If you solve for $