Deriving the parametric form of line

Discussion in 'Homework Help' started by zulfi100, Dec 20, 2013.

  1. zulfi100

    Thread Starter Active Member

    Jun 7, 2012
    384
    1
    Hi,
    Can somebody help me to derive the paramteric eq of line:
    Eq of Line: ax+by+c=0
    x= -cu/a
    y= c(u-1)/b
    I dont know what is 'u'?
    I got this from:
    www.cse.msu.edu/~cse472/lectures/14.pdf‎

    Somebody plz guide me.

    Zulfi.
     
  2. studiot

    AAC Fanatic!

    Nov 9, 2007
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    u is the parameter.

    What happens if you first put y=0 into the original equation?

    you get x = -c/a

    Now multiply this by any number (that's what a parameter is)

    you get x = -(c/a)u

    Try to see if you can now figure out for y.
     
  3. zulfi100

    Thread Starter Active Member

    Jun 7, 2012
    384
    1
    Hi,
    Thanks for your help. When I am trying by your method , i am getting the same value of y as that of x i.e.
    y=-cu/b

    However, when i put the value of x into eq:
    ax+by +c =0

    i have:
    a(-cu/a) +by + c=0

    & y= c(u-1)/b

    Why I cant use your method. Thanks for your help.

    Zulfi.
     
  4. studiot

    AAC Fanatic!

    Nov 9, 2007
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    There you are, you have figured it out.

    :)

    Try to understand what happens when you first put x=0, that is why we do this.

    Of course y is not only zero it can be any number, including 0.
    It is convenient to pick a value that eliminates y from the original equation and gives us a formula for x for that value of y only

    writing x= -cu/a

    gives us a whole family of lines as u varies and so x also varies.

    These are all parallel and have the same slope as the original, but only one of these is the one we want.

    In order to select the correct one we must reintroduce y.

    Now that x can be any value since u can be any value we cannot simply choose x = 0 like we did with y.

    We have to substitute for x in terms of u (as you did).

    This will then give us a correct formula for y in terms of u, that allows for possible values of x.

    In fact you can see that the second formula is also a family of lines that intersects the first family.

    You should draw a few out for yourself to see the effect.
     
  5. WBahn

    Moderator

    Mar 31, 2012
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    Since this is a line, we know that both x and y will change linearly, right?

    That means that our parameteric equations will be of the form:

    x = m1*u + b1
    y = m2*u + b2

    Now you can just pick any four different points on the line and solve for m1, b1, m2, and b2. But that's a pain and a hassle that is not needed.

    So let's pick a couple of special points and see if we can make things a lot simpler.

    Q1) There is a value of u that will result in x being zero. Call this value Ux. What is it?

    Q2) There is a value of u that will result in y being zero. Call this value Uy. What is it?

    Q3) Using the original equation for the line, what will y be at when u=Ux?

    Q4) Using the original equation for the line, what will x be at when u=Uy?

    Play with that a while and see how far you can get.
     
    Last edited: Dec 22, 2013
  6. zulfi100

    Thread Starter Active Member

    Jun 7, 2012
    384
    1
    Hi,
    Thanks for your response. I cant understand your answers for 1, 2, 3, & 4 but i would try and reply you as soon as possible.

    Zulfi.
     
  7. WBahn

    Moderator

    Mar 31, 2012
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    Ignore those answers. The bottom of the post (which I've since deleted) was what I was first typing and partway through I decided to take a diffrerent approach. The two sets of questions are completely unrelated.

    Aside: This is why I never type stuff, like obscene remarks or personal comments that I might be thinking, into a post or an e-mail. Too much of a chance that sooner or later I will accidentally hit Send without removing them first.
     
  8. zulfi100

    Thread Starter Active Member

    Jun 7, 2012
    384
    1
    Hi,
    Thanks for your time.
    Okay i am trying.
    x = m1*u + b1
    y = m2*u + b2

    1. There is a value of u that will result in x being zero. Call this value Ux. What is it?
    Ux= -b1/m1

    2.There is a value of u that will result in y being zero. Call this value Uy. What is it?
    Uy= -b2/m2

    Q3) Using the original equation for the line, what will y be at when u=Ux?
    y=m2 * u + b2
    y= m2 * (-b1/m1) + b2


    Q4) Using the original equation for the line, what will x be at when u=Uy?
    x = m1*u + b1
    x= m1 * (-b2/m2) + b1


    I have done this as far as i can think.
    Plz guide what you want to tell.
    Zulfi.
     
  9. studiot

    AAC Fanatic!

    Nov 9, 2007
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    519
    Did you look at my post#4?
     
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