Lösung 2.2:5d

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K (Robot: Automated text replacement (-{{Displayed math +{{Abgesetzte Formel))
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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
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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{{Displayed math||<math>k_{2} = -\frac{1}{k_{1}} = -\frac{1}{2}</math>}}
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{{Abgesetzte Formel||<math>k_{2} = -\frac{1}{k_{1}} = -\frac{1}{2}</math>}}
since the line <math>y=2x+5</math> has a slope <math>k_{1}=2</math>
since the line <math>y=2x+5</math> has a slope <math>k_{1}=2</math>
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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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{{Abgesetzte Formel||<math>y=-\frac{1}{2}x+m</math>}}
with ''m'' as an unknown constant.
with ''m'' as an unknown constant.
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Because the point (2,4) should lie on the line, (2,4) must satisfy the equation of the line,
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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{{Abgesetzte Formel||<math>4=-\frac{1}{2}\cdot 2+m\,,</math>}}
i.e. <math>m=5</math>. The equation of the line is <math>y=-\frac{1}{2}x+5</math>.
i.e. <math>m=5</math>. The equation of the line is <math>y=-\frac{1}{2}x+5</math>.

Version vom 08:28, 22. Okt. 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

\displaystyle k_{2} = -\frac{1}{k_{1}} = -\frac{1}{2}

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

\displaystyle y=-\frac{1}{2}x+m

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,

\displaystyle 4=-\frac{1}{2}\cdot 2+m\,,

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