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Math Help - Lines with the same slope.

  1. #1
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    Lines with the same slope.

    I have a problem in proving this statement:

    If two lines have the same slope then they are either parallel or incidental (the same line).

    Thank You!!
    Last edited by abutazil352; January 8th 2012 at 03:54 PM.
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  2. #2
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    Re: Lines with the same slope.

    y=3x+3 and y=3x-10 are parallel.

    but

    y=3x+3 and y=3x+3 are the same.

    Therefore the difference in these examples is the influence of the constant term.
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  3. #3
    Super Member Quacky's Avatar
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    Re: Lines with the same slope.

    Suggestion: let l_1 have the equation y=mx+c
    and let l_2 have the equation y=mx+k. These are two lines which have the same slope.

    Now solve for the points of intersection. Parallel lines will never intersect, so we can deduce that the only point of interception is where c=k and the lines are incidental; otherwise, they are parallel.
    Last edited by Quacky; January 8th 2012 at 04:11 PM.
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    Re: Lines with the same slope.

    thank you. i can't view the some parts of your reply though. im new here.
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    Re: Lines with the same slope.

    thank you.
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  6. #6
    Super Member Quacky's Avatar
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    Re: Lines with the same slope.

    Quote Originally Posted by abutazil352 View Post
    thank you. i can't view the some parts of your reply though. im new here.
    I assume you were referring to the use of latex in my post? Where are you having a problem?
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  7. #7
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    Re: Lines with the same slope.

    Quote Originally Posted by Quacky View Post
    I assume you were referring to the use of latex in my post? Where are you having a problem?


    Don't you just img.top {vertical-align:15%;}
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    math?
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  8. #8
    MHF Contributor MarkFL's Avatar
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    Re: Lines with the same slope.

    The slope m of a line may be converted to an angle θ using:

    \theta=\tan^{-1}(m)

    Suppose we have two slopes m_1 and m_2.

    If the slopes are parallel or perpendicular, we may state:

    \theta_1-\theta_2=\theta_2-\theta_1

    \tan^{-1}(m_1)-\tan^{-1}(m_2)=\tan^{-1}(m_2)-\tan^{-1}(m_1)

    2\tan^{-1}(m_1)=2\tan^{-1}(m_2)

    Taking the tangent of both sides and applying the double-angle identity for tangent and dividing through by 2, we have:

    \frac{m_1}{1-m_1^2}=\frac{m_2}{1-m_2^2}

    m_1(1-m_2^2)=m_2(1-m_1^2)

    m_1-m_1m_2^2-m_2+m_2m_1^2=0

    (m_1-m_2)(m_1m_2+1)=0

    The root m_1=m_2 corresponds to parallel slopes.

    The root m_1m_2=-1 corresponds to perpendicular slopes.
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