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calculusxy

  • one year ago

MEDAL!!! Is it possible for two lines with positive slopes to be perpendicular to each other?

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  1. calculusxy
    • one year ago
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    I think I got it. The slope needs to be negative reciprocals of the other. So like 5x needs to be -1/5x?

  2. calculusxy
    • one year ago
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    Am I correct?

  3. calculusxy
    • one year ago
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    @Michele_Laino

  4. anonymous
    • one year ago
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    No. Theres no way for a line to cross another and form right angles unless one has a negative slope and one has a positive slope. "If two lines are perpendicular and neither one is vertical, then one of the lines has a positive slope, and the other has a negative slope. Also, the absolute values of their slopes are reciprocals." http://www.cliffsnotes.com/math/geometry/coordinate-geometry/slopes-parallel-and-perpendicular-lines

  5. Michele_Laino
    • one year ago
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    if two perpendicular lines have their slopes , say m_1, and m_2 both positive, then the product of such slopes is also positive, namely: \[\large {m_1}{m_2} > 0\]

  6. Michele_Laino
    • one year ago
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    it is a contradiction, since we know that the product of such slopes has to check this condition: \[\large {m_1}{m_2} = - 1\]

  7. calculusxy
    • one year ago
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  8. calculusxy
    • one year ago
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    So I need to make a perpendicular line through the -2,3 point. How can I do that?

  9. Michele_Laino
    • one year ago
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    it is simple the slope of the requested line, is: \[\large m = - \frac{1}{5}\]

  10. calculusxy
    • one year ago
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    I have the slope intercept form for -2,3 as y=5x + 2

  11. Michele_Laino
    • one year ago
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    so you have to apply this equation: \[\large y - 3 = - \frac{1}{5}\left( {x + 2} \right)\]

  12. calculusxy
    • one year ago
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    So now do I need to do: -1/5x + 2= y?

  13. calculusxy
    • one year ago
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    Is your and my equations the same thing? @Michele_Laino

  14. Michele_Laino
    • one year ago
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    I think that they are different equations, since after a simplification, I can rewrite my equation as follows: \[\large y = - \frac{x}{5} + \frac{{13}}{5}\]

  15. calculusxy
    • one year ago
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  16. calculusxy
    • one year ago
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    Would that line work?

  17. calculusxy
    • one year ago
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    My line doesn't work. It is not perpendicular.

  18. Michele_Laino
    • one year ago
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    I think not, since that line is not perpendicular to both the parallel lines

  19. Michele_Laino
    • one year ago
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    please try with this line: \[\large y = - \frac{x}{5} + \frac{{13}}{5}\]

  20. calculusxy
    • one year ago
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    @Michele_Laino How did you get 13/5?

  21. Michele_Laino
    • one year ago
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    here are the missing steps: \[\large \begin{gathered} y - 3 = - \frac{x}{5} - \frac{2}{5} \hfill \\ \hfill \\ y = - \frac{x}{5} - \frac{2}{5} + 3 \hfill \\ \hfill \\ y = - \frac{x}{5} + \frac{{15 - 2}}{5} \hfill \\ \hfill \\ y = - \frac{x}{5} + \frac{{13}}{5} \hfill \\ \end{gathered} \]

  22. calculusxy
    • one year ago
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    I still do not get how you got 13/5. These are my steps: \[y - 3 = -\frac{ 1 }{ 5 } (x+2)\] \[y - 3=-\frac{ 1 }{ 5 }x-\frac{ 2 }{ 5 }\] \[y=-\frac{ 1 }{ 5 }x-\frac{ 2 }{ 5 } + 3\] \[y=-\frac{ 1 }{ 5 }x+\frac{ 3 }{ 5 }\]

  23. calculusxy
    • one year ago
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    I am really sorry fir being redundant. :(

  24. calculusxy
    • one year ago
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    *for

  25. Michele_Laino
    • one year ago
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    hint: \[ \Large - \frac{2}{5} + 3 = \frac{{ - 2 + \left( {3 \times 5} \right)}}{5} = ...?\]

  26. calculusxy
    • one year ago
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    How did you get -2+(3 x 5)?

  27. Michele_Laino
    • one year ago
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    since the least common multiple of 5 and 1 is 5 \[\Large - \frac{2}{5} + 3 = - \frac{2}{5} + \frac{3}{1}\]

  28. Michele_Laino
    • one year ago
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    and multipling both numerator and denominator of the second fraction, by 5, we get: \[\large \frac{2}{5} + \frac{3}{1} = - \frac{2}{5} + \frac{{3 \times 5}}{{1 \times 5}} = - \frac{2}{5} + \frac{{15}}{5} = \frac{{ - 2 + 15}}{5}\]

  29. calculusxy
    • one year ago
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    Oh okay!!!!! I totally get it now :)

  30. Michele_Laino
    • one year ago
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    :)

  31. calculusxy
    • one year ago
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    So for x I can input any number now?

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