anonymous
  • anonymous
please look at my drawing @phi @KingGeorge
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
|dw:1360298829035:dw| find this vector
anonymous
  • anonymous
your x axis coordinate is incorrect
anonymous
  • anonymous
it should be (4,0)

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anonymous
  • anonymous
yes , thanks
anonymous
  • anonymous
That is a beautiful triangle you have there.
anonymous
  • anonymous
then use mid point formula your gonna get, (2,1) as coordinates of bisector n then your vector is , V= 2i +j
KingGeorge
  • KingGeorge
modphys doesn't want the mid point. He want the vector that's perpendicular.
anonymous
  • anonymous
well thats perpendicular bisector that means dividing equally that means mid point
KingGeorge
  • KingGeorge
What I would do, is write the line in the form \(y=-x/2+2\). The slope of the intersecting line will be \[-\frac{1}{-\frac{1}{2}}=2.\] So you have a line that looks like \(y=2x\) as the line that intersects. Set them equal to each other to get \(2x=-x/2+2\implies 5x/2=2\implies x=1\).
KingGeorge
  • KingGeorge
Plug that in again, \(x=1\implies y=2(1)=2\), so the point of intersection is \((1,2)\).
KingGeorge
  • KingGeorge
Which in turn means the vector is \[\vec{i}+2\vec{j}\]
anonymous
  • anonymous
I got same answer but here what my book did
1 Attachment
anonymous
  • anonymous
1 Attachment
KingGeorge
  • KingGeorge
If \(x=.8\), then the vector is \[.8\vec{i}+1.4\vec{j}\] Taking the dot product of that with the vector \[4\vec{i}-2\vec{j},\] we get \[.8\cdot4-2\cdot1.4=0.4\neq0\]so that solution is wrong.
RadEn
  • RadEn
|dw:1360299909016:dw| just check the magnitude of vector A : (4*2)/sqrt(4^2+2^2) = 8/sqrt(20) = 8/2sqrt(5) = 4/5 * sqrt(5) 0.8 and 1.6 has the same magnitude
phi
  • phi
Just as in your book, we can say the direction vector B (from pt (0,2) to (4,0) is (4,-2) let vector A be perpendicular to B. A dot B = 0 (x,y) dot (4,-2) =0 we get 4x = 2y pick a nice number for x, like 1. then y =2 vector A = (1,2) is perpendicular to vector B make A unit length, by dividing by its length. |A|= sqrt(5) Au = (1,2)/sqrt(5) now to get the length to scale Au so it reaches the line between (0,2) and (4,0), we notice that the vector (4,0) projected onto Au has the correct length Au dot (4,0) = 4/sqrt(5) The vector we want is 4/sqrt(5) * (1,2)/sqrt(5) = 4/5 * (1,2) or (0.8, 1.6)

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