Lösung 4.1:4b

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If we use the distance formula
If we use the distance formula
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{{Displayed math||<math>d=\sqrt{(x-a)^2+(y-b)^2}</math>}}
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<math>d=\sqrt{\left( x-a \right)^{2}+\left( y-b \right)^{2}}</math>
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to determine the distance between the points <math>(x,y) = (-2,5)</math> and <math>(a,b) = (3,-1)</math>, we get
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{{Displayed math||<math>\begin{align}
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to determine the distance between the points
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d &= \sqrt{(-2-3)^2+(5-(-1))^2}\\[5pt]
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<math>\left( x \right.,\left. y \right)=\left( -2 \right.,\left. 5 \right)</math>
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&= \sqrt{(-5)^2+6^2}\\[5pt]
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and
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&= \sqrt{25+36}\\[5pt]
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<math>\left( a \right.,\left. b \right)=\left( 3 \right.,\left. -1 \right)</math>, we get
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&= \sqrt{61}\,\textrm{.}
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\end{align}</math>}}
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<math>\begin{align}
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& d=\sqrt{\left( -2-3 \right)^{2}+\left( 5-\left( -1 \right) \right)^{2}} \\
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& =\sqrt{\left( -5 \right)^{2}+6^{2}}=\sqrt{25+36}=\sqrt{61} \\
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\end{align}</math>
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Version vom 10:30, 7. Okt. 2008

If we use the distance formula

Vorlage:Displayed math

to determine the distance between the points \displaystyle (x,y) = (-2,5) and \displaystyle (a,b) = (3,-1), we get

Vorlage:Displayed math

Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-TU-Berlin/index.php/L%C3%B6sung_4.1:4b“