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Linear Graph Line

  1. Feb 13, 2007 #1
    1. The problem statement, all variables and given/known data
    The line l has equation y = 3x + 4
    Cans someone xplain how [tex]y = -\frac{1}{3}x-4[/tex] is perpendicular to line l(bisects l at 90°)
    2. Relevant equations

    y = mx + c
     
    Last edited: Feb 13, 2007
  2. jcsd
  3. Feb 13, 2007 #2

    cristo

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    What condition must the gradients of two perpendicular lines satisfy?
     
  4. Feb 13, 2007 #3
    negtive gradient.

    y = mx + c where m is a negitive number?
     
  5. Feb 13, 2007 #4

    HallsofIvy

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    No, remember that the slope of a line is the tangent of the angle the line makes with the x-axis. If two lines are y= m1x+ b and y= m2x+ c, then the two lines are parallel if and only if m1= m2 and parallel if and only if (m1)(m2)= -1.

    Those can both be derived from properties of the tangent function.
     
  6. Feb 13, 2007 #5

    cristo

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    Just correcting a typo: you mean perpendicular if and only if (m1)(m2)= -1.
     
  7. Feb 13, 2007 #6
    cant it be

    perpendicular if and only if (m1)(m2)= -0.5? Or any other number there? Was that just an example? If not what is so special about -1?

    (m1)(m2)= -1
    so 3 * -1/3 = -1

    cheerz, Big help
     
  8. Feb 14, 2007 #7

    Integral

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    Graph a set of perpendicular intersecting lines.

    Now using the graphs, compute the slopes, can you now see why the relationship is [itex] m_1 m_2 = -1 [/itex]

    You may need to look at several sets of lines to see the relationships.
     
  9. Feb 14, 2007 #8

    HallsofIvy

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    Thomas49th, are you saying you cannot read our responses?
     
  10. Feb 14, 2007 #9
    aha, ingenius. It works. Thanks a load :cool:
     
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