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Flächeninhalt und Umfang der eingeschlossenen Fläche berechnen?

y1 = -x+1

y2 =-1/2*x-1

y3 = 2*x+4

 18.06.2015

Beste Antwort 

 #5
avatar+26318 
+5

Flächeninhalt und Umfang der eingeschlossenen Fläche berechnen?

y1 = -x+1

y2 =-1/2*x-1

y3 = 2*x+4

Gegebene Geradengleichungen der Form y = mx + b  oder y - mx = b

$$\small{\text{$
\begin{array}{llll}
(1) & y = -x+1 & m_1 = -1 & b_1 = 1 \\
(2) & y = -\frac{1}{2}x-1 & m_2 = -\frac{1}{2} & b_2 = -1 \\
(3) & y = 2x+4 & m_3 = 2 & b_3 = 4 \\
\end{array}
$}}$$

 

I. Berechnung der Schnittpunkte:

1. Schnitt $$\small{\text{$S_1$}}$$ Gerade (1) mit Gerade (2 ):

$$\small{\text{$
\begin{array}{rcl}
1\cdot y_s - m_1 \cdot x_s &=& b_1 \\
1\cdot y_s - m_2 \cdot x_s &=& b_2 \\
\\
\hline
\\
y_s = \dfrac
{\begin{vmatrix} b_1 & -m_1\\ b_2 &-m_2 \end{vmatrix}
}
{\begin{vmatrix} 1 & -m_1\\ 1 &-m_2 \end{vmatrix}
}
=\dfrac
{\begin{vmatrix} 1 & -(-1)\\ -1 & -(-\frac{1}{2}) \end{vmatrix}
}
{\begin{vmatrix} 1 & -(-1)\\ 1 & -(-\frac{1}{2}) \end{vmatrix}
}
=\dfrac
{\begin{vmatrix} 1 & 1\\ -1 & \frac{1}{2} \end{vmatrix}
}
{\begin{vmatrix} 1 & 1\\ 1 & \frac{1}{2} \end{vmatrix}
}
=\dfrac
{ 1\cdot \frac{1}{2} - (-1)\cdot 1 }
{ 1\cdot \frac{1}{2} - 1\cdot 1 }
=\dfrac
{ \frac{1}{2} + 1 }
{ \frac{1}{2} - 1 }
=\dfrac
{ \frac{3}{2} }
{ -\frac{1}{2} }
= -3\\\\
x_s = \dfrac
{\begin{vmatrix} 1 & b_1\\ 1 & b_2 \end{vmatrix}
}
{\begin{vmatrix} 1 & -m_1\\ 1 &-m_2 \end{vmatrix}
}
=\dfrac
{\begin{vmatrix} 1 & 1\\ 1 & -1\end{vmatrix}
}
{\begin{vmatrix} 1 & -(-1)\\ 1 & -(-\frac{1}{2}) \end{vmatrix}
}
=\dfrac
{\begin{vmatrix} 1 & 1\\ 1 & -1\end{vmatrix}
}
{\begin{vmatrix} 1 & 1\\ 1 & \frac{1}{2} \end{vmatrix}
}
=\dfrac
{ 1\cdot (-1) - 1\cdot 1 }
{ 1\cdot \frac{1}{2} - 1\cdot 1 }
=\dfrac
{ -1-1 }
{ \frac{1}{2} - 1 }
=\dfrac
{ -2 }
{ -\frac{1}{2} }
= 4
\end{array}
$}}$$

 

 

2. Schnitt $$\small{\text{$S_2$}}$$ Gerade (2) mit Gerade (3):

$$\small{\text{$
\begin{array}{rcl}
1\cdot y_s - m_2 \cdot x_s &=& b_2 \\
1\cdot y_s - m_3 \cdot x_s &=& b_3 \\
\\
\hline
\\
y_s = \dfrac
{\begin{vmatrix} b_2 & -m_2\\ b_3 &-m_3 \end{vmatrix}
}
{\begin{vmatrix} 1 & -m_2\\ 1 &-m_3 \end{vmatrix}
}
=\dfrac
{\begin{vmatrix} -1 & -\frac{1}{2}\\ 4 & 2\end{vmatrix}
}
{\begin{vmatrix} 1 & -(-\frac{1}{2})\\ 1 & -2\end{vmatrix}
}
=\dfrac
{\begin{vmatrix} -1 & -\frac{1}{2}\\ 4 & 2\end{vmatrix}
}
{\begin{vmatrix} 1 & \frac{1}{2} \\ 1 & -2\end{vmatrix}
}
=\dfrac
{ -1\cdot 2 - 4 \cdot(- \frac{1}{2}) }
{ 1\cdot (- 2) - 1 \cdot \frac{1}{2} }
=\dfrac
{ -2+2}
{ -2 - \frac{1}{2} }
=\dfrac
{ 0 }
{ -\frac{5}{2} }
= 0\\\\
x_s = \dfrac
{\begin{vmatrix} 1 & b_2\\ 1 & b_3 \end{vmatrix}
}
{\begin{vmatrix} 1 & -m_2\\ 1 &-m_3 \end{vmatrix}
}
=\dfrac
{\begin{vmatrix} 1 & -1\\ 1 & 4\end{vmatrix}
}
{\begin{vmatrix} 1 & -(-\frac{1}{2})\\ 1 & -2\end{vmatrix}
}
=\dfrac
{\begin{vmatrix} 1 & -1\\ 1 & 4\end{vmatrix}
}
{\begin{vmatrix} 1 & \frac{1}{2} \\ 1 & -2\end{vmatrix}
}
=\dfrac
{ 1\cdot 4 - 1\cdot (- 1) }
{ 1\cdot (- 2) - 1 \cdot \frac{1}{2} }
=\dfrac
{ 4+1 }
{ -2 - \frac{1}{2} }
=\dfrac
{ 5 }
{ -\frac{5}{2} }
= -2
\end{array}
$}}$$

