ist nicht massgetreu.

Die allgemeine Geradengleichung lautet:$y=m\cdot x+t\backslash quad\; y=\backslash textcolor\{ff6600\}\{m\}\backslash cdot\; x+\backslash textcolor\{009999\}\{t\}$

Setze $CC$ und $DD$ in die Geradengleichung ein.

$(C):-1m+t=-1(C):-1m\backslash \; +\backslash \; t\backslash \; =\backslash \; -1$

$(D):4m+t=1\Rightarrow t=1-4m(D):\backslash \; \backslash \; \backslash \; 4m\backslash \; +\backslash \; t\backslash \; =\backslash \; \backslash \; \backslash \; \backslash \; 1\backslash quad\backslash Rightarrow\backslash quad\backslash \; t\backslash \; =\backslash \; 1-4m\backslash$

Setze $mm$ in$(D)(D)$ ein und berechne $tt$.

$t=1-4\cdot {\displaystyle \frac{2}{5}}\Rightarrow t={\displaystyle \frac{5}{5}}-{\displaystyle \frac{8}{5}}=-{\displaystyle \frac{3}{5}}\Rightarrow t=-{\displaystyle \frac{3}{5}}t\backslash \; =\backslash \; 1-4\backslash cdot\backslash dfrac\{2\}\{5\}\backslash quad\backslash Rightarrow\backslash quad\; t\backslash \; =\backslash \; \backslash dfrac\{5\}\{5\}-\backslash dfrac\{8\}\{5\}=-\backslash dfrac\{3\}\{5\}\backslash quad\backslash Rightarrow\backslash quad\; t\backslash \; =\backslash \; -\backslash dfrac\{3\}\{5\}$

Die Funktionsgleichung der Geraden $g_{3}g\_3\backslash$ lautet:

$g_3:y={\displaystyle \frac{2}{5}}x-{\displaystyle \frac{3}{5}}g\_\{3\}:\backslash \; y\backslash \; =\backslash \; \backslash dfrac\{2\}\{5\}x-\backslash dfrac\{3\}\{5\}$

Berechnung des Schnittpunktes $TT$.

Stelle die Geradengleichung $g_4\backslash \; g\_\{4\}\backslash$nach $y\backslash \; y\backslash$um und setze $g_1=g4g\_\{1\}=g\{4\}$.

$g_4:\text{ –}\mathrm{0,5}x=\text{–}5\text{ – }y\Rightarrow y=\mathrm{0,5}x-5g\_\{4\}:\; \backslash \; –0\{,\}5x\; =\; –5\backslash \; –\backslash \; y\backslash qquad\backslash Rightarrow\backslash qquad\; y=0\{,\}5x-5$

Setze$g_1=g_4\backslash \; g\_\{1\}=g\_\{4\}$

Setze $x=8x=8\backslash$in $g_{1}g\_1\backslash$oder$g_4\backslash \; g\_4\backslash$ein und berechne $y\mathrm{.}y.$

$xx$ eingesetzt in $g_{1}g\_1\backslash$ergibt:

$-\mathrm{0,5}\cdot 8+3=y\Rightarrow y=-1\backslash \; -0\{,\}5\backslash cdot8+3=y\backslash quad\backslash Rightarrow\backslash quad\; y=-1$

Die Koordinaten des Schnittpunktes sind dann:

$T(8\mathrm{\mid }-1)T(8|-1)$

Die allgemeine Geradengleichung lautet:$y=m\cdot x+t\backslash quad\; y=\backslash textcolor\{ff6600\}\{m\}\backslash cdot\; x+\backslash textcolor\{009999\}\{t\}$

Den $yy$-Achsenabschnitt kannst Du aus der Skizze ablesen:$t=1\backslash \; t=1$

Setze die Koordinaten des Schnittpunkts der Geraden$g_5\backslash \; g\_\{5\}\backslash$mit der $xx$-Achse in die Geradengleichung ein, und berechne die Steigung $mm$.

$y=mx+t0=-4m+1\mathrm{\mid }+(4m)4m=1\mathrm{\mid }:(4)\Rightarrow m={\displaystyle \frac{1}{4}}\backslash \; y=\backslash \; \backslash \; \backslash \; mx+t\backslash \backslash \backslash \; 0=-4m+1\backslash \; |+(4m)\backslash \backslash \backslash \; 4m=1\backslash \; |:(4)\backslash qquad\backslash Rightarrow\backslash qquad\; m=\backslash dfrac\{1\}\{4\}$

Die Funktionsgleichung der Geraden $g_{5}g\_5\backslash$ lautet:

$g_5:y={\displaystyle \frac{1}{4}}x+1g\_\{5\}:y=\backslash dfrac\{1\}\{4\}x+1$

Der Tangens im rechwinkligen Dreieck ist definiert als:

$\tan (\alpha )={\displaystyle \frac{\text{Gegenkathete}}{\text{Ankathete}}}\backslash quad\backslash tan(\backslash alpha)=\backslash dfrac\{\backslash text\{Gegenkathete\}\}\{\backslash text\{Ankathete\}\}$

$\tan (\alpha )={\displaystyle \frac{1}{4}}\Rightarrow \alpha ={\mathrm{14,04}}^{\circ }\backslash quad\backslash tan(\backslash alpha)=\backslash dfrac\{1\}\{4\}\backslash qquad\backslash Rightarrow\backslash qquad\; \backslash alpha=14\{,\}04^\{\backslash circ\}$

Der Winkel $\alpha \backslash alpha\backslash$beträgt:$\alpha \approx {14}^{\circ }\backslash hspace\{10mm\}\backslash alpha\backslash approx14^\{\backslash circ\}$