Results 1 to 3 of 3

Math Help - A simple question...

  1. #1
    Newbie
    Joined
    Feb 2009
    Posts
    18

    A simple question...

    Determine the slope of a line perpendicular to each of the following:
    a. y = 3x - 5
    b. 13x - 7y - 11 = 0

    Please help me with this problem, thanks
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member
    Joined
    Mar 2008
    Posts
    182
    Okay, I'll take part a, then you can try b..
    a. y = 3x - 5

    If a line L is perpendicular to M, then the slope of M * slope of L = -1
    So first we need to find the slope of our line, y = 3x - 5.
    The general equation of a line is expressed as y = mx + c where m is the slope, and c is some constant. So in our case c = -5, and m = 3.
    So the slope of the line a = 3.

    Now for a slope to be perpendicular,
    slope of a*slope of new line = -1.
    3 * slope of new line = -1
    slope of new line = -1/3.

    So the slope of the perpendicular line is -1/3.

    The method for b is the same...
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Feb 2009
    Posts
    18
    Thanks a lot!

    I know how to solve part b now.

    Simply find the slope of 13x-7y-11=0

    simplify the function first, then we got y=13/7x-11/7, the slope is 13/7

    Since slope of b * slope of new line = -1

    Therefore, the slope of a line perpendicular to 13x-7y-11=0 is -7/13.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] Really simple question too hard for a simple mind
    Posted in the Statistics Forum
    Replies: 2
    Last Post: October 5th 2010, 08:03 AM
  2. Easy Question - Simple Interest, Simple Discount
    Posted in the Business Math Forum
    Replies: 0
    Last Post: September 21st 2010, 08:22 PM
  3. Replies: 4
    Last Post: May 4th 2010, 09:08 AM
  4. Simple Integral Question (I think it's simple)
    Posted in the Calculus Forum
    Replies: 7
    Last Post: February 13th 2010, 02:37 PM
  5. Replies: 1
    Last Post: May 30th 2009, 02:59 PM

Search Tags


/mathhelpforum @mathhelpforum