 

3. Schnitt $$\small{\text{$S_3$}}$$ Gerade (3) mit Gerade (1):

$$\small{\text{$
\begin{array}{rcl}
1\cdot y_s - m_3 \cdot x_s &=& b_3 \\
1\cdot y_s - m_1 \cdot x_s &=& b_1 \\
\\
\hline
\\
y_s = \dfrac
{\begin{vmatrix} b_3 & -m_3\\ b_1 &-m_1 \end{vmatrix}
}
{\begin{vmatrix} 1 & -m_3\\ 1 &-m_1 \end{vmatrix}
}
=\dfrac
{\begin{vmatrix} 4 & -2\\ 1 & -(-1)\end{vmatrix}
}
{\begin{vmatrix} 1 & -2\\ 1 & -(-1)\end{vmatrix}
}
=\dfrac
{\begin{vmatrix} 4 & -2\\ 1 & 1\end{vmatrix}
}
{\begin{vmatrix} 1 & -2\\ 1 & 1\end{vmatrix}
}
=\dfrac
{ 4\cdot 1 - 1 \cdot (-2) }
{ 1\cdot 1 - 1 \cdot (-2) }
=\dfrac
{ 4+2}
{ 1+2 }
=\dfrac
{ 6 }
{ 3 }
= 2\\\\
x_s = \dfrac
{\begin{vmatrix} 1 & b_3\\ 1 & b_1 \end{vmatrix}
}
{\begin{vmatrix} 1 & -m_3\\ 1 &-m_1 \end{vmatrix}
}
=\dfrac
{\begin{vmatrix} 1 & 4\\ 1 & 1\end{vmatrix}
}
{\begin{vmatrix} 1 & -2\\ 1 & -(-1)\end{vmatrix}
}
=\dfrac
{\begin{vmatrix} 1 & 4\\ 1 & 1\end{vmatrix}
}
{\begin{vmatrix} 1 & -2\\ 1 & 1\end{vmatrix}
}
=\dfrac
{ 1\cdot 1 - 1\cdot 4 }
{ 1\cdot 1 - 1 \cdot (-2) }
=\dfrac
{ 1-4 }
{ 1+2 }
=\dfrac
{ -3 }
{ 3 }
= -1
\end{array}
$}}$$

 

Die 3 Schnittpunkte zusammengefasst:

$$\small{\text{$
\begin{array}{|l|r|c|}
\hline
& x & y \\
\hline
S_1 &4 & -3 \\
S_2 &-2 & 0 \\
S_3 &-1 & 2 \\
\hline
\end{array}
$}}$$

 

Die Strecke von $$\small{\text{$S_1 $ nach $ S_2$}}$$:

$$\small{\text{$ \begin{array}{lrc} \overline{S_1S_2} =\sqrt{ [(4)-(-2)]^2 + [0-(-3)]^2 } =\sqrt{ 6^2 + 3^2 } =\sqrt{ 36 + 9 } =\sqrt{ 45 } =6,70820393250 \end{array} $}}$$

 

Die Strecke von $$\small{\text{$S_2 $ nach $ S_3$}}$$:

$$\small{\text{$ \begin{array}{lrc} \overline{S_2S_3} =\sqrt{ [(-1)-(-2)]^2 + (2-0)^2 } =\sqrt{ (-1+2)^2 + 2^2 } =\sqrt{ 1^2 + 4 } =\sqrt{ 5 } =2.23606797750 \end{array} $}}$$

 

Die Strecke von $$\small{\text{$S_3 $ nach $ S_1$}}$$:

$$\small{\text{$ \begin{array}{lrc} \overline{S_2S_3} =\sqrt{ [4-(-1)]^2 + (-3-2)^2 } =\sqrt{ (4+1)^2 + (-5)^2 } =\sqrt{ 5^2 + 5^2 } =\sqrt{ 50 } =7.07106781187 \end{array} $}}$$

 

Der Umfang der eingeschlossenen Fläche ( Dreieck) beträgt: $$\small{\text{$\overline{S_1S_2}+\overline{S_2S_3}+\overline{S_2S_3} =6,70820393250+2,23606797750+7,07106781187 =16,0153397219 $}}$$

 

