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Parallelogram homework help

  1. Nov 18, 2008 #1
    1. The problem statement, all variables and given/known data
    This isn't really homework, but a not assigned problem out of my mathbook which is kinda confusing me..

    #74. We have two pairs of parallel lines in R^2 defined by the linear equations below:

    a1x + b1y = r1
    a1x + b1y = s1
    a2x + b2y = r2
    a2x + b2y = s2

    We assume that these lines enclose a parallelogram P. Find the very simplest formula for the area of P in terms of a1, b1, a2, b2, r1, r2, s1, s2.

    2. Relevant equations



    3. The attempt at a solution
    Not sure how to attempt this problem because there are 4 equations for two lines?
     
  2. jcsd
  3. Nov 18, 2008 #2

    Dick

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    Re: Parallelogram?

    There are FOUR lines. There are two pairs of parallel lines. The first two don't intersect and the last two don't intersect.
     
  4. Nov 18, 2008 #3
    Re: Parallelogram?

    Then the area P would be:

    P = ||r1 x s1||
    or..
    P = ||r2 x s2||

    Correct?
     
    Last edited: Nov 18, 2008
  5. Nov 18, 2008 #4

    Dick

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    Re: Parallelogram?

    r1 and s1 are numbers, not vectors. You can't take their cross product. You need to find some points on the intersection of the lines and take their difference to get vectors.
     
  6. Nov 20, 2008 #5
    Re: Parallelogram?

    I think I understand how to do it...

    Find the intersection of the points:
    a1x + b1y = r1 and a2x + b2y = r2
    a1x + b1y = r1 and a2x + b2y = s2
    a1x + b1y = s1 and a2x + b2y = r2
    a1x + b1y = s1 and a2x + b2y = s2

    Get the (x, y) for each of those points

    Then subtract two sets by your selected new origin getting a and b.
    Then put those into ||a x b||

    But the question is asking me to put it into a formula; any help?
     
  7. Nov 20, 2008 #6

    Dick

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  8. Nov 21, 2008 #7
    Re: Parallelogram?

    Couldn't I just take the determinant of a1, b1, a2, b2?

    Area = |a1b2 - a2b1|
     
  9. Nov 21, 2008 #8

    Dick

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    Re: Parallelogram?

    Sure you can take the determinant, but what makes you think that has anything to do with AREA? It's obviously wrong. The a's and b's alone only tell you about the angles of the sides. The r's and s's determine how far apart they are.
     
  10. Nov 21, 2008 #9

    HallsofIvy

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    Re: Parallelogram?

    It is true that if two intersecting sides of a parallelogram can be represented by [itex]\vec{v}[/itex] and [itex]\vec{u}[/itex] then the area of the rectangle is given by the length of [itex]\vec{u}\times\vec{v}[/itex].

    It is also true that the area of a rectangular solid with sides intersecting at one corner given by vectors [itex]\vec{u}[/itex], [itex]\vec{v}[/itex], [itex]\vec{w}[/itex] is given by the "triple product", [itex]|\vec{u}\cdot(\vec{v}\times \vec{w})|[/itex] which is equal to the determinant formed with the vectors as columns or rows.
     
  11. Nov 21, 2008 #10
    Re: Parallelogram?

    a1x + b1y = r1 and
    a2x + b2y = r2

    X1 = (r1b2-b1r2)/(a1b2-b1a2)
    Y1 = (a1r2-r1a2)/(a1b2-b1a2)

    a1x + b1y = r1 and
    a2x + b2y = s2

    X2 = (r1b2-b1s2)/(a1b2-b1a2)
    Y2 = (a1s2-r1a2)/(a1b2-b1a2)

    a1x + b1y = s1 and
    a2x + b2y = r2

    X3 = (s1b2-b1r2)/(a1b2-b1a2)
    Y3 = (a1r2-s1a2)/(a1b2-b1a2)

    a1x + b1y = s1 and
    a2x + b2y = s2

    X4 = (s1b2-b1s2)/(a1b2-b1a2)
    Y4 = (a1s2-s1a2)/(a1b2-b1a2)

    Using a1x + b1y = r1 and a2x + b2y = r2 as midpoint:

    AX = X2 - X1
    AY = Y2 - Y1

    BX = X3 - X1
    BY = Y3 - Y1

    Area = AXBY - BXAY

    So, the simplest formula for P would be:

    P = ((r1b2-b1s2)/(a1b2-b1a2) - (r1b2-b1r2)/(a1b2-b1a2))((a1r2-s1a2)/(a1b2-b1a2) - (a1r2-r1a2)/(a1b2-b1a2)) - ((s1b2-b1r2)/(a1b2-b1a2) - (r1b2-b1r2)/(a1b2-b1a2))((a1s2-r1a2)/(a1b2-b1a2) - (a1r2-r1a2)/(a1b2-b1a2))
     
    Last edited: Nov 21, 2008
  12. Nov 21, 2008 #11

    Dick

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    Re: Parallelogram?

    If that is correct, then it's one answer. Have you checked it. I really doubt that it's the simplest. Since this isn't homework, I hope you aren't neglecting your real homework to pour time into this. Gotta confess, I actually don't know the answer.
     
  13. Nov 21, 2008 #12
    Re: Parallelogram?

    It hasn't been simplified, so it isn't in the simplest form. I am just checking if I did the correct steps as I have nothing to compare against to see if the answer is correct.

    I am not neglecting my homework to do this.
     
  14. Nov 21, 2008 #13

    Dick

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    Re: Parallelogram?

    Ok, it looks like you are doing the right sort of stuff. If you really want to check if it's correct, why don't you make some sample parallelograms with real numbers and check it out? I'm sorry to seem lazy here and not checking it myself. But, in fact, I am lazy. Sorry. The lord helps those who help themselves.
     
  15. Nov 23, 2008 #14
    Re: Parallelogram?

    Can anyone else confirm if I did this problem correctly?
     
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