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hunter111

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

  • 11 months ago
  • 11 months ago

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  1. goformit100
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    Use distance formula.

    • 11 months ago
  2. hunter111
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    I don't see how that would help

    • 11 months ago
  3. goformit100
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    |dw:1367651958829:dw|

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

    • 11 months ago
  5. e.mccormick
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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

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

    • 11 months ago
  7. alicealc
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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

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

    • 11 months ago
  9. e.mccormick
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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.

    • 11 months ago
  10. hunter111
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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 months ago
  11. e.mccormick
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    Well, in this case, it isn't but... let me see.... I may have an easy reference already made.

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

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

    • 11 months ago
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  14. e.mccormick
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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.

    • 11 months ago
  15. hunter111
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    OOOOO

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

    • 11 months ago
  17. hunter111
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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

    • 11 months ago
  18. dan815
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    |dw:1367653361531:dw|

    • 11 months ago
  19. e.mccormick
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    YES!

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

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

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

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

    • 11 months ago
  24. dan815
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    also another way

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

    • 11 months ago
  26. dan815
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    |dw:1367653577281:dw|

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

    • 11 months ago
  28. dan815
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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

    • 11 months ago
  29. e.mccormick
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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.

    • 11 months ago
  30. dan815
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    basically it has to do with |dw:1367653732887:dw|

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

    • 11 months ago
  32. hunter111
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    I definitely will

    • 11 months ago
  33. dan815
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    18.02 lecture 2 i think

    • 11 months ago
  34. dan815
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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

    • 11 months ago
  35. hunter111
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    Okay, thanks again!

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

    • 11 months ago
  37. e.mccormick
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    Have fun!

    • 11 months ago
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