Die Fläche des Dreiecks nach Heron:

$$\small{\text{$
\boxed{~~ A =\sqrt{ s\cdot(s-a)\cdot(s-b)\cdot(s-c)} \qquad
s = (a + b + c)/2 ~~}
$}}\\\\
\small{\text{$
\begin{array}{lcl}
a= \overline{S_1S_2} = 6,70820393250\\
b= \overline{S_2S_3} = 2,23606797750\\
c= \overline{S_3S_1} = 7,07106781187 \\\\
s = \dfrac{16,0153397219}{2}= 8,00766986093\\
\end{array}
$}}\\\\\\
\small{\text{$
\begin{array}{lcl}
A =\sqrt{ 8,00766986093\cdot(8,00766986093-6,70820393250)\cdot(8,00766986093-2,23606797750)\cdot(8,00766986093-7,07106781187)}\\
A =\sqrt{ 8,00766986093\cdot 1,29946592843 \cdot 5,77160188343\cdot 0,93660204906}\\
A =\sqrt{ 56.2499999994 }\\
A=7,5
\end{array}
$}}\\\\$$

Der Flächeninhalt der eingeschlossenen Fläche ( Dreieck) beträgt 7,5

 

Probe der Fläche:

$$\small{\text{$ \begin{array}{|l|r|c|} \hline & x & y \\ \hline S_1 &4 & -3 \\ S_2 &-2 & 0 \\ S_3 &-1 & 2 \\ \hline \end{array}
$}}\\\\
\small{\text{$
\boxed{
~~
A = \dfrac{1}{2}\cdot \begin{vmatrix}
1 & 1 & 1 \\
x_{s_1} & x_{s_2} & x_{s_3} \\
y_{s_1} & y_{s_2} & y_{s_3} \\
\end{vmatrix}
~~
}
$}}\\\\
\small{\text{$
\begin{array}{rcl}
A &=&\dfrac{1}{2}\cdot \begin{vmatrix}
1 & 1 & 1 \\
4 & -2 & -1 \\
-3 & 0 & 2 \\
\end{vmatrix} \\\\
&=&\frac{1}{2} \cdot
[ 1\cdot (-2) \cdot 2 + 4\cdot 0 \cdot 1 + (-3)\cdot 1 \cdot (-1)
- (-3)\cdot (-2) \cdot 1 - 0\cdot (-1) \cdot 1 - 4\cdot 1 \cdot 2
]\\\\
&=&\frac{1}{2} \cdot
[ 1\cdot (-2) \cdot 2 + (-3)\cdot 1 \cdot (-1)
- (-3)\cdot (-2) \cdot 1 - 4\cdot 1 \cdot 2
]\\\\
&=&\frac{1}{2} \cdot
[ -4 + 3 - 6 \cdot 1 - 8]\\\\
&=&\frac{1}{2} \cdot
[ -15 ]\\\\
&=& - 7,5 \\\\
|A| &=& 7,5 \qquad \mathrm{okay}
\end{array}
$}}$$

 19.06.2015
 #1
avatar
0
y (1) = -x+1
y (2) =-1/2*x-1
y (3) = 2*x+4

Erstmal würde ich die Schnittpunkte berechnen:

$$\underset{\,\,\,\,{\textcolor[rgb]{0.66,0.66,0.66}{\rightarrow {\mathtt{y, x}}}}}{{solve}}{\left(\begin{array}{l}{\mathtt{y}}={\mathtt{\,-\,}}{\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}}\\
{\mathtt{y}}={\frac{{\mathtt{1}}}{{\mathtt{2}}}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,-\,}}{\mathtt{1}}\end{array}\right)} \Rightarrow \left\{ \begin{array}{l}{\mathtt{y}} = {\mathtt{\,-\,}}{\frac{{\mathtt{1}}}{{\mathtt{3}}}}\\
{\mathtt{x}} = {\frac{{\mathtt{4}}}{{\mathtt{3}}}}\\
\end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{y}} = -{\mathtt{0.333\: \!333\: \!333\: \!333\: \!333\: \!3}}\\
{\mathtt{x}} = {\mathtt{1.333\: \!333\: \!333\: \!333\: \!333\: \!3}}\\
\end{array} \right\}$$

$$\underset{\,\,\,\,{\textcolor[rgb]{0.66,0.66,0.66}{\rightarrow {\mathtt{y, x}}}}}{{solve}}{\left(\begin{array}{l}{\mathtt{y}}={\frac{{\mathtt{1}}}{{\mathtt{2}}}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,-\,}}{\mathtt{1}}\\
{\mathtt{y}}={\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{4}}\end{array}\right)} \Rightarrow \left\{ \begin{array}{l}{\mathtt{y}} = {\mathtt{\,-\,}}{\frac{{\mathtt{8}}}{{\mathtt{3}}}}\\
{\mathtt{x}} = {\mathtt{\,-\,}}{\frac{{\mathtt{10}}}{{\mathtt{3}}}}\\
\end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{y}} = -{\mathtt{2.666\: \!666\: \!666\: \!666\: \!666\: \!7}}\\
{\mathtt{x}} = -{\mathtt{3.333\: \!333\: \!333\: \!333\: \!333\: \!3}}\\
\end{array} \right\}$$

$$\underset{\,\,\,\,{\textcolor[rgb]{0.66,0.66,0.66}{\rightarrow {\mathtt{y, x}}}}}{{solve}}{\left(\begin{array}{l}{\mathtt{y}}={\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{4}}\\
{\mathtt{y}}={\mathtt{\,-\,}}{\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}}\end{array}\right)} \Rightarrow \left\{ \begin{array}{l}{\mathtt{y}} = {\mathtt{2}}\\
{\mathtt{x}} = -{\mathtt{1}}\\
\end{array} \right\}$$

Lasst mich nachdenken...

