Straight line: Why one slope is different from other If I have two points $A(x_A, y_A)$ and $B(x_B, y_B)$, in order to find the slope I need:
$m_1 = \frac{y_\color{red}B - y_\color{red}A}{x_B - x_A}$
Ok, but I also discovered that $-\frac{a}{b}$ is also the slope, and to find these coefficients I need:
$a = y_\color{red}A - y_\color{red}B $
$b = x_B - x_A $
$m_2 = -\frac{a}{b}$
Note the reds. This is what change from one to other. ie,
$m_1 = -m_2$
Isn't better just remove that minus sign from $-\frac{a}{b}$ rewriting $a = y_\color{red}A - y_\color{red}B $ to $a = y_\color{blue}B - y_\color{blue}A$ like the first one? Because thats what produce the minus sign. Then,
$m_1= m_2$
Despite that, I see -a/b everywhere. Why this is not simplified? Is there any reason I'm missing?
 A: The important thing to consider is this: where do $a$ and $b$ come from? There must be an equation for the line that includes $a$ and $b,$ otherwise it's completely arbitrary. For example, if the line is given by the equation $ax+by+c=0$ (or some such), and we happen to know that $b\ne 0,$ then we find that the slope is indeed $-\frac{a}{b}.$
So, if we are given, say, the points $(-1,5)$ and $(3,3),$ then the slope of the line passing through them is $$\frac{3-5}{3-(-1)}=-\frac12.$$ So, if we put $a=1,b=2,$ then we would like for both points to satisfy the equation $x+2y+c=0$ for some $c.$ Substitution shows that $c=-9$ does the job, and so the line through the two points is given by the equation $x+2y-9=0.$
We could also (for example) have taken $a=-2,b=-4,$ in which case we would have found $c=18,$ so that $-2x-4y+18=0$ is yet another equation for that line. More generally, whatever (necessarily non-zero) value we take for $a,$ we will have $b=2a$ and $c=-9a.$
However, in this context, we could not have just taken $-\frac12=\frac{a}{b}.$ Why is that? Well, that would mean that $b=-2a$ for some $a\ne 0,$ and we'd want both points to satisfy $$ax+by+c=0\\ax-2ay+c=0$$ Substitution of the point $(-1,5)$ shows us that $c=11a,$ while substitution of the point $(3,3)$ shows us that $c=3a.$ But the only way we can have $3a=11a$ for a real number $a$ is if $a=0,$ which is impossible in this context. Hence, no such line can pass through both given points. That is why there is the (seemingly) unnecessary $-$ sign in there.
