Lösung 2.2:5b

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Because the straight line is to have a gradient of
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<math>-3</math>, its equation can be written as
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<math>y=-3x+m</math>
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where
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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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<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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Version vom 09:13, 18. Sep. 2008

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


\displaystyle y=-3x+m


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


\displaystyle -2=-3\centerdot 1+m


which gives that \displaystyle m=1.

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


Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-TU-Berlin/index.php/L%C3%B6sung_2.2:5b“