anonymous
  • anonymous
Given: points A(2; 1), B(3; 2), C(4; 4) andD(5; 2). IsABCDa parallelogram
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
goformit100
  • goformit100
Use distance formula.
anonymous
  • anonymous
I don't see how that would help
goformit100
  • goformit100
|dw:1367651958829:dw|

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

e.mccormick
  • e.mccormick
How it would help is because of the properties of parallelograms.
e.mccormick
  • 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
  • anonymous
I kinda see what you're saying, but this section is on vectors
anonymous
  • anonymous
if vector AB = vector DC and vector AD = vector BC, then it's a parallelogram. vector AB = point B - point A
anonymous
  • anonymous
This is what my prof. put as the answer ABCD is a Parrallogram if vector AB+VECTOR AD=VECTOR AC
e.mccormick
  • 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
  • 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
  • e.mccormick
Well, in this case, it isn't but... let me see.... I may have an easy reference already made.
anonymous
  • anonymous
right, for this problem the answer is that it isn't a Parrallogram
e.mccormick
  • e.mccormick
OK, this was a draft, but it has it in there.
1 Attachment
e.mccormick
  • 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
  • anonymous
OOOOO
e.mccormick
  • e.mccormick
Yah, I know, most people don't have a rough draft of vector additon sitting around in a PDF. LOL.
anonymous
  • 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
  • dan815
|dw:1367653361531:dw|
e.mccormick
  • e.mccormick
YES!
dan815
  • dan815
or u can do it thru dot produts and see the angles inbetween match up or not
anonymous
  • anonymous
Brilliant! Thanks! :) Is there a way for me to save this conversation
e.mccormick
  • e.mccormick
/nod /nod Which might be a better test for random points.
anonymous
  • anonymous
Interesting, i didn't even think of applying the dot product
dan815
  • dan815
also another way
e.mccormick
  • e.mccormick
Oh no! It is math! There is more than one way to skin a math problem! Hehe.
dan815
  • dan815
|dw:1367653577281:dw|
dan815
  • dan815
just giving u a couple solutions cuz i dunno if u learnt the proof for cross product area yet
dan815
  • dan815
check out the MIT 18.02 first or 2nd lecture for a nice explaination of cross product area and its proof
e.mccormick
  • 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
  • dan815
basically it has to do with |dw:1367653732887:dw|
dan815
  • dan815
just watch the lecture, if u are interested in how its proved
anonymous
  • anonymous
I definitely will
dan815
  • dan815
18.02 lecture 2 i think
dan815
  • 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
  • anonymous
Okay, thanks again!
dan815
  • dan815
your welcome man, have a good one
e.mccormick
  • e.mccormick
Have fun!

Looking for something else?

Not the answer you are looking for? Search for more explanations.