Results 1 to 13 of 13

Math Help - Parallel/Perpendicular Lines

  1. #1
    Member
    Joined
    Apr 2009
    Posts
    96

    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....
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Sep 2008
    From
    West Malaysia
    Posts
    1,261
    Thanks
    1
    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) .
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor red_dog's Avatar
    Joined
    Jun 2007
    From
    Medgidia, Romania
    Posts
    1,252
    Thanks
    5
    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}
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Apr 2009
    Posts
    96
    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?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Apr 2009
    Posts
    96
    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?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor

    Joined
    Aug 2006
    Posts
    20,160
    Thanks
    2165
    Awards
    1
    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.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Member
    Joined
    Apr 2009
    Posts
    96
    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 .
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor

    Joined
    Aug 2006
    Posts
    20,160
    Thanks
    2165
    Awards
    1
    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.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Member
    Joined
    Apr 2009
    Posts
    96
    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?
    Follow Math Help Forum on Facebook and Google+

  10. #10
    MHF Contributor

    Joined
    Aug 2006
    Posts
    20,160
    Thanks
    2165
    Awards
    1
    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.
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Member
    Joined
    Apr 2009
    Posts
    96
    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.
    Follow Math Help Forum on Facebook and Google+

  12. #12
    Member
    Joined
    Apr 2009
    Posts
    96
    Follow Math Help Forum on Facebook and Google+

  13. #13
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,947
    Thanks
    6
    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}.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Parallel and perpendicular lines?
    Posted in the Algebra Forum
    Replies: 5
    Last Post: July 10th 2011, 04:38 PM
  2. Parallel/Perpendicular Lines
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: December 16th 2008, 08:45 AM
  3. Parallel and Perpendicular lines
    Posted in the Algebra Forum
    Replies: 5
    Last Post: April 18th 2008, 08:49 PM
  4. Parallel and Perpendicular Lines?
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: November 1st 2007, 09:23 AM
  5. Parallel and Perpendicular Lines
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: March 21st 2007, 12:13 AM

Search Tags


/mathhelpforum @mathhelpforum