Showing two ordered pairs are parallel

[CaptainSFS]

1. The problem statement, all variables and given/known data
We say two ordered pairs p1=(x1,y1) and p2=(x2,y2) are perpendicular if and only if x1x2 +y1y2=0.

For any three ordered pairs p1, p2, and p3, if p1 is perpendicular to p2 and p2 is perpendicular to p3 then how would I show that p1 is parallel to p3?

I believe I need to somehow show that p1 and p3 have a similar scalar, but I do not know how to show that work. So if someone could show me, that would be awesome.

2. Relevant equations



3. The attempt at a solution
haha, i tried setting the two perpendicular equations equal to each other, but I don't know if that amounted to anything.

[HallsofIvy]

1. The problem statement, all variables and given/known data
We say two ordered pairs p1=(x1,y1) and p2=(x2,y2) are perpendicular if and only if x1x2 +y1y2=0.
I don't understand. I would tend to associate "ordered pairs" (ordered pairs of numbers or points?) with a point and points are not "parallel" or "pependicular". What are you definitions of "ordered pairs", "parallel", and "perpendicular" in this situation?

For any three ordered pairs p1, p2, and p3, if p1 is perpendicular to p2 and p2 is perpendicular to p3 then how would I show that p1 is parallel to p3?

I believe I need to somehow show that p1 and p3 have a similar scalar, but I do not know how to show that work. So if someone could show me, that would be awesome.

2. Relevant equations



3. The attempt at a solution
haha, i tried setting the two perpendicular equations equal to each other, but I don't know if that amounted to anything.

[Dick]

Ok, so you have p1x*p2x=-p1y*p2y and p3x*p2x=-p3y*p2y. What do you conclude if you divide those equations by each other? You should probably also worry about what happens if some of those terms happen to be zero, but I'll leave that case up to you.

[CaptainSFS]

Hey thanks for your help. that makes sense because they will always equal each other then if they are parallel. i was along the right track i think, but for some reason it makes sense. haha. thanks again for your help.

I have one more question though. Why are you allowed to divide the first equation by the second?

[Dick]

Hey thanks for your help. that makes sense because they will always equal each other then if they are parallel. i was along the right track i think, but for some reason it makes sense. haha. thanks again for your help.

I have one more question though. Why are you allowed to divide the first equation by the second?

If a=b and c=d and c and d are not zero then a/c=b/d. Right? It's sort of in the nature of equality.