 19.06.2015
 #2
avatar
0

Nun zum Umfang:

P(y1/y2)=(4/3;-1/3)

P(y2/y3)=(-10/3;-8/3)

P(y3/y1)=(-1/2)

$${\mathtt{Strecke}} = {\sqrt{{\left[{\mathtt{x2}}{\mathtt{\,-\,}}{\mathtt{x1}}\right]}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\left[{\mathtt{y2}}{\mathtt{\,-\,}}{\mathtt{y1}}\right]}^{{\mathtt{2}}}}}$$

Umfang = Strecke P(y1/y2) P(y2/y3) + Strecke P(y2/y3) P(y3/y1) + Strecke P(y3/y1) P(y1/y2) =

$${\mathtt{Umfang}} = {\sqrt{{\left[\left({\frac{{\mathtt{4}}}{{\mathtt{3}}}}\right){\mathtt{\,\small\textbf+\,}}\left({\frac{{\mathtt{10}}}{{\mathtt{3}}}}\right)\right]}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\left[{\mathtt{\,-\,}}\left({\frac{{\mathtt{1}}}{{\mathtt{3}}}}\right){\mathtt{\,\small\textbf+\,}}\left({\frac{{\mathtt{8}}}{{\mathtt{3}}}}\right)\right]}^{{\mathtt{2}}}}}{\mathtt{\,\small\textbf+\,}}{\sqrt{{\left[{\mathtt{\,-\,}}\left({\frac{{\mathtt{10}}}{{\mathtt{3}}}}\right){\mathtt{\,\small\textbf+\,}}\left({\mathtt{1}}\right)\right]}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\left[{\mathtt{\,-\,}}\left({\frac{{\mathtt{8}}}{{\mathtt{3}}}}\right){\mathtt{\,-\,}}\left({\mathtt{2}}\right)\right]}^{{\mathtt{2}}}}}{\mathtt{\,\small\textbf+\,}}{\sqrt{{\left[{\mathtt{\,-\,}}\left({\mathtt{1}}\right){\mathtt{\,-\,}}\left({\frac{{\mathtt{4}}}{{\mathtt{3}}}}\right)\right]}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\left[\left({\mathtt{2}}\right){\mathtt{\,\small\textbf+\,}}\left({\frac{{\mathtt{1}}}{{\mathtt{3}}}}\right)\right]}^{{\mathtt{2}}}}} \Rightarrow {\mathtt{Umfang}} = {\mathtt{13.734\: \!815\: \!540\: \!536\: \!239\: \!6}}$$

Schaut bitte mal drüber ob alles stimmt.

 19.06.2015
 #3
avatar
0

Nun zur Fläche:

$${\mathtt{Flaeche}} = {\frac{{\mathtt{1}}}{{\mathtt{4}}}}{\mathtt{\,\times\,}}{\sqrt{{\left[{{\mathtt{a}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{{\mathtt{b}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{{\mathtt{c}}}^{{\mathtt{2}}}\right]}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{2}}{\mathtt{\,\times\,}}\left[{{\mathtt{a}}}^{{\mathtt{4}}}{\mathtt{\,\small\textbf+\,}}{{\mathtt{b}}}^{{\mathtt{4}}}{\mathtt{\,\small\textbf+\,}}{{\mathtt{c}}}^{{\mathtt{4}}}\right]}}$$

a = Strecke P(y1/y2) P(y2/y3) = $${\sqrt{{\left[\left({\frac{{\mathtt{4}}}{{\mathtt{3}}}}\right){\mathtt{\,\small\textbf+\,}}\left({\frac{{\mathtt{10}}}{{\mathtt{3}}}}\right)\right]}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\left[{\mathtt{\,-\,}}\left({\frac{{\mathtt{1}}}{{\mathtt{3}}}}\right){\mathtt{\,\small\textbf+\,}}\left({\frac{{\mathtt{8}}}{{\mathtt{3}}}}\right)\right]}^{{\mathtt{2}}}}}$$

b = Strecke P(y2/y3) P(y3/y1) = $${\sqrt{{\left[{\mathtt{\,-\,}}\left({\frac{{\mathtt{10}}}{{\mathtt{3}}}}\right){\mathtt{\,\small\textbf+\,}}\left({\mathtt{1}}\right)\right]}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\left[{\mathtt{\,-\,}}\left({\frac{{\mathtt{8}}}{{\mathtt{3}}}}\right){\mathtt{\,-\,}}\left({\mathtt{2}}\right)\right]}^{{\mathtt{2}}}}}$$

C = Strecke P(y3/y1) P(y1/y2) = $${\sqrt{{\left[{\mathtt{\,-\,}}\left({\mathtt{1}}\right){\mathtt{\,-\,}}\left({\frac{{\mathtt{4}}}{{\mathtt{3}}}}\right)\right]}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\left[\left({\mathtt{2}}\right){\mathtt{\,\small\textbf+\,}}\left({\frac{{\mathtt{1}}}{{\mathtt{3}}}}\right)\right]}^{{\mathtt{2}}}}}$$

 

