I'm in precalculus and totally new to this 'proofs with axioms' concept, and it's stumping me beyond belief. I get the concept, I just don't understand how to start or where to implement the proofs.

The problem states: Give a careful proof of $|a-b| = |b-a|$. Justify each step.

This is how I attempted…

assume $a,b ∈ \mathbb{R}$  

$(a,b) \geq 0$, only $0$ when $a=b$ -absolute value property 1  

Lets say $|b-a| = c$, so  

$|a-b|=c$  

$|a||-b|=c$ (absolute value property 3)  

$|a||+b|=c$ (absolute value property 2)  

$|b||a|=c$ (commutative for multiplication axiom //would this be addition?)  

$|b||-a|=c$ (absolute value property 2)  

$|b-a|=c$ (absolute value property 3)  

$|b-a|=|b-a|$ ?? 

I'm confused because the hint for the homework says I will need to use one of the distributive rules, the definition of absolute value, and some of the absolute value properties. 
..I have no idea what the point would be to use the distributive rule for this. Add another factor? For the sake of what?

Here are the absolute value properties…

1. $|a| \geq 0$
2. $|a| = |-a|$
3. $|ab| = |a||b|$
4. $|a/b| = |a|/|b|$
5. $|a+b|<= |a|+|b|$