Results 1 to 3 of 3

Math Help - Need help with linear algebra?

  1. #1
    Newbie
    Joined
    Sep 2009
    Posts
    9

    Need help with linear algebra?

    Using Ax+By+C=0 (line 1), determine the point which line 2 touches line 1 where line 2 is perpendicular to line 1, and line 2 touches the origin.

    I got some weird result: ( -(A*C)/(B^2+A^2), (A^2 *C -C)/(B^3+A^2 *B) )
    Sorry for my computer language.

    Gut feeling tells me it's wrong. Anyone concur?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    11,751
    Thanks
    484
    Quote Originally Posted by ISeriouslyNeedHelp View Post
    Using Ax+By+C=0 (line 1), determine the point which line 2 touches origin and line 1 where line 2 is perpendicular to line 1

    I got some weird result: ( -(A*C)/(B^2+A^2), (A^2 *C -C)/(B^3+A^2 *B) )
    Sorry for my computer language.

    Gut feeling tells me it's wrong. Anyone concur?
    x-value is fine ... y-value (meh)

    Ax+By+C=0

    By = -Ax - C

    y = -\frac{A}{B}x - \frac{C}{B}

    slope of this line is m = -\frac{A}{B}

    a normal line has a slope that is the opposite reciprocal ...

    m_p = \frac{B}{A}


    equation of the line with slope m_p that passes thru the origin ...

    y = \frac{B}{A}x


    intersection ... y = y

    \frac{B}{A}x = -\frac{A}{B}x - \frac{C}{B}

    multiply every term by AB to clear the fractions ...

    B^2x = -A^2x - AC

    A^2 x + B^2 x = -AC

    x = -\frac{AC}{A^2 + B^2}

    y = \frac{B}{A} \left( -\frac{AC}{A^2 + B^2} \right) = -\frac{BC}{A^2 + B^2}
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Sep 2009
    Posts
    9
    Thanks. Very big help.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 1
    Last Post: August 1st 2011, 10:00 PM
  2. Replies: 2
    Last Post: December 6th 2010, 03:03 PM
  3. Replies: 7
    Last Post: August 30th 2009, 10:03 AM
  4. Linear Algebra Question - Linear Operators
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: March 3rd 2009, 04:33 AM
  5. Replies: 3
    Last Post: June 2nd 2007, 10:08 AM

Search Tags


/mathhelpforum @mathhelpforum