Solution 2.2:5b
From Förberedande kurs i matematik 1
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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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| + | {{Displayed math||<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'') = (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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| + | which gives that <math>m=1</math>. | ||
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| + | The answer is thus that the equation of the line is <math>y=-3x+1</math>. | ||
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[[Image:1_2_2_5_b_ss1.jpg|center|300px]] | [[Image:1_2_2_5_b_ss1.jpg|center|300px]] | ||
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Current revision
Because the straight line is to have a slope of \displaystyle -3, its equation can be written as
| \displaystyle y=-3x+m\,, |
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,
| \displaystyle -2=-3\cdot 1+m\,, |
which gives that \displaystyle m=1.
The answer is thus that the equation of the line is \displaystyle y=-3x+1.
