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Math Help - Points on line?

  1. #1
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    Question Points on line?

    I want to know if there's a way to find out if several points form a straight line. I could of course use y=kx+m but i don't wanna solve it that way. I could check if the points have the same y-value but different x-values. But how do I check if the line is a diagonal?
    e.g let's say I have theese points: P1(2,1), P2(4,3), P3(6,5), P4(8,7). I can easily visualize 'em and see that they form a straight line, but how can I solve it without drawing them?
    Last edited by crimshaft; February 3rd 2011 at 07:22 AM.
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    \vec{P_1P_3},\vec{P_1P_4} are proportional vectors to \vec{P_1P_2}


    Fernando Revilla
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    Ok, and then I have to do gauss-jordan Elimination? Hm... seems there's no easy way around this Thanks Fernando
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    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by crimshaft View Post
    Ok, and then I have to do gauss-jordan Elimination? Hm... seems there's no easy way around this

    You needn't Gauss Jordan Elimination and you needn't Lax Milgran Theorem either .

    Fernando Revilla
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    Quote Originally Posted by FernandoRevilla View Post
    You needn't Gauss Jordan Elimination and you needn't Lax Milgran Theorem either .

    Fernando Revilla
    What do you propose? How do I otherwise check if they're proportional to eachother? What method should I read about?
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    Quote Originally Posted by crimshaft View Post
    e.g let's say I have theese points: P1(2,1), P2(4,3), P3(6,5), P4(8,7). I can easily visualize 'em and see that they form a straight line, but how can I solve it without drawing them?
    P_2-P_1=<2,2>~\&~P_2-P_3=<-2,-2> because those are multiples of each other P_1,~P_2,~\&~P_3 are collinear.
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  7. #7
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    Ah ok. But i'm searching for a way to know if the points sum up to a straight line that is if they all lie on the same line. If I used that solution wouldn't it indicate that two paralell lines lie on the same line?
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  8. #8
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by crimshaft View Post
    What do you propose? How do I otherwise check if they're proportional to eachother? What method should I read about?

    (a,b) is proportional to (c,d)\neq (0,0) iff ad-bc=0 .

    Fernando Revilla
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  9. #9
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    Sry, my bad. I get it now. Though I might use the one point formula in the end. Think it will lead to a faster algorithm. Thanks!
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  10. #10
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    Similar but perhaps a little faster: If all the points lie on a single line, the slopes will be the same:
    \frac{y_1- y_0}{x_1- x_0}= \frac{y_2- y_0}{x_2- x_0}= \frac{y_3- y_0}{x_3- x_0}= \cdot\cdot\cdot= \frac{y_n- y_0}{x_n- x_0}
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