Results 1 to 4 of 4

Math Help - Lines

  1. #1
    Banned
    Joined
    Jan 2007
    Posts
    315

    Lines

    Prove that if two nonvertical lines have slopes whose product is -1, then the lines are perpendicular.

    HINT GIVEN:

    Use the converse of the Pythagorean Theorem.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by symmetry View Post
    Prove that if two nonvertical lines have slopes whose product is -1, then the lines are perpendicular.

    HINT GIVEN:

    Use the converse of the Pythagorean Theorem.
    The converse of the Pythagorean theorem is that if triangle ABC has
    sides a,b,c with the usual labelling, and a^2+b^2=c^2, then the angle
    at C is right.

    The lines meet at some point, call it (a,b), and let them have slopes m (!=0)
    and 1/m.

    Additional points on the two lines are (a-1, b-m) and (a-1, b+1/m).

    Now we have three points, we have a triangle and can calculate the
    lengths of the sides, and we will find the square of the side connecting
    (a-1, b-m) and (a-1, b+1/m) is equal to the sum of the squares of the
    other two sides.

    The case where one of the slopes is 0 can be handled as a special case
    and I hope you see that in that case the result is obvious.

    RonL
    Last edited by CaptainBlack; January 16th 2007 at 08:56 PM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,857
    Thanks
    739
    Hello, symmetry!

    An interesting problem . . . I had to "invent" my own approach.
    I'm certain others have shorter, more elegant proofs.


    Prove that if two nonvertical lines have slopes whose product is -1,
    then the lines are perpendicular.

    \text{Let }m_1 = \frac{q}{p}\:\text{ and }\:m_2 = -\frac{p}{q}

    Then: . m_1\cdot m_2 \:=\:\left(\frac{1}{p}\right)\left(-\frac{p}{q}\right) \:=\:-1
    . . We have two slopes whose product is -1.


    We know that: slope \:=\:\frac{rise}{run}

    So m_1 = \frac{q}{p} looks like this:
    Code:
                      *
                  *   |q
              *       |
          * - - - - - *
                  p

    And m_2 = -\frac{p}{q} looks like this;
    Code:
             q
          * - - *
                |
            *   |
                | -p
              * |
                |
                *

    Sketch them on the same graph.
    Code:
          |           F
          |           * (p,q)
          |    a  *   |
          |   *       |
        O * - - + - - + -
     (0,0)|     |
          | *   |
          |     |
          | b * |
          |     |
          |     * (q,-p)
          |     G

    Now consider the triangle FOG.

    Let a \:= \:OF \:=\:\sqrt{(p-0)^2 + (q-0)^2} \:=\:\sqrt{p^2+q^2}

    Let b \:=\:OG\:=\:\sqrt{(q-0)^2+(\text{-}p-0)^2} \:=\:\sqrt{p^2+q^2}

    Let c \:=\:GF \:=\:\sqrt{(p-q)^2 + (q+p)^2} \:=\:\sqrt{2p^2 + 2q^2}


    We note that: . a^2 + b^2 \:=\:\left(\sqrt{p^2+q^2}\right)^2 + \left(\sqrt{p^2 + q^2}\right)^2 \:=\:2p^2+2q^2

    . . . and that: . c^2\:=\:\left(\sqrt{2p^2+2q^2}\right)^2\:=\:2p^2+2  q^2


    Since a^2 + b^2\:=\:c^2, then \Delta FOG is a right triangle with \angle FOG = 90^o.

    Follow Math Help Forum on Facebook and Google+

  4. #4
    Banned
    Joined
    Jan 2007
    Posts
    315

    ok

    I honor both replies but soroban did a fantastic job using the correct math symbols with LAtex.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Lines of best fit
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: September 24th 2011, 12:38 PM
  2. Replies: 1
    Last Post: April 18th 2010, 12:52 AM
  3. Replies: 1
    Last Post: March 14th 2010, 02:22 PM
  4. Replies: 2
    Last Post: January 25th 2009, 09:41 PM
  5. Lines
    Posted in the Pre-Calculus Forum
    Replies: 3
    Last Post: August 3rd 2007, 12:43 AM

Search Tags


/mathhelpforum @mathhelpforum