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hunter111
 one year ago
Given: points A(2; 1), B(3; 2), C(4; 4) andD(5; 2). IsABCDa parallelogram
hunter111
 one year ago
Given: points A(2; 1), B(3; 2), C(4; 4) andD(5; 2). IsABCDa parallelogram

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goformit100
 one year ago
Best ResponseYou've already chosen the best response.1Use distance formula.

hunter111
 one year ago
Best ResponseYou've already chosen the best response.1I don't see how that would help

goformit100
 one year ago
Best ResponseYou've already chosen the best response.1dw:1367651958829:dw

e.mccormick
 one year ago
Best ResponseYou've already chosen the best response.0How it would help is because of the properties of parallelograms.

e.mccormick
 one year ago
Best ResponseYou've already chosen the best response.0What 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

hunter111
 one year ago
Best ResponseYou've already chosen the best response.1I kinda see what you're saying, but this section is on vectors

alicealc
 one year ago
Best ResponseYou've already chosen the best response.0if vector AB = vector DC and vector AD = vector BC, then it's a parallelogram. vector AB = point B  point A

hunter111
 one year ago
Best ResponseYou've already chosen the best response.1This is what my prof. put as the answer ABCD is a Parrallogram if vector AB+VECTOR AD=VECTOR AC

e.mccormick
 one year ago
Best ResponseYou've already chosen the best response.0AH, 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.

hunter111
 one year ago
Best ResponseYou've already chosen the best response.1I'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.

e.mccormick
 one year ago
Best ResponseYou've already chosen the best response.0Well, in this case, it isn't but... let me see.... I may have an easy reference already made.

hunter111
 one year ago
Best ResponseYou've already chosen the best response.1right, for this problem the answer is that it isn't a Parrallogram

e.mccormick
 one year ago
Best ResponseYou've already chosen the best response.0OK, this was a draft, but it has it in there.

e.mccormick
 one year ago
Best ResponseYou've already chosen the best response.0If 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.

e.mccormick
 one year ago
Best ResponseYou've already chosen the best response.0Yah, I know, most people don't have a rough draft of vector additon sitting around in a PDF. LOL.

hunter111
 one year ago
Best ResponseYou've already chosen the best response.1so 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

dan815
 one year ago
Best ResponseYou've already chosen the best response.0or u can do it thru dot produts and see the angles inbetween match up or not

hunter111
 one year ago
Best ResponseYou've already chosen the best response.1Brilliant! Thanks! :) Is there a way for me to save this conversation

e.mccormick
 one year ago
Best ResponseYou've already chosen the best response.0/nod /nod Which might be a better test for random points.

hunter111
 one year ago
Best ResponseYou've already chosen the best response.1Interesting, i didn't even think of applying the dot product

e.mccormick
 one year ago
Best ResponseYou've already chosen the best response.0Oh no! It is math! There is more than one way to skin a math problem! Hehe.

dan815
 one year ago
Best ResponseYou've already chosen the best response.0just giving u a couple solutions cuz i dunno if u learnt the proof for cross product area yet

dan815
 one year ago
Best ResponseYou've already chosen the best response.0check out the MIT 18.02 first or 2nd lecture for a nice explaination of cross product area and its proof

e.mccormick
 one year ago
Best ResponseYou've already chosen the best response.0I 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.

dan815
 one year ago
Best ResponseYou've already chosen the best response.0basically it has to do with dw:1367653732887:dw

dan815
 one year ago
Best ResponseYou've already chosen the best response.0just watch the lecture, if u are interested in how its proved

dan815
 one year ago
Best ResponseYou've already chosen the best response.018.02 lecture 2 i think

dan815
 one year ago
Best ResponseYou've already chosen the best response.0in simple 2D terms, cross product is basically like applying the dotproduct for the new rotated vector now that the angle between them is 90theta and inbetween them

dan815
 one year ago
Best ResponseYou've already chosen the best response.0your welcome man, have a good one
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