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Solution 2.2:5e

From Förberedande kurs i matematik 1

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Current revision (12:54, 24 September 2008) (edit) (undo)
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The line should go through the points
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The line should go through the points (5,0) and (0,-8) which must therefore satisfy the equation of the line <math>y=kx+m</math>, i.e.
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<math>\left( 5 \right.,\left. 0 \right)</math>
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and
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<math>\left( 0 \right.,\left. -8 \right)</math>
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which must therefore satisfy the equation of the line
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<math>y=kx+m</math>, i.e.
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{{Displayed math||<math>0=k\cdot 5+m\qquad\text{and}\qquad -8 = k\cdot 0+m\,\textrm{.}</math>}}
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<math>0=k\centerdot 5+m</math>
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From the second equation, we get <math>m=-8</math> and substituting this into the first equation gives
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{{Displayed math||<math>0=5k-8\quad\Leftrightarrow\quad k={8}/{5}\,\textrm{.}</math>}}
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From the other equation, we get
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The slope of the line is 8/5.
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<math>m=-8</math>
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and substituting this into the first equation gives
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<math>0=5k-8\ \Leftrightarrow \ k={8}/{5}\;</math>
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<center>[[Image:S1_2_2_5_e.jpg|400px]]</center>
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The gradient of the line is
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<math>{8}/{5}\;</math>.
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{{NAVCONTENT_START}}
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[[Image:S1_2_2_5_e.jpg|400px]]
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<!--<center> [[Image:2_2_5e.gif]] </center>-->
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{{NAVCONTENT_STOP}}
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Current revision

The line should go through the points (5,0) and (0,-8) which must therefore satisfy the equation of the line y=kx+m, i.e.

0=k5+mand−8=k0+m.

From the second equation, we get m=−8 and substituting this into the first equation gives

0=5k−8k=85.

The slope of the line is 8/5.


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