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Given: points A(2; 1), B(3; 2), C(4; 4) andD(5; 2). IsABCDa parallelogram
 11 months ago
 11 months ago
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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goformit100Best ResponseYou've already chosen the best response.1
Use distance formula.
 11 months ago

hunter111Best ResponseYou've already chosen the best response.1
I don't see how that would help
 11 months ago

goformit100Best ResponseYou've already chosen the best response.1
dw:1367651958829:dw
 11 months ago

e.mccormickBest ResponseYou've already chosen the best response.0
How it would help is because of the properties of parallelograms.
 11 months ago

e.mccormickBest ResponseYou've already chosen the best response.0
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

hunter111Best ResponseYou've already chosen the best response.1
I kinda see what you're saying, but this section is on vectors
 11 months ago

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

hunter111Best ResponseYou've already chosen the best response.1
This is what my prof. put as the answer ABCD is a Parrallogram if vector AB+VECTOR AD=VECTOR AC
 11 months ago

e.mccormickBest ResponseYou've already chosen the best response.0
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

hunter111Best ResponseYou've already chosen the best response.1
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

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

hunter111Best ResponseYou've already chosen the best response.1
right, for this problem the answer is that it isn't a Parrallogram
 11 months ago

e.mccormickBest ResponseYou've already chosen the best response.0
OK, this was a draft, but it has it in there.
 11 months ago

e.mccormickBest ResponseYou've already chosen the best response.0
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

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

hunter111Best ResponseYou've already chosen the best response.1
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

dan815Best ResponseYou've already chosen the best response.0
or u can do it thru dot produts and see the angles inbetween match up or not
 11 months ago

hunter111Best ResponseYou've already chosen the best response.1
Brilliant! Thanks! :) Is there a way for me to save this conversation
 11 months ago

e.mccormickBest ResponseYou've already chosen the best response.0
/nod /nod Which might be a better test for random points.
 11 months ago

hunter111Best ResponseYou've already chosen the best response.1
Interesting, i didn't even think of applying the dot product
 11 months ago

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

dan815Best ResponseYou've already chosen the best response.0
just giving u a couple solutions cuz i dunno if u learnt the proof for cross product area yet
 11 months ago

dan815Best ResponseYou've already chosen the best response.0
check out the MIT 18.02 first or 2nd lecture for a nice explaination of cross product area and its proof
 11 months ago

e.mccormickBest ResponseYou've already chosen the best response.0
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

dan815Best ResponseYou've already chosen the best response.0
basically it has to do with dw:1367653732887:dw
 11 months ago

dan815Best ResponseYou've already chosen the best response.0
just watch the lecture, if u are interested in how its proved
 11 months ago

dan815Best ResponseYou've already chosen the best response.0
18.02 lecture 2 i think
 11 months ago

dan815Best ResponseYou've already chosen the best response.0
in 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
 11 months ago

hunter111Best ResponseYou've already chosen the best response.1
Okay, thanks again!
 11 months ago

dan815Best ResponseYou've already chosen the best response.0
your welcome man, have a good one
 11 months ago
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