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anonymous
 3 years ago
please look at my drawing
@phi
@KingGeorge
anonymous
 3 years ago
please look at my drawing @phi @KingGeorge

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1360298829035:dw find this vector

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0your x axis coordinate is incorrect

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0That is a beautiful triangle you have there.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0then use mid point formula your gonna get, (2,1) as coordinates of bisector n then your vector is , V= 2i +j

KingGeorge
 3 years ago
Best ResponseYou've already chosen the best response.1modphys doesn't want the mid point. He want the vector that's perpendicular.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0well thats perpendicular bisector that means dividing equally that means mid point

KingGeorge
 3 years ago
Best ResponseYou've already chosen the best response.1What 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
 3 years ago
Best ResponseYou've already chosen the best response.1Plug that in again, \(x=1\implies y=2(1)=2\), so the point of intersection is \((1,2)\).

KingGeorge
 3 years ago
Best ResponseYou've already chosen the best response.1Which in turn means the vector is \[\vec{i}+2\vec{j}\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I got same answer but here what my book did

KingGeorge
 3 years ago
Best ResponseYou've already chosen the best response.1If \(x=.8\), 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 \[.8\cdot42\cdot1.4=0.4\neq0\]so that solution is wrong.

RadEn
 3 years ago
Best ResponseYou've already chosen the best response.0dw: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
 3 years ago
Best ResponseYou've already chosen the best response.0Just 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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