butterflyprincess
Major MATH HELP! D: ** Pic included!



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STEPHEN1998
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wat

butterflyprincess
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I need some help with this problem

STEPHEN1998
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ok

STEPHEN1998
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wats ur answer

butterflyprincess
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I dont have onne, I need help : I attached a link of the problem

STEPHEN1998
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i did

STEPHEN1998
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hold on


zordoloom
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Do you still need help?

butterflyprincess
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Yes please (:

STEPHEN1998
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ya

zordoloom
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Okay, Do you know how to even start working on this problem?


zordoloom
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So are you working with vectors?

butterflyprincess
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indeed I am!

butterflyprincess
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Im just tottally lost on this problem

zordoloom
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Ok, so have you done vector proofs before?

butterflyprincess
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no neverO=

zordoloom
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ok, so ob+ba=oa

zordoloom
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0=(0,0) A=(x1,y1) B=(x2,y2)

zordoloom
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Vector BA=AB=(x1x2, y1y2)


zordoloom
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Good so far?

zordoloom
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All what you have to do now is find vecotor OA and vecotr OB

butterflyprincess
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O: oh really but is there like another pic you have to draw/

zordoloom
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no


zordoloom
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dw:1352869007184:dw

zordoloom
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so vector BA is equal to AB
so when you subtract the points elementwise
you get x1x2 and y1y2

zordoloom
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dw:1352869116355:dw

butterflyprincess
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Oh right, now I find Oa Ob ?How can I do thta again?

zordoloom
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dw:1352869175679:dw

zordoloom
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now you do the same for vector ob

zordoloom
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dw:1352869297947:dw

butterflyprincess
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So OB is OB = OB= <0x2 , 0y2 > ?

zordoloom
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no, it would be <OB> = BO = <x2,y2>

zordoloom
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So, now you need to subtract <OB> from <OA>. to prove that it equals <AB>

butterflyprincess
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ohhh right!

zordoloom
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zordoloom
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look at the attachment, the circled ones.

zordoloom
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How would you prove the?

zordoloom
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*them?

zordoloom
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They are equal.

zordoloom
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so, by the subtraction of the vectors, you prove that <BA> = <OA><OB>

butterflyprincess
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OH theyre equal!!
Its < x1x2 , y1y2 > !

zordoloom
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pretty much.

zordoloom
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Now to solve part 2, set point c equal to (x3, y3)

zordoloom
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then I would solve for x3,y3 in terms of the other known quantities, x1,y1 and x2,y2 using the given ratio. the rest is just algebra.

butterflyprincess
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So all what xOA is saying is to ditribute the x?

zordoloom
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Ya

butterflyprincess
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Oh okay yy

butterflyprincess
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dw:1352870110521:dw


zordoloom
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yes.

butterflyprincess
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Cool thanks ((:

hartnn
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<OC>=CO

hartnn
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<BC>/<CA> =x/y
so,
(CB)/(AC) =x/y
now can, you isolate C from here ?
try it...

butterflyprincess
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you want me to isolare c from (CB)/(AC) =x/y
right?

hartnn
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yes,

hartnn
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u see, how i got, (CB)/(AC) =x/y
from given condition, <BC>/<CA> =x/y

butterflyprincess
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Wait how would you do that because C is being subtracted

hartnn
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u mean how to isolate C ?
crossmultiply....

butterflyprincess
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yea youd get y(cb) = x(ac) ...

hartnn
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try to isolate C from that...

butterflyprincess
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so I distribute..first then

butterflyprincess
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yccb = xa  xc

hartnn
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yes, distribute then bring all terms with 'c' together, then factor out c

hartnn
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its ycyb on the left...

butterflyprincess
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So C= xaxc / (yb)

hartnn
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there, u were not able to isolate C, notice, c is there on right side also

hartnn
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ycyb = xa  xc
yc+xc = xa+yb
got this step ?

hartnn
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now can u isolate c ?

butterflyprincess
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I thought ititd be yccb = xa  xc since u cross mult tho..

butterflyprincess
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dw:1352871594989:dw

hartnn
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dw:1352871643318:dw


hartnn
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dw:1352871704757:dw

butterflyprincess
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yesyes i did that xD

hartnn
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what does left side simplify to ?

hartnn
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knowing that its given that x+y=1

butterflyprincess
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C= xa+yb / y+x

hartnn
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what is x+y= ?

butterflyprincess
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Wait but does C equal to that O:

hartnn
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yes.

butterflyprincess
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oh okay lol Um y+x = .. ?

hartnn
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its given to be =1
see the question

butterflyprincess
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Oh cool so it would just be xa+yb

hartnn
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yes, c=xa+yb
now u know what to do next ?

butterflyprincess
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PLug it to...

hartnn
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we know that C=(x3,y3)
and O=(0,0)
so can you find <OC> = ?

butterflyprincess
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<x30,y30>

hartnn
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which is ?


hartnn
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so <OC> =C
so we can replace C by <OC>
right ?

butterflyprincess
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ohhyeauhhh! ool!

hartnn
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in same way, we can replace A by ... ?
B by ... ?

butterflyprincess
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A with x1,y1??

hartnn
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A with <OA>

hartnn
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<OC> =C
so we can replace C by <OC>
<OA> =A
so we can replace A by <OA>
<OB> =B
so we can replace B by <OB>
right ?


hartnn
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so put all those in
c=xa+y
and see whether u get what needs to be proven..

butterflyprincess
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dw:1352872625517:dw

butterflyprincess
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Do u mean plug in the value sliket his

hartnn
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dw:1352872688272:dw

butterflyprincess
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OH THATS what you ment I tots get that!

hartnn
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all steps clear ? any doubts ?

butterflyprincess
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Wait and how did u kno that OC=c and OA =A and OB = B ?

hartnn
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C = (x3,y3)
<OC> =CO = (x3,y3)(0,0) = (x30,y30) = (x3,y3) = C
same for A and b

hartnn
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got it ?

butterflyprincess
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yea thanks (: Do you think we can do another in like 10 min?? I I need to sleep asap since I need to wake up early tomorrow (x

hartnn
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same type ? ....actually even i need to go in few minutes...

hartnn
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lets me see the question atleast

butterflyprincess
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OK but && Oh and alsoo how did you know that ( C B)/(AC) = x/y

hartnn
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it is given that BC /CA = x/y
now
BC=CB
CA=AC

hartnn
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remember for any vector XY, we can write it as YX

butterflyprincess
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oh okay why is it flipped again?

hartnn
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flipped again ? means ?

butterflyprincess
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Oh I mean like why isnt it like BC= Bc instead its CB

hartnn
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its the conventional way of representation,
but if you take BC=BC , you will get same exact final answer....

hartnn
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so its no difference, just a way of representation

butterflyprincess
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Oh right k thnx!
lemme post the other problem I dont think we have enough tim :C