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Math Help - Parallel/Perpendicular Lines

  1. #1
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    Parallel/Perpendicular Lines

    Determine the equation of the line that passes through the point (2,1) and is:
    a. Parallel to the line 3x-2y=4
    b. Perpendicular to the line 3x-2y=4

    First, I rearranged 3x-2y=4 to be in slope-intercept form:
    y=3/2x-2

    For part b, I took the slope and put it into point-slope form:
    y-1=3/2(x-2)
    and got the same thing as slope-intercept form.

    I think I'm doing something wrong....
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  2. #2
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    Quote Originally Posted by BeSweeet View Post
    Determine the equation of the line that passes through the point (2,1) and is:
    a. Parallel to the line 3x-2y=4
    b. Perpendicular to the line 3x-2y=4

    First, I rearranged 3x-2y=4 to be in slope-intercept form:
    y=3/2x-2

    For part b, I took the slope and put it into point-slope form:
    y-1=3/2(x-2)
    and got the same thing as slope-intercept form.

    I think I'm doing something wrong....

    HI

    (a) m=3/2 , since it is parallel , then substitute into the equation
    y-y1=m(x-x1)

    (b) m=-2/3 because it is perpendicular (m1m2=-1) , then substitute into the same equation as (a) .
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  3. #3
    MHF Contributor red_dog's Avatar
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    The point (2,1) is on the line, so the question a) doesn't make sense.

    For b), the slope of the perpendicular line is -\frac{2}{3}
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  4. #4
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    I have to determine the equation of the line. Would the equation just be in slope-intercept form once I do the thing for point-slope?
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    Ok, I think I got it!

    For parallel lines, just find the slope, and stick that into point-slope form.

    For perpendicular lines, find the slope, and find what other number it will multiply by to equal -1, then stick that into point-slope form.

    I just have one [stupid] question: How about did you getting -2/3?
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  6. #6
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    Consider the line \ell: Ax+By+C=0
    Any line parallel to \ell has this form: Ax+By+D=0.

    Any line perpendicular to \ell has this form: -Bx+Ay+E=0.
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  7. #7
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    Quote Originally Posted by Plato View Post
    Consider the line \ell: Ax+By+C=0
    Any line parallel to \ell has this form: Ax+By+D=0.

    Any line perpendicular to \ell has this form: -Bx+Ay+E=0.
    Well that just made things even worse .
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  8. #8
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    Quote Originally Posted by BeSweeet View Post
    Well that just made things even worse .
    Why do you say that?
    Example: Write lines parallel and perpendicular to 3x-4y+7=0 through the point (0,1).

    Answer:
    Parallel: 3x-4y+D=0
    Perpendicular: 4x+3y+E=0

    Now all you have to do is substitute x=0~\&~y=1 to find D~\&~E.
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  9. #9
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    Quote Originally Posted by Plato View Post
    Why do you say that?
    Example: Write lines parallel and perpendicular to 3x-4y+7=0 through the point (0,1).

    Answer:
    Parallel: 3x-4y+D=0
    Perpendicular: 4x+3y+E=0

    Now all you have to do is substitute x=0~\&~y=1 to find D~\&~E.
    That's really confusing. Where did C go? Did you subtract it from both sides? ( 3x-4y=-7).

    Isn't it easier (for parallel) to just find the slope, and stick it into point-slope form to get my equation, and for perpendicular, find the slope that is equal to -1 (when multiplied by the other number), stick it into point-slope form, and get my equation?
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  10. #10
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    In your case, websites such as this seem to do more harm than good.
    You need a sit-down, one-to-one, tutorial with a live instructor.
    You have some fundamental misunderstandings.
    Please plan to do that.
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  11. #11
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    For most of my other questions, they were answered and were eventually figured out and I understood them.

    I understand this problem by what I've said here, but not by what you said.
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  13. #13
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    Quote Originally Posted by BeSweeet View Post
    Ok, I think I got it!

    For parallel lines, just find the slope, and stick that into point-slope form.

    For perpendicular lines, find the slope, and find what other number it will multiply by to equal -1, then stick that into point-slope form.

    I just have one [stupid] question: How about did you getting -2/3?
    The gradient of the line 3x-2y=4 is 3/2.

    And you should know that if two lines are perpendicular, then the product of their gradients is equal to -1: m_{1} m_{2} = -1.

    So \frac{3}{2} \, m_{2} = -1 \Rightarrow m_2 = - \frac{2}{3}.
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