Lösung 4.1:4a

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If we draw in the points in a coordinate system, we can see the line between the points as the hypotenuse in an imaginary right-angled triangle, where the opposite and adjacent are parallel with the
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<math>x</math>
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- and
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<math>y</math>
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-axes.
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{{NAVCONTENT_START}}
{{NAVCONTENT_START}}
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[[Image:4_1_4_a-1(2)_1.gif|center]]
[[Image:4_1_4_a-1(2)_1.gif|center]]
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In this triangle, it is easy to measure the lengths of the opposite and the adjacent, which are simply the distances between the points in the
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<math>x</math>
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-
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<math>y</math>
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-directions.
[[Image:4_1_4_a-1(2)_2.gif|center]]
[[Image:4_1_4_a-1(2)_2.gif|center]]
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<center> [[Image:4_1_4a-1(2).gif]] </center>
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{{NAVCONTENT_STOP}}
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{{NAVCONTENT_START}}
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<center> [[Image:4_1_4a-2(2).gif]] </center>
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{{NAVCONTENT_STOP}}
{{NAVCONTENT_STOP}}
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Using Pythagoras' theorem, we can then calculate the length of the hypotenuse, which is also the distance between the points:
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<math>\begin{align}
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& d=\sqrt{\left( \Delta x \right)^{2}+\left( \Delta y \right)^{2}}=\sqrt{4^{2}+3^{2}} \\
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& =\sqrt{16+9}=\sqrt{25}=5 \\
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\end{align}</math>
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NOTE: In general, the distance between two points
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<math>\left( x \right.,\left. y \right)</math>
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and
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<math>\left( a \right.,\left. b \right)</math>
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is given by the formula
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<math>d=\sqrt{\left( x-a \right)^{2}+\left( y-b \right)^{2}}</math>

Version vom 09:56, 27. Sep. 2008

If we draw in the points in a coordinate system, we can see the line between the points as the hypotenuse in an imaginary right-angled triangle, where the opposite and adjacent are parallel with the \displaystyle x - and \displaystyle y -axes.

In this triangle, it is easy to measure the lengths of the opposite and the adjacent, which are simply the distances between the points in the \displaystyle x - \displaystyle y -directions.


Using Pythagoras' theorem, we can then calculate the length of the hypotenuse, which is also the distance between the points:


\displaystyle \begin{align} & d=\sqrt{\left( \Delta x \right)^{2}+\left( \Delta y \right)^{2}}=\sqrt{4^{2}+3^{2}} \\ & =\sqrt{16+9}=\sqrt{25}=5 \\ \end{align}


NOTE: In general, the distance between two points \displaystyle \left( x \right.,\left. y \right) and \displaystyle \left( a \right.,\left. b \right) is given by the formula


\displaystyle d=\sqrt{\left( x-a \right)^{2}+\left( y-b \right)^{2}}

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