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Points on a line

  1. Jul 23, 2009 #1
    1. The problem statement, all variables and given/known data
    Problem 1.2.10 from Linear Algebra and Its Applications by Gilbert Strang:

    Under what condition on [tex]y_{1}, y_{2}, y_{3}[/tex] do the points [tex](0,y_{1}), (1,y_{2}), (2,y_{3})[/tex] lie on a straight line?

    2. Relevant equations

    [tex] y = ax + b [/tex]

    3. The attempt at a solution

    If [tex]y_{1}=0[/tex],
    then
    [tex]y_{2}=y_{1}+1[/tex]
    and
    [tex]y_{3}=y_{1}+2[/tex]

    This is true, but I guess it's possible to create a lot of straight lines if starting out at the origin.
    My attempt at a solution even agrees with the solutions given by Strang.
    Am I not reading the problem correctly?
    I feel kind of silly for asking this, as it's really basic..
     
  2. jcsd
  3. Jul 23, 2009 #2

    Office_Shredder

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    You provide a possible solution, but there are a lot more. What they're really striving for:
    Any two points on the plane will give you a unique line. So given y1 and y2, a line will pass through your first two points, and only one such line will. What are the conditions for the third line to lie on the line also? Hint: Think about the slope of the line
     
  4. Jul 23, 2009 #3

    CompuChip

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    So apparently for the special case y1 = 0 you chose, your solution agrees with the general solution from the book. Now try it for general y1 (i.e. what happens if you shift y1 by a constant c?)
     
  5. Jul 23, 2009 #4
    Slope of the first line:

    [tex] a = \frac{y_{2}-y_{1}}{x_{2}-x_{1}} [/tex]

    Slope of second line:

    [tex] a = \frac{y_{3}-y_{1}}{x_{3}-x_{1}} [/tex]

    The slopes of the two lines must be equal:

    [tex] \frac{y_{2}-y_{1}}{y_{3}-y{1}} = \frac{x_{2}-x_{1}}{x_{3}-x_{1}} [/tex]

    I think that;

    [tex] y_{2}-y_{1} = x_{2}-x_{1} [/tex]
    and
    [tex] y_{3}-y_{1} = x_{3}-x_{1} [/tex]

    are valid for all [tex] y_{1} [/tex]

    Thanks for helping me out guys!
     
  6. Jul 23, 2009 #5

    CompuChip

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    Yes, now plug in the numbers for xi: you should get the same equations as in your earlier post.
     
  7. Jul 23, 2009 #6
    Indeed sir,

    [tex] y_{2}-y_{1} = 1 - 0 = 1 [/tex]
    [tex] y_{3}-y_{1} = 2 - 0 = 2 [/tex]
     
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