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

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Use distance formula.
I don't see how that would help

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Other answers:

How it would help is because of the properties of parallelograms.
What goformit100 is talking about should be somewhere in your book. A list of information like this one:
I kinda see what you're saying, but this section is on vectors
if vector AB = vector DC and vector AD = vector BC, then it's a parallelogram. vector AB = point B - point A
This is what my prof. put as the answer ABCD is a Parrallogram if vector AB+VECTOR AD=VECTOR AC
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.
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.
Well, in this case, it isn't but... let me see.... I may have an easy reference already made.
right, for this problem the answer is that it isn't a Parrallogram
OK, this was a draft, but it has it in there.
1 Attachment
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.
Yah, I know, most people don't have a rough draft of vector additon sitting around in a PDF. LOL.
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
or u can do it thru dot produts and see the angles inbetween match up or not
Brilliant! Thanks! :) Is there a way for me to save this conversation
/nod /nod Which might be a better test for random points.
Interesting, i didn't even think of applying the dot product
also another way
Oh no! It is math! There is more than one way to skin a math problem! Hehe.
just giving u a couple solutions cuz i dunno if u learnt the proof for cross product area yet
check out the MIT 18.02 first or 2nd lecture for a nice explaination of cross product area and its proof
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.
basically it has to do with |dw:1367653732887:dw|
just watch the lecture, if u are interested in how its proved
I definitely will
18.02 lecture 2 i think
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
Okay, thanks again!
your welcome man, have a good one
Have fun!

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