 19.06.2015
 #4
avatar
0

Ach die Formel will einfach nicht, sie sollte so aussehen:

Flaeche = 1/4*sqrt([(sqrt([(4/3)-(-10/3)]^2+[(-1/3)-(-8/3)]^2)^2)+(sqrt([(-10/3)-(-1)]^2+[(-8/3)-(2)]^2)^2)+(sqrt([(-1)-(4/3)]^2+[(2)-(-1/3)]^2)^2)]^2-2[(sqrt([(4/3)-(-10/3)]^2+[(-1/3)-(-8/3)]^2)^4)+(sqrt([(-10/3)-(-1)]^2+[(-8/3)-(2)]^2)^4)+(sqrt([(-1)-(4/3)]^2+[(2)-(-1/3)]^2)^4)])

 19.06.2015
 #5
avatar+26318 
+5
Beste Antwort

Flächeninhalt und Umfang der eingeschlossenen Fläche berechnen?

y1 = -x+1

y2 =-1/2*x-1

y3 = 2*x+4

Gegebene Geradengleichungen der Form y = mx + b  oder y - mx = b

$$\small{\text{$
\begin{array}{llll}
(1) & y = -x+1 & m_1 = -1 & b_1 = 1 \\
(2) & y = -\frac{1}{2}x-1 & m_2 = -\frac{1}{2} & b_2 = -1 \\
(3) & y = 2x+4 & m_3 = 2 & b_3 = 4 \\
\end{array}
$}}$$

 

I. Berechnung der Schnittpunkte:

1. Schnitt $$\small{\text{$S_1$}}$$ Gerade (1) mit Gerade (2 ):

$$\small{\text{$
\begin{array}{rcl}
1\cdot y_s - m_1 \cdot x_s &=& b_1 \\
1\cdot y_s - m_2 \cdot x_s &=& b_2 \\
\\
\hline
\\
y_s = \dfrac
{\begin{vmatrix} b_1 & -m_1\\ b_2 &-m_2 \end{vmatrix}
}
{\begin{vmatrix} 1 & -m_1\\ 1 &-m_2 \end{vmatrix}
}
=\dfrac
{\begin{vmatrix} 1 & -(-1)\\ -1 & -(-\frac{1}{2}) \end{vmatrix}
}
{\begin{vmatrix} 1 & -(-1)\\ 1 & -(-\frac{1}{2}) \end{vmatrix}
}
=\dfrac
{\begin{vmatrix} 1 & 1\\ -1 & \frac{1}{2} \end{vmatrix}
}
{\begin{vmatrix} 1 & 1\\ 1 & \frac{1}{2} \end{vmatrix}
}
=\dfrac
{ 1\cdot \frac{1}{2} - (-1)\cdot 1 }
{ 1\cdot \frac{1}{2} - 1\cdot 1 }
=\dfrac
{ \frac{1}{2} + 1 }
{ \frac{1}{2} - 1 }
=\dfrac
{ \frac{3}{2} }
{ -\frac{1}{2} }
= -3\\\\
x_s = \dfrac
{\begin{vmatrix} 1 & b_1\\ 1 & b_2 \end{vmatrix}
}
{\begin{vmatrix} 1 & -m_1\\ 1 &-m_2 \end{vmatrix}
}
=\dfrac
{\begin{vmatrix} 1 & 1\\ 1 & -1\end{vmatrix}
}
{\begin{vmatrix} 1 & -(-1)\\ 1 & -(-\frac{1}{2}) \end{vmatrix}
}
=\dfrac
{\begin{vmatrix} 1 & 1\\ 1 & -1\end{vmatrix}
}
{\begin{vmatrix} 1 & 1\\ 1 & \frac{1}{2} \end{vmatrix}
}
=\dfrac
{ 1\cdot (-1) - 1\cdot 1 }
{ 1\cdot \frac{1}{2} - 1\cdot 1 }
=\dfrac
{ -1-1 }
{ \frac{1}{2} - 1 }
=\dfrac
{ -2 }
{ -\frac{1}{2} }
= 4
\end{array}
$}}$$

 

 

