Results 1 to 4 of 4
Like Tree3Thanks
  • 3 Post By BobP

Math Help - Straight Line through a Fixed Point

  1. #1
    Newbie
    Joined
    May 2013
    From
    India
    Posts
    16

    Straight Line through a Fixed Point

    I am not able to solve this problem

    Two fixed straight line OX and OY are cut by a variable line in the points A and B respectively and P and Q are feet of perpendiculars drawn from A and B upon the lines OBY and OAX. Show that if AB pass through a fixed point, then PQ will also pass through a fixed point.

    What I am doing is
    I am considering OX and OY as axes.

    Thanx for advance

    Straight Line through a Fixed Point-img_0117.jpg
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member
    Joined
    Nov 2013
    From
    Australia
    Posts
    191
    Thanks
    39

    Re: Straight Line through a Fixed Point

    Hi Varunkanpur,
    I don't understand your question. Maybe it is just me.

    Does someone in the forum understand this question?
    If so, could they explain it to me please?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member
    Joined
    Jun 2009
    Posts
    660
    Thanks
    133

    Re: Straight Line through a Fixed Point

    It yields to a co-ordinate geometry approach, I can't see anything better.

    Let OX be the horizontal co-ordinate axis and let OY be the line with equation y=mx, (i.e., OY need not be the vertical co-ordinate axis).

    Let the straight line AB, (A and B being the intercepts with OX and OY respectively) pass through the fixed point S(s,t) and have the variable gradient m'.

    The equation of AB will be y=m'x+(t-m's).

    Substituting y=0 gets us the co-ordinates of A,  \left ( \frac{m's-t}{m'},0 \right ).

    Solving simultaneously y=mx with the equation of AB produces the co-ordinates of B, \left(\frac{m's-t}{m'-m},\frac{m(m's-t)}{m'-m}\right).

    For Q, drop a vertical from B onto OX, it will have co-ordinates \left(\frac{m's-t}{m'-m},0\right).

    For P, drop a vertical from A onto OY. If P has co-ordinates (v,w), then since AP is perpendicular to OY, \frac{w-0}{v-(m's-t)/m'}=-\frac{1}{m}, so

    mm'w=m's-t-m'v, and since w=mv,

    v=\frac{m's-t}{m'(m^{2}+1)} and w=\frac{m(m's-t)}{m'(m^{2}+1)}.

    Having the co-ordinates of P and Q, the equation of the line PQ can be calculated,

    y=x\frac{m-m'}{mm'+1}+\frac{m's-t}{mm'+1}.

    It can now be verified that for any value of the variable gradient m', the line always passes through the point with
    co-ordinates

    \left(\frac{s+mt}{m^{2}+1},\frac{ms-t}{m^{2}+1}\right).
    Thanks from johng, varunkanpur and Melody2
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    May 2013
    From
    India
    Posts
    16

    Re: Straight Line through a Fixed Point

    I thank you from answering my question, BobP
    It help me a lot. I was struck at this question.

    Again Thanx
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 10
    Last Post: September 30th 2013, 10:46 PM
  2. Fixed Point
    Posted in the Advanced Math Topics Forum
    Replies: 0
    Last Post: December 17th 2012, 09:38 AM
  3. Replies: 1
    Last Post: October 18th 2011, 11:45 AM
  4. Replies: 2
    Last Post: June 8th 2010, 04:59 PM
  5. Replies: 3
    Last Post: February 20th 2008, 10:17 AM

Search Tags


/mathhelpforum @mathhelpforum