Aus Online Mathematik Brückenkurs 1

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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 <math>x</math> - and <math>y</math> -axes. {{NAVCONTENT_START}} [[Image:4_1_4_a-1(2)_1.gif|center]] 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 <math>x</math> - <math>y</math> -directions. [[Image:4_1_4_a-1(2)_2.gif|center]] {{NAVCONTENT_STOP}} Using Pythagoras' theorem, we can then calculate the length of the hypotenuse, which is also the distance between the points: <math>\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}</math> NOTE: In general, the distance between two points <math>\left( x \right.,\left. y \right)</math> and <math>\left( a \right.,\left. b \right)</math> is given by the formula <math>d=\sqrt{\left( x-a \right)^{2}+\left( y-b \right)^{2}}</math>

• 4.1.4a - Solution - A line segment between the points (1,1) and (5,4)
• 4.1.4a - Solution - A triangle with the line segment between the points (1,1) and (5,4) as the hypotenuse
• Vorlage:Abgesetzte Formel

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