2. Schnitt $$\small{\text{$S_2$}}$$ Gerade (2) mit Gerade (3):

$$\small{\text{$
\begin{array}{rcl}
1\cdot y_s - m_2 \cdot x_s &=& b_2 \\
1\cdot y_s - m_3 \cdot x_s &=& b_3 \\
\\
\hline
\\
y_s = \dfrac
{\begin{vmatrix} b_2 & -m_2\\ b_3 &-m_3 \end{vmatrix}
}
{\begin{vmatrix} 1 & -m_2\\ 1 &-m_3 \end{vmatrix}
}
=\dfrac
{\begin{vmatrix} -1 & -\frac{1}{2}\\ 4 & 2\end{vmatrix}
}
{\begin{vmatrix} 1 & -(-\frac{1}{2})\\ 1 & -2\end{vmatrix}
}
=\dfrac
{\begin{vmatrix} -1 & -\frac{1}{2}\\ 4 & 2\end{vmatrix}
}
{\begin{vmatrix} 1 & \frac{1}{2} \\ 1 & -2\end{vmatrix}
}
=\dfrac
{ -1\cdot 2 - 4 \cdot(- \frac{1}{2}) }
{ 1\cdot (- 2) - 1 \cdot \frac{1}{2} }
=\dfrac
{ -2+2}
{ -2 - \frac{1}{2} }
=\dfrac
{ 0 }
{ -\frac{5}{2} }
= 0\\\\
x_s = \dfrac
{\begin{vmatrix} 1 & b_2\\ 1 & b_3 \end{vmatrix}
}
{\begin{vmatrix} 1 & -m_2\\ 1 &-m_3 \end{vmatrix}
}
=\dfrac
{\begin{vmatrix} 1 & -1\\ 1 & 4\end{vmatrix}
}
{\begin{vmatrix} 1 & -(-\frac{1}{2})\\ 1 & -2\end{vmatrix}
}
=\dfrac
{\begin{vmatrix} 1 & -1\\ 1 & 4\end{vmatrix}
}
{\begin{vmatrix} 1 & \frac{1}{2} \\ 1 & -2\end{vmatrix}
}
=\dfrac
{ 1\cdot 4 - 1\cdot (- 1) }
{ 1\cdot (- 2) - 1 \cdot \frac{1}{2} }
=\dfrac
{ 4+1 }
{ -2 - \frac{1}{2} }
=\dfrac
{ 5 }
{ -\frac{5}{2} }
= -2
\end{array}
$}}$$

 

3. Schnitt $$\small{\text{$S_3$}}$$ Gerade (3) mit Gerade (1):

$$\small{\text{$
\begin{array}{rcl}
1\cdot y_s - m_3 \cdot x_s &=& b_3 \\
1\cdot y_s - m_1 \cdot x_s &=& b_1 \\
\\
\hline
\\
y_s = \dfrac
{\begin{vmatrix} b_3 & -m_3\\ b_1 &-m_1 \end{vmatrix}
}
{\begin{vmatrix} 1 & -m_3\\ 1 &-m_1 \end{vmatrix}
}
=\dfrac
{\begin{vmatrix} 4 & -2\\ 1 & -(-1)\end{vmatrix}
}
{\begin{vmatrix} 1 & -2\\ 1 & -(-1)\end{vmatrix}
}
=\dfrac
{\begin{vmatrix} 4 & -2\\ 1 & 1\end{vmatrix}
}
{\begin{vmatrix} 1 & -2\\ 1 & 1\end{vmatrix}
}
=\dfrac
{ 4\cdot 1 - 1 \cdot (-2) }
{ 1\cdot 1 - 1 \cdot (-2) }
=\dfrac
{ 4+2}
{ 1+2 }
=\dfrac
{ 6 }
{ 3 }
= 2\\\\
x_s = \dfrac
{\begin{vmatrix} 1 & b_3\\ 1 & b_1 \end{vmatrix}
}
{\begin{vmatrix} 1 & -m_3\\ 1 &-m_1 \end{vmatrix}
}
=\dfrac
{\begin{vmatrix} 1 & 4\\ 1 & 1\end{vmatrix}
}
{\begin{vmatrix} 1 & -2\\ 1 & -(-1)\end{vmatrix}
}
=\dfrac
{\begin{vmatrix} 1 & 4\\ 1 & 1\end{vmatrix}
}
{\begin{vmatrix} 1 & -2\\ 1 & 1\end{vmatrix}
}
=\dfrac
{ 1\cdot 1 - 1\cdot 4 }
{ 1\cdot 1 - 1 \cdot (-2) }
=\dfrac
{ 1-4 }
{ 1+2 }
=\dfrac
{ -3 }
{ 3 }
= -1
\end{array}
$}}$$

 

Die 3 Schnittpunkte zusammengefasst:

$$\small{\text{$
\begin{array}{|l|r|c|}
\hline
& x & y \\
\hline
S_1 &4 & -3 \\
S_2 &-2 & 0 \\
S_3 &-1 & 2 \\
\hline
\end{array}
$}}$$

 

Die Strecke von $$\small{\text{$S_1 $ nach $ S_2$}}$$:

$$\small{\text{$ \begin{array}{lrc} \overline{S_1S_2} =\sqrt{ [(4)-(-2)]^2 + [0-(-3)]^2 } =\sqrt{ 6^2 + 3^2 } =\sqrt{ 36 + 9 } =\sqrt{ 45 } =6,70820393250 \end{array} $}}$$

 

Die Strecke von $$\small{\text{$S_2 $ nach $ S_3$}}$$:

$$\small{\text{$ \begin{array}{lrc} \overline{S_2S_3} =\sqrt{ [(-1)-(-2)]^2 + (2-0)^2 } =\sqrt{ (-1+2)^2 + 2^2 } =\sqrt{ 1^2 + 4 } =\sqrt{ 5 } =2.23606797750 \end{array} $}}$$

 

Die Strecke von $$\small{\text{$S_3 $ nach $ S_1$}}$$:

$$\small{\text{$ \begin{array}{lrc} \overline{S_2S_3} =\sqrt{ [4-(-1)]^2 + (-3-2)^2 } =\sqrt{ (4+1)^2 + (-5)^2 } =\sqrt{ 5^2 + 5^2 } =\sqrt{ 50 } =7.07106781187 \end{array} $}}$$

 

