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the vector would have to be a scalar of (4,-1) in order to be parallel to it

4 = 8t
-1 = 2t

it can also be in the opposit direction and be parallel

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-1/4 is the slope we want to obtain right?

what is it possible?

are you talking about the first one or second?

b/c the second one I got -1/4 and I can't square root it

the second one cant be parallel; there is no hope of a negative slope with it

and the first one I can't get a value that's works because one of them needs to be negative?

okay so the second one doesn't work I can't explain it

I just can't completely find t when I tried

t^2 is always positive; no way to get a negative slope

ohh that's makes more sense thanks!

youre welcome :)

@amistre64: why has the vector be a scalar of (4,-1), it can be any scalar(4x,-1x) isn't it?

a scalar of (4,-1) simply means that any scalar will do

t<8,2> != x<4,-1> no matter what you do to it right?

they are orthoganal vector

maybe not ortho; but no the same slope :)

sry. what does <> mean?

<> denotes a vector as opposed to () which is a point

<3,5> is a vector; (3,5) is a point

Oh, so for no value of t are any of the two vecs parallel, isnt it?

correct

for \[t \in R\]

the question asked shows point notation; but is assumed to be vectors :)

thanks :)

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