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Find the line that is not parallel to the other three.

  1. $x = (2,0) + t(1,1)$

  2. $x = (1,2) + t(-1,1)$

  3. $x = (1,1) + t(1,-1)$

  4. $x = (1,3) + t(-1,1)$

Not sure how to approach this as I only know how to find parallel vectors. So how can I find parallel lines? Apparently, the line that is not parallel to the other three is option 1, which I believe is because of the point (1, 1) not being a scalar multiple of the others, but what about the other point?

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    $\begingroup$ You are on the right track. Lines are parallel iff their direction vectors are parallel. Direction vectors are parallel iff they are scalar multiples of each other. $\endgroup$ – trancelocation Sep 12 at 4:04
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From another perspective: $$1) \quad x = (2,0) + t(1,1) \Rightarrow (x,y)=(2+t,t) \Rightarrow y=x-2 \quad \quad \quad \ \\ 2) \quad x = (1,2) + t(-1,1) \Rightarrow (x,y)=(1-t,2+t) \Rightarrow y=-x+3\\ 3) \quad x = (1,1) + t(1,-1) \Rightarrow (x,y)=(1+t,1-t) \Rightarrow y=-x+2\\ 4) \quad x = (1,3) + t(-1,1) \Rightarrow (x,y)=(1-t,3+t) \Rightarrow y=-x+4\\$$ The lines are parallel except the first as their slopes ($-1$) are equal.

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So parallel lines have the same "slope"

How can we generalize that? Well, for line 1, we go up $1$ on y-axis for every $1$ we go up on $x$-axis.

For the other three, we go down $1$ on the y-axis for every $1$ we go up on $y$-axis.

SO #1 is not parallel.

We can also see this because $(-1,1)$, $(1,-1)$, and $(-1,1)$ are all multiples of each other, so $(1,1)$ is the oddball.

Remember parallel means same slope so we're not concerned with the first term in each line expression.

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You have given a pair of straight lines in each line of text. The two vectors are given by $$( u,v)+ t(a,b) $$

Parallelism depends on slope equality. So we can disregard first bracket in each case.

The relative slope we can define is then simply

$$\frac {dy/dt}{dx/dt}=b/a $$

Among the four choices the first has a relative slope $+1$ and the other three have $-1$

So the first choice is correct for parallelism.

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Not sure how to approach this as I only know how to find parallel vectors.

Well, that is what this is really all about - checking if two vectors are parallel.

If you have two lines like

  1. $(x, y) = (c1, d1) + t(a1,b1)$

  2. $(x, y) = (c2, d2) + t(a2,b2)$

then $(a1,b1)$ and $(a2,b2)$ are direction vectors for the lines. If those direction vectors are parallel then the lines are parallel.

So simply calculate the determinant like

$a1 * b2 - b1*a2$

If result is zero the lines are parallel.

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