Der Umfang der eingeschlossenen Fläche ( Dreieck) beträgt: $$\small{\text{$\overline{S_1S_2}+\overline{S_2S_3}+\overline{S_2S_3} =6,70820393250+2,23606797750+7,07106781187 =16,0153397219 $}}$$

 

Die Fläche des Dreiecks nach Heron:

$$\small{\text{$
\boxed{~~ A =\sqrt{ s\cdot(s-a)\cdot(s-b)\cdot(s-c)} \qquad
s = (a + b + c)/2 ~~}
$}}\\\\
\small{\text{$
\begin{array}{lcl}
a= \overline{S_1S_2} = 6,70820393250\\
b= \overline{S_2S_3} = 2,23606797750\\
c= \overline{S_3S_1} = 7,07106781187 \\\\
s = \dfrac{16,0153397219}{2}= 8,00766986093\\
\end{array}
$}}\\\\\\
\small{\text{$
\begin{array}{lcl}
A =\sqrt{ 8,00766986093\cdot(8,00766986093-6,70820393250)\cdot(8,00766986093-2,23606797750)\cdot(8,00766986093-7,07106781187)}\\
A =\sqrt{ 8,00766986093\cdot 1,29946592843 \cdot 5,77160188343\cdot 0,93660204906}\\
A =\sqrt{ 56.2499999994 }\\
A=7,5
\end{array}
$}}\\\\$$

Der Flächeninhalt der eingeschlossenen Fläche ( Dreieck) beträgt 7,5

 

Probe der Fläche:

$$\small{\text{$ \begin{array}{|l|r|c|} \hline & x & y \\ \hline S_1 &4 & -3 \\ S_2 &-2 & 0 \\ S_3 &-1 & 2 \\ \hline \end{array}
$}}\\\\
\small{\text{$
\boxed{
~~
A = \dfrac{1}{2}\cdot \begin{vmatrix}
1 & 1 & 1 \\
x_{s_1} & x_{s_2} & x_{s_3} \\
y_{s_1} & y_{s_2} & y_{s_3} \\
\end{vmatrix}
~~
}
$}}\\\\
\small{\text{$
\begin{array}{rcl}
A &=&\dfrac{1}{2}\cdot \begin{vmatrix}
1 & 1 & 1 \\
4 & -2 & -1 \\
-3 & 0 & 2 \\
\end{vmatrix} \\\\
&=&\frac{1}{2} \cdot
[ 1\cdot (-2) \cdot 2 + 4\cdot 0 \cdot 1 + (-3)\cdot 1 \cdot (-1)
- (-3)\cdot (-2) \cdot 1 - 0\cdot (-1) \cdot 1 - 4\cdot 1 \cdot 2
]\\\\
&=&\frac{1}{2} \cdot
[ 1\cdot (-2) \cdot 2 + (-3)\cdot 1 \cdot (-1)
- (-3)\cdot (-2) \cdot 1 - 4\cdot 1 \cdot 2
]\\\\
&=&\frac{1}{2} \cdot
[ -4 + 3 - 6 \cdot 1 - 8]\\\\
&=&\frac{1}{2} \cdot
[ -15 ]\\\\
&=& - 7,5 \\\\
|A| &=& 7,5 \qquad \mathrm{okay}
\end{array}
$}}$$

heureka 19.06.2015
 #6
avatar+12515 
0

Omi67 19.06.2015
 #7
avatar
0

Danke Heureka für die Korrektur. (Habe ein Minuszeichen in der zweiten Gleichung übersehen.)

In der Klasse 10 ober 11 rechnet man übrigens so:

$${\mathtt{F}} = {\frac{{\mathtt{1}}}{{\mathtt{2}}}}{\mathtt{\,\times\,}}\left(\left[{\mathtt{x1}}{\mathtt{\,\times\,}}\left({\mathtt{y2}}{\mathtt{\,-\,}}{\mathtt{y3}}\right)\right]{\mathtt{\,\small\textbf+\,}}\left[{\mathtt{x2}}{\mathtt{\,\times\,}}\left({\mathtt{y3}}{\mathtt{\,-\,}}{\mathtt{y1}}\right)\right]{\mathtt{\,\small\textbf+\,}}\left[{\mathtt{x3}}{\mathtt{\,\times\,}}\left({\mathtt{y1}}{\mathtt{\,-\,}}{\mathtt{y2}}\right)\right]\right)$$

$${\mathtt{Flaeche}} = {\left|{\frac{{\mathtt{1}}}{{\mathtt{2}}}}{\mathtt{\,\times\,}}\left(\left[{\mathtt{4}}{\mathtt{\,\times\,}}\left({\mathtt{0}}{\mathtt{\,-\,}}{\mathtt{2}}\right)\right]{\mathtt{\,-\,}}\left[{\mathtt{2}}{\mathtt{\,\times\,}}\left({\mathtt{2}}{\mathtt{\,\small\textbf+\,}}{\mathtt{3}}\right)\right]{\mathtt{\,-\,}}\left[{\mathtt{1}}{\mathtt{\,\times\,}}\left({\mathtt{\,-\,}}{\mathtt{3}}{\mathtt{\,\small\textbf+\,}}{\mathtt{0}}\right)\right]\right)\right|} \Rightarrow {\mathtt{Flaeche}} = {\mathtt{7.5}}$$

