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

- anonymous

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

- schrodinger

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- goformit100

Use distance formula.

- anonymous

I don't see how that would help

- goformit100

|dw:1367651958829:dw|

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## More answers

- e.mccormick

How it would help is because of the properties of parallelograms.

- e.mccormick

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

- anonymous

I kinda see what you're saying, but this section is on vectors

- anonymous

if vector AB = vector DC and vector AD = vector BC, then it's a parallelogram.
vector AB = point B - point A

- anonymous

This is what my prof. put as the answer
ABCD is a Parrallogram if vector AB+VECTOR AD=VECTOR AC

- e.mccormick

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.

- anonymous

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.

- e.mccormick

Well, in this case, it isn't but... let me see.... I may have an easy reference already made.

- anonymous

right, for this problem the answer is that it isn't a Parrallogram

- e.mccormick

OK, this was a draft, but it has it in there.

##### 1 Attachment

- e.mccormick

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.

- anonymous

OOOOO

- e.mccormick

Yah, I know, most people don't have a rough draft of vector additon sitting around in a PDF. LOL.

- anonymous

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

- dan815

|dw:1367653361531:dw|

- e.mccormick

YES!

- dan815

or u can do it thru dot produts and see the angles inbetween match up or not

- anonymous

Brilliant! Thanks! :)
Is there a way for me to save this conversation

- e.mccormick

/nod /nod Which might be a better test for random points.

- anonymous

Interesting, i didn't even think of applying the dot product

- dan815

also another way

- e.mccormick

Oh no! It is math! There is more than one way to skin a math problem! Hehe.

- dan815

|dw:1367653577281:dw|

- dan815

just giving u a couple solutions cuz i dunno if u learnt the proof for cross product area yet

- dan815

check out the MIT 18.02 first or 2nd lecture for a nice explaination of cross product area and its proof

- e.mccormick

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.

- dan815

basically it has to do with |dw:1367653732887:dw|

- dan815

just watch the lecture, if u are interested in how its proved

- anonymous

I definitely will

- dan815

18.02 lecture 2 i think

- dan815

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

- anonymous

Okay, thanks again!

- dan815

your welcome man, have a good one

- e.mccormick

Have fun!

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