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

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I don't see how that would help

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

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

e.mccormick
 3 years 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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I kinda see what you're saying, but this section is on vectors

anonymous
 3 years 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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0This is what my prof. put as the answer ABCD is a Parrallogram if vector AB+VECTOR AD=VECTOR AC

e.mccormick
 3 years 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.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I'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
 3 years 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.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0right, for this problem the answer is that it isn't a Parrallogram

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

e.mccormick
 3 years 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
 3 years 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.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so 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
 3 years 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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Brilliant! Thanks! :) Is there a way for me to save this conversation

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Interesting, i didn't even think of applying the dot product

e.mccormick
 3 years 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
 3 years 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
 3 years 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
 3 years 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
 3 years ago
Best ResponseYou've already chosen the best response.0basically it has to do with dw:1367653732887:dw

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

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