Lösung 2.2:5b

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Because the straight line is to have a gradient of
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Because the straight line is to have a slope of <math>-3</math>, its equation can be written as
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<math>-3</math>, its equation can be written as
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{{Displayed math||<math>y=-3x+m\,,</math>}}
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<math>y=-3x+m</math>
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where ''m'' is a constant. If the line is also to pass through the point (''x'',''y'')&nbsp;= (1,-2), the point must satisfy the equation of the line,
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{{Displayed math||<math>-2=-3\cdot 1+m\,,</math>}}
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where
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which gives that <math>m=1</math>.
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<math>m</math>
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is a constant. If the line is also to pass through the point
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<math>\left( x \right.,\left. y \right)=\left( 1 \right.,\left. -2 \right)</math>, the point
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must satisfy the equation of the line
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The answer is thus that the equation of the line is <math>y=-3x+1</math>.
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<math>-2=-3\centerdot 1+m</math>
 
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which gives that
 
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<math>m=1</math>.
 
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The answer is thus that the equation of the line is
 
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<math>y=-3x+1</math>.
 
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{{NAVCONTENT_START}}
 
[[Image:1_2_2_5_b_ss1.jpg|center|300px]]
[[Image:1_2_2_5_b_ss1.jpg|center|300px]]
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{{NAVCONTENT_STOP}}
 

Version vom 12:28, 24. Sep. 2008

Because the straight line is to have a slope of \displaystyle -3, its equation can be written as

Vorlage:Displayed math

where m is a constant. If the line is also to pass through the point (x,y) = (1,-2), the point must satisfy the equation of the line,

Vorlage:Displayed math

which gives that \displaystyle m=1.

The answer is thus that the equation of the line is \displaystyle y=-3x+1.


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