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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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 ''x''- and ''y''-axes, respectively.
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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}}
 
[[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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<math>y</math>
 
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-directions.
 
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[[Image:4_1_4_a-1(2)_2.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 ''x''- and ''y''-directions, respectively.
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{{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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{| align="center"
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|align="center"|[[Image:4_1_4_a-1(2)_2.gif|center]]
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|-
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|align="center"|<small>∆''x''&nbsp;=&nbsp;5&nbsp;-&nbsp;1&nbsp;=&nbsp;4 &nbsp;and&nbsp; ∆''y''&nbsp;=&nbsp;4&nbsp;-&nbsp;1&nbsp;=&nbsp;3</small>
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|}
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<math>\begin{align}
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Using the Pythagorean theorem, we can then calculate the length of the hypotenuse, which is also the distance between the points:
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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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{{Displayed math||<math>\begin{align}
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d &= \sqrt{(\Delta x)^2 + (\Delta y)^2} = \sqrt{4^2+3^2}\\[5pt]
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&= \sqrt{16+9} = \sqrt{25} = 5\,\textrm{.}
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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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Note: In general, the distance between two points <math>(x,y)</math> and <math>(a,b)</math> is given by the formula
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<math>d=\sqrt{\left( x-a \right)^{2}+\left( y-b \right)^{2}}</math>
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{{Displayed math||<math>d = \sqrt{(x-a)^2 + (y-b)^2}\,\textrm{.}</math>}}

Version vom 10:40, 3. Okt. 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 x- and y-axes, respectively.



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 x- and y-directions, respectively.


∆x = 5 - 1 = 4  and  ∆y = 4 - 1 = 3


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

Vorlage:Displayed math


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

Vorlage:Displayed math

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