Finding the line that is not parallel with others Find the line that is not parallel to the other three.


*

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

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

*$x = (1,1) + t(1,-1)$

*$x = (1,3) + t(-1,1)$
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?
 A: 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.
A: 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.
A: 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.
A: 
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


*

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

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