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Proving that this equation goes through the points

  1. Mar 25, 2009 #1
    1. The problem statement, all variables and given/known data
    Use a vector solution to show that a scalar equation for the line through the points P1(x1,y1) and P2(x2,y2) is y-y1/x-x2=y2-y1/x2-x1


    2. Relevant equations
    Find a vector which is normal to the line and then use the dot product of this vector and P1P


    3. The attempt at a solution
    I tried using numbers but I got lost. I don't what exactly this question is asking. Would I have to find the slope and then come up with a solution from that?
     
  2. jcsd
  3. Mar 25, 2009 #2

    tiny-tim

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    Hi unknown101! :smile:
    (is that the right answer?)

    Hint: call a point on the line P.

    What do you know about the vectors PP1 and P1P2 ? :wink:
     
  4. Mar 26, 2009 #3
    OK so I call a point P. I don't know what you mean about those two vectors. Would I have to find the m which is the slop and then n which is the normal point. Looking at my notes none of this makes sense:confused:
     
  5. Mar 26, 2009 #4

    tiny-tim

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    vector maths

    ah … doesn't look as if your teacher has introduced you to the joys of vector maths, as opposed to coordinate maths.

    Vector maths tries to avoid using coordinates …

    for example, (a + b)2 = a2 + 2a.b + b2 is proved by simple algebra, and is a lot easier than the coordinate proof! :wink:

    In vector maths, you can only really use three combinations …

    a.b, axb, and ka (where k is a constant) …

    which one(s) do you think would help in proving that PP1 is parallel to P1P2 ? :smile:
     
  6. Mar 26, 2009 #5
    Would I use a.b to prove that is it parallel?
     
  7. Mar 26, 2009 #6

    tiny-tim

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    It's easier to use a = kb
     
  8. Mar 26, 2009 #7
    So if I use a=kb will I have to find the constant. Looking at the equation I'm trying to prove, do I have to to find out what y and x( the x and y without the number)?
     
  9. Mar 27, 2009 #8

    tiny-tim

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    Hint: write a = kb in coordinates, and see if you can eliminate k. :wink:
     
  10. Mar 27, 2009 #9
    I wrote a=kb in coordinates. I don't if I did it correctly.
    x1y1=k(x2y2)
    x1k1/x2y2=k

    Is that all i need to do?
     
  11. Mar 27, 2009 #10

    tiny-tim

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    you're making it look like one equation …

    it's (x1 , y1) = k(x2 , y2), which is two equations …

    so write them out and eliminate k :smile:
     
  12. Mar 27, 2009 #11
    So I did that I got (x1, y1)/x2, y2)=k
     
  13. Mar 27, 2009 #12
    So this is the final answer?
     
  14. Mar 27, 2009 #13

    tiny-tim

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    uhhh?

    that doesn't even make sense …

    you can't divide a vector by another vector

    write two equations
     
  15. Mar 27, 2009 #14
    You said I should write out 2 equations:
    (x1,y1)=k(x2,y2)
    x1,y1=kx2,ky2
    Is that right?
     
  16. Mar 27, 2009 #15

    tiny-tim

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    Yes … but it would be clearer if you wrote it explicitly as two equations …

    anyway, now eliminate k :smile:
     
  17. Mar 27, 2009 #16
    By writing as two equations do you mean as in...
    1.(x1,y1)
    2.(kx2,ky2)

    Eliminate k. The only way I can think for eliminating k is
    kx2,ky2=0
    kx2,ky2/k=o/k
    x2,y2=0

    I don't know if I'm doing this right.
     
  18. Mar 27, 2009 #17

    tiny-tim

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    They aren't equations …

    an equation is something with an = sign in the middle
    ?? this doesn't make any sense at all

    write (x1 , y1) = k(x2 , y2) as two equations …

    that's two completely separate sentences, each with an = in the middle
     
  19. Mar 27, 2009 #18
    Ok I kind of understand it now.
    This is what I have so far
    x1,y1=kx2,ky2
    (x1,y1)/(k,k)=(kx2,ky2)/(k,k)
    (x1,y1)/(k,k)=(x2,y2)
     
  20. Mar 27, 2009 #19
    Is that right?
     
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