// Numbas version: finer_feedback_settings {"name": "Geradengleichungen", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"name": "Geradengleichungen", "tags": [], "metadata": {"description": "

Find the equation of a straight line which has a given gradient $m$ and passes through the given point $(a,b)$.

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Original: https://numbas.mathcentre.ac.uk/question/11698/equation-of-a-straight-line/ by Newcastle University Mathematics and Statistics,

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translated to German, added Parts b), c), d), e)

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Verschiedene Fragen zu Geradengleichungen.

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a)

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Die Geradengleichung hat die Form $y=mx+b$.

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Dabei ist $m$ die Steigung. Wir können $b$ (den \"$y$-Achenabschnitt\") ausrechnen, indem wir ausnutzen, dass der Punkt mit Koordinaten $x=\\var{a}$ und $y=\\var{b}$ auf der Geraden liegt, dass also diese Werte für $x$ und $y$ die Geradengleichung erfüllen müssen.

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Wir erhalten (hier wird $\\times$ statt $\\cdot$ für die Multiplikation geschrieben):
\\[ \\begin{eqnarray} \\var{b}&=&\\simplify[std]{({b-d}/{a-c}){a}+b} \\Rightarrow\\\\ b&=&\\simplify[std]{{b}-({b-d}/{a-c}){a}={(b*c-a*d)}/{(c-a)}} \\end{eqnarray} \\]

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Die Geradengleichung ist mithin
\\[y = \\simplify[std]{({b-d}/{a-c})x+{b*c-a*d}/{c-a}}\\]

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b)

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Die Geradengleichung hat die Form $y=mx+b$. Dass $P_1$ und $P_2$ beide auf der Geraden liegen, bedeutet, dass

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\\[\\begin{align} \\var{bp1x} m + b & = \\var{bp1y}\\\\ \\var{bp2x} m + b & = \\var{bp2y}\\end{align}\\]

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Wenn wir die beiden Gleichungen voneinander abziehen und nach $m$ auflösen, erhalten wir

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\\[ m = \\frac{\\simplify[basic]{{bp2y}-{bp1y}}}{\\simplify[basic]{{bp2x}-{bp1x}}} = \\simplify{({bp2y}-{bp1y})/({bp2x}-{bp1x})}. \\]

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Dies können wir in die erste Gleichung einsetzen, um $b$ auszurechnen. Das Ergebnis ist

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\\[ y = \\simplify[std]{{bm}*x+{bb}}.\\]

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c)

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Für die gesuchten Koordinaten $x$ und $y$ müssen beide Geradengleichungen erfüllt sein, es gilt also

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\\[\\simplify[std]{{cm1} x + {cb1}} = y = \\simplify[std]{{cm2} x + {cb2}}.\\]

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Diese Gleichung können wir nach $x$ auflösen und erhalten

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\\[ x = \\simplify[std]{{(cb1-cb2)/(cm2-cm1)}}\\quad\\text{und}\\quad y = \\simplify[std]{{(cb1-cb2)/(cm2-cm1)*cm1+cb1}}.\\]

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d)

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Zuerst stellen wir wie in Teil b) die Geradengleichungen für $g$ und $h$ auf:

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\\[ g\\colon y = \\simplify[std]{{dmg} x + {dbg}},\\]

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\\[ h\\colon y = \\simplify[std]{{dmh} x + {dbh}}.\\]

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Dann können wir mit demselben Vorgehen wie in Teil c) den Schnittpunkt dieser beiden Geraden berechnen. Das Ergebnis ist der Punkt

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\\[\\left(\\simplify[std]{{dx}}, \\simplify[std]{{dy}}\\right).\\]

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e)

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Die Geradengleichung für die Gerade durch $P_1$ und $P_2$ ist

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\\[ y = \\simplify[std]{{em}x+{eb}}. \\]

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Für die Koordinaten des Schnittpunkts gilt also $y = \\simplify[std]{{em}x+{eb}}$ und $x=\\var{ea}$, der Schnittpunkt ist folglich $(\\var{ea}, \\simplify[std]{{em*ea+eb}})$.

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Bemerkung: Die Gerade, die durch die Gleichung $x=\\var{ea}$ ist eine senkrechte Gerade. Diese Gerade ist nicht der Funktionsgraph einer linearen Funktion, und die Gerade lässt sich daher nicht durch eine Gleichung der Form $y=mx+b$ beschreiben.

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Geben Sie die Gleichung der Geraden mit

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Geben Sie Ihre Antwort in der Form $y=mx+b$ für geeignete Werte von $m$ und $b$ an.

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Geben Sie $m$ und $b$ als Bruchzahlen oder als ganze Zahlen an.

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Für einen Hinweis klicken Sie Zeige Tipps (es wird 1 Punkt abgezogen).

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$y=\\;\\phantom{{}}$[[0]]

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Die Geradengleichung hat die Form $y=mx+b$.

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Dabei ist $m$ die Steigung. Sie können $b$ (den \"$y$-Achenabschnitt\") ausrechnen, indem Sie ausnutzen, dass der Punkt mit Koordinaten $x=\\var{a}$ und $y=\\var{b}$ auf der Geraden liegt, dass also diese Werte für $x$ und $y$ die Geradengleichung erfüllen müssen.

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Geben Sie die Gleichung der Geraden durch die beiden folgenden Punkte an: $P_1 = (\\var{bp1x}, \\var{bp1y})$, $P_2 = (\\var{bp2x}, \\var{bp2y})$.

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$y=\\;\\phantom{{}}$[[0]]

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Geben Sie den Schnittpunkt $(x,y)$ der folgenden beiden Geraden an:

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$g\\colon y=\\simplify[std]{ {cm1}*x + {cb1}}$, $h\\colon y=\\simplify[std]{ {cm2}*x + {cb2}}$.

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$x=$ [[0]]

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$y=$ [[1]]

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Sei $g$ die Gerade durch die Punkt $P_1 = (\\var{dp1x}, \\var{dp1y})$ und $P_2=(\\var{dp2x}, \\var{dp2y})$, und sei $h$ die Gerade durch die Punkte $P_3 = (\\var{dp3x}, \\var{dp3y})$ und $P_4 = (\\var{dp4x}, \\var{dp4y})$. Geben Sie den Schnittpunkt $(x, y)$ von $g$ und $h$ an.

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$x=$ [[0]]

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$y=$ [[1]]

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Die Geradengleichungen für $g$ und $h$ sind

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\\[ g\\colon y = \\simplify[std]{{dmg} x + {dbg}},\\]

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\\[ h\\colon y = \\simplify[std]{{dmh} x + {dbh}}.\\]

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Geben Sie den Schnittpunkt $(x,y)$ der Geraden $g$ durch die Punkte $P_1 = (\\var{ep1x}, \\var{ep1y})$ und $P_2=(\\var{ep2x}, \\var{ep2y})$ mit der Geraden $x=\\var{ea}$ an.

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$x=$ [[0]]

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$y=$ [[1]]

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Die Geradengleichung für die Gerade durch $P_1$ und $P_2$ ist

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\\[y = \\simplify[std]{{em}*x+{eb}}.\\]

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