# Show two unique points lie on a unique line.

1. Sep 7, 2013

1. The problem statement, all variables and given/known data
Suppose (a,b) and (c,d) are unique points in $\mathbb{R}^{2}$ that satisfy Ax+By+C=0. Suppose also that the solutions of Ax+By+C=0 form a line. Then the line that (a,b) and (c,d) lie on is unique.

2. Relevant equations
I showed in a previous part of the problem that there exists real numbers A,B,C not all equal to 0 that satisfy Aa+Bb+C=0 and Ac+Bd+C=0. I also showed that if A',B',C' were also nontrivial solutions that satisfied the two equations then A',B',C' were scalar multiples of A,B,C.

3. The attempt at a solution
Suppose that (a,b) and (c,d) are unique points that satisfy Ax+By+C=0. Suppose also that the set of solutions of Ax+By+C=0 forms a line that is not unique so there exists another equation such that A'x+B'y+C'=0 but we previously showed that A',B',C' must be scalar multiples of A,B,C so the two lines are actually the same. Hence the line that (a,b) and (c,d) lie on is unique.

I thought that since A',B',C' were scalar multiples of A,B,C then the new equation was just a stretched version of the older line so they must be the same line. Is this logic correct?

2. Sep 7, 2013

### Zondrina

Assuming you've shown this :

Then yes, the trick would be to assume that the line was not unique and that there are two lines. Then showing the two lines are the same shows that the original line must be unique.

3. Sep 7, 2013

Does A',B',C' being scalar multiples of A,B,C show that we are dealing with the same lines or should I be further justifying this?

4. Sep 7, 2013

### Zondrina

If you aren't changing the actual location of the coordinates $(a,b)$ and $(c,d)$, then the scalar line is a stretched version of the original line. So therefore they are the same line and pass through the same points, it's just one is longer than the other.

5. Sep 7, 2013

Ah! Cool. Thanks.

6. Sep 7, 2013

### HallsofIvy

Well, you wouldn't say "one is longer than the other" because lines, as opposed to line segments, so not have a "length".

7. Sep 7, 2013