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hunter111

  • one year ago

Given: points A(2; 1), B(3; 2), C(4; 4) andD(5; 2). IsABCDa parallelogram

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  1. goformit100
    • one year ago
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    Use distance formula.

  2. hunter111
    • one year ago
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    I don't see how that would help

  3. goformit100
    • one year ago
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    |dw:1367651958829:dw|

  4. e.mccormick
    • one year ago
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    How it would help is because of the properties of parallelograms.

  5. e.mccormick
    • one year ago
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    What goformit100 is talking about should be somewhere in your book. A list of information like this one: http://www.regentsprep.org/Regents/math/geometry/GP9/LParallelogram.htm

  6. hunter111
    • one year ago
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    I kinda see what you're saying, but this section is on vectors

  7. alicealc
    • one year ago
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    if vector AB = vector DC and vector AD = vector BC, then it's a parallelogram. vector AB = point B - point A

  8. hunter111
    • one year ago
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    This is what my prof. put as the answer ABCD is a Parrallogram if vector AB+VECTOR AD=VECTOR AC

  9. e.mccormick
    • one year ago
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    AH, in vectors. OK, yes, if they define the sides of the parallelogram. So if you can prove that they are one vector added on to the other, and vice verca, then they define it.

  10. hunter111
    • one year ago
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    I'm on the verge of seeing what you're saying. But, I don't see how proving that by proving that when we add 2 of the vectors it equals the 3rd one (which is given), indicates that the set of given points is a Parrallogram.

  11. e.mccormick
    • one year ago
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    Well, in this case, it isn't but... let me see.... I may have an easy reference already made.

  12. hunter111
    • one year ago
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    right, for this problem the answer is that it isn't a Parrallogram

  13. e.mccormick
    • one year ago
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    OK, this was a draft, but it has it in there.

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  14. e.mccormick
    • one year ago
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    If you look at the first page, the vectors can add up on either side to meet at the far corner. In this problem of yours, they add one way, but not the other. Well, they don't reach one of the points through addition.

  15. hunter111
    • one year ago
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    OOOOO

  16. e.mccormick
    • one year ago
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    Yah, I know, most people don't have a rough draft of vector additon sitting around in a PDF. LOL.

  17. hunter111
    • one year ago
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    so if we have to vectors and add them they form Parrallogram no matter what. But this problem is asking specifically if the Parrallogram they form has a vertices on the indicated points? essentially

  18. dan815
    • one year ago
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    |dw:1367653361531:dw|

  19. e.mccormick
    • one year ago
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    YES!

  20. dan815
    • one year ago
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    or u can do it thru dot produts and see the angles inbetween match up or not

  21. hunter111
    • one year ago
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    Brilliant! Thanks! :) Is there a way for me to save this conversation

  22. e.mccormick
    • one year ago
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    /nod /nod Which might be a better test for random points.

  23. hunter111
    • one year ago
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    Interesting, i didn't even think of applying the dot product

  24. dan815
    • one year ago
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    also another way

  25. e.mccormick
    • one year ago
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    Oh no! It is math! There is more than one way to skin a math problem! Hehe.

  26. dan815
    • one year ago
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    |dw:1367653577281:dw|

  27. dan815
    • one year ago
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    just giving u a couple solutions cuz i dunno if u learnt the proof for cross product area yet

  28. dan815
    • one year ago
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    check out the MIT 18.02 first or 2nd lecture for a nice explaination of cross product area and its proof

  29. e.mccormick
    • one year ago
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    I was also thinking if they formed a basis for the span of linar combinations it could also be an indicator.... but they might not be parts of the same parallelogram and still have that be true.

  30. dan815
    • one year ago
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    basically it has to do with |dw:1367653732887:dw|

  31. dan815
    • one year ago
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    just watch the lecture, if u are interested in how its proved

  32. hunter111
    • one year ago
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    I definitely will

  33. dan815
    • one year ago
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    18.02 lecture 2 i think

  34. dan815
    • one year ago
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    in simple 2D terms, cross product is basically like applying the dotproduct for the new rotated vector now that the angle between them is 90-theta and inbetween them

  35. hunter111
    • one year ago
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    Okay, thanks again!

  36. dan815
    • one year ago
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    your welcome man, have a good one

  37. e.mccormick
    • one year ago
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    Have fun!

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