please look at my drawing

@phi
@KingGeorge

Question

please look at my drawing

@phi 
@KingGeorge

anonymous · Answer

|dw:1360298829035:dw|

find this vector

anonymous · Answer

your x axis coordinate  is incorrect

anonymous · Answer

it should be (4,0)

anonymous · Answer

yes , thanks

anonymous · Answer

That is a beautiful triangle you have there.

anonymous · Answer

then use mid point formula your gonna get, (2,1) as coordinates of bisector n then your vector is , V= 2i +j

kinggeorge · Answer

modphys doesn't want the mid point. He want the vector that's perpendicular.

anonymous · Answer

well thats perpendicular bisector that means dividing equally that means mid point

kinggeorge · Answer

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 · Answer

Plug that in again, $x=1\implies y=2(1)=2$, so the point of intersection is $(1,2)$.

kinggeorge · Answer

Which in turn means the vector is $$\vec{i}+2\vec{j}$$

anonymous · Answer

I got same answer but here what my book did

anonymous · Answer

attachment

kinggeorge · Answer

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 · Answer

|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