Lösung 2.2:5d

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If two non-vertical lines are perpendicular to each other, their gradients
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If two non-vertical lines are perpendicular to each other, their slopes <math>k_{1}</math> and <math>k_{2}</math> satisfy the relation <math>k_{1}k_{2}=-1</math>, and from this we have that the line we are looking for must have a slope that is given by
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<math>k_{1}</math>
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and
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<math>k_{2}</math>
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satisfy the relation
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<math>k_{1}k_{2}=-1</math>, and from this we have that the line we are looking for must have a gradient that is given by
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{{Displayed math||<math>k_{2} = -\frac{1}{k_{1}} = -\frac{1}{2}</math>}}
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since the line <math>y=2x+5</math> has a slope <math>k_{1}=2</math>
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<math>k_{2}=-\frac{1}{k_{1}}=-\frac{1}{2}</math>
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(the coefficient in front of ''x'').
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since the line
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<math>y=2x+5</math>
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has a gradient
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<math>k_{1}=2</math>
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(the coefficient in front of
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<math>x</math>
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).
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The line we are looking for can thus be written in the form
The line we are looking for can thus be written in the form
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{{Displayed math||<math>y=-\frac{1}{2}x+m</math>}}
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<math>y=-\frac{1}{2}x+m</math>
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with ''m'' as an unknown constant.
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with
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<math>m</math>
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as an unknown constant.
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Because the point
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<math>\left( 2 \right.,\left. 4 \right)</math>
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should lie on the line,
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<math>\left( 2 \right.,\left. 4 \right)</math>
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must satisfy the equation of the line,
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<math>4=-\frac{1}{2}\centerdot 2+m</math>
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i.e.
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<math>m=5</math>. The equation of the line is .
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<math>y=-\frac{1}{2}x+5</math>
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Because the point (2,4) should lie on the line, (2,4) must satisfy the equation of the line,
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{{Displayed math||<math>4=-\frac{1}{2}\cdot 2+m\,,</math>}}
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i.e. <math>m=5</math>. The equation of the line is <math>y=-\frac{1}{2}x+5</math>.
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{{NAVCONTENT_START}}
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<center>[[Image:2_2_5d-2(2).gif]]</center>
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<center> [[Image:2_2_5d-2(2).gif]] </center>
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Version vom 12:44, 24. Sep. 2008

If two non-vertical lines are perpendicular to each other, their slopes \displaystyle k_{1} and \displaystyle k_{2} satisfy the relation \displaystyle k_{1}k_{2}=-1, and from this we have that the line we are looking for must have a slope that is given by

Vorlage:Displayed math

since the line \displaystyle y=2x+5 has a slope \displaystyle k_{1}=2 (the coefficient in front of x).

The line we are looking for can thus be written in the form

Vorlage:Displayed math

with m as an unknown constant.

Because the point (2,4) should lie on the line, (2,4) must satisfy the equation of the line,

Vorlage:Displayed math

i.e. \displaystyle m=5. The equation of the line is \displaystyle y=-\frac{1}{2}x+5.


Image:2_2_5d-2(2).gif
Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-TU-Berlin/index.php/L%C3%B6sung_2.2:5d“