 19.06.2015
 #8
avatar+26318 
0

Die Fläche A eines  Dreiecks mit Determinante berechnen:

Gegeben sind drei Punkte :

$$\small{\text{$
S_1(x_1|y_1) = (4|-3) \qquad S_2 (x_2|y_2)=(-2|0) \qquad S_3(x_3|y_3)=(-1|2)
$}}$$

$$\small{\text{$
\begin{array}{|l|l|l|} \hline & x & y \\
\hline
S_1 & x_1=4 & y_1=-3 \\
S_2 & x_2=-2 & y_2= 0 \\
S_3 & x_3=-1 & y_3=2 \\
\hline
\end{array} $}}\\\\$$

$$\small{\text{$
A = \dfrac{1}{2}\cdot
\begin{vmatrix}
x_1 & y_1 & 1 \\
x_2 & y_2 & 1 \\
x_3 & y_3 & 1 \\
\end{vmatrix}
$ }}\\\\$$

Auflösung nach der 1. Spalte:

$$\small{\text{$
\begin{array}{lll}
A &=& \dfrac{1}{2}\cdot
\begin{vmatrix}
\textcolor[rgb]{1,0,0}{x_1} & y_1 & 1 \\
\textcolor[rgb]{1,0,0}{x_2} & y_2 & 1 \\
\textcolor[rgb]{1,0,0}{x_3} & y_3 & 1 \\
\end{vmatrix}
=\frac12\cdot \left(
\textcolor[rgb]{1,0,0}{x_1}\cdot
\begin{vmatrix}
y_2 & 1 \\
y_3 & 1 \\
\end{vmatrix}
-\textcolor[rgb]{1,0,0}{x_2}\cdot
\begin{vmatrix}
y_1 & 1 \\
y_3 & 1 \\
\end{vmatrix}
+\textcolor[rgb]{1,0,0}{x_3}\cdot
\begin{vmatrix}
y_1 & 1 \\
y_2 & 1 \\
\end{vmatrix}
\right)\\\\
&=& \frac12\cdot \left(~~ [~x_1\cdot
(y_2-y_3)~]-[~x_2\cdot
(y_1-y_3)~]+[~x_3\cdot
(y_1-y_2)~] ~~\right)\\\\
&=&
\mathbf{ \frac12\cdot \left(~~ [~x_1\cdot
(y_2-y_3)~]+[~x_2\cdot
(y_3-y_1)~]+[~x_3\cdot
(y_1-y_2)~] ~~\right)
} \\ \\
&=& \frac12\cdot \left(~~ [~4\cdot
(0-2)~]-[~2\cdot
(2-(-3))~]-[~1\cdot
(-3-0)~] ~~\right)\\\\
&=& \frac12\cdot \left(~~ [~4\cdot
(-2)~]-[~2\cdot
(2+3))~]-[~1\cdot
(-3)~] ~~\right)\\\\
&=& \frac12\cdot \left(~~ -8-10+3 ~~\right)\\\\
&=& -7,5\\\\
A&=& 7,5
\end{array}
$ }}\\\\$$

 

Auflösung nach der 2. Spalte:

$$\small{\text{$
\begin{array}{lll}
A &=& \dfrac{1}{2}\cdot
\begin{vmatrix}
x_1 & \textcolor[rgb]{1,0,0}{y_1} & 1 \\
x_2 & \textcolor[rgb]{1,0,0}{y_2} & 1 \\
x_3 & \textcolor[rgb]{1,0,0}{y_3} & 1 \\
\end{vmatrix}
=\frac12\cdot \left(
-\textcolor[rgb]{1,0,0}{y_1}\cdot
\begin{vmatrix}
x_2 & 1 \\
x_3 & 1 \\
\end{vmatrix}
+\textcolor[rgb]{1,0,0}{y_2}\cdot
\begin{vmatrix}
x_1 & 1 \\
x_3 & 1 \\
\end{vmatrix}
-\textcolor[rgb]{1,0,0}{y_3}\cdot
\begin{vmatrix}
x_1 & 1 \\
x_2 & 1 \\
\end{vmatrix}
\right)\\\\
&=& \frac12\cdot
\left(~~
-[~y_1\cdot (x_2-x_3)~]
+[~y_2\cdot (x_1-x_3)~]
-[~y_3\cdot (x_1-x_2)~]
~~\right)\\\\
&=& \mathbf{
\frac12\cdot
\left(~~
[~y_1\cdot (x_3-x_2)~]
+[~y_2\cdot (x_1-x_3)~]
+[~y_3\cdot (x_2-x_1)~]
~~\right)
}\\\\
&=& \frac12\cdot
\left(~~
[~-3\cdot (-1-(-2))~]
+[~0\cdot (4-(-1))~]
+[~2\cdot (-2-4)~]
~~\right)\\\\
&=& \frac12\cdot
\left(~~
[~-3\cdot 1~]
+[~2\cdot (-6)~]
~~\right)\\\\
&=& \frac12\cdot
\left(~~
-3 - 12
~~\right)\\\\
&=& \frac12\cdot
\left(~~
-15~~\right)\\\\
&=& -7,5\\\\
A&=& 7,5
\end{array}
$ }}\\\\$$

 

 22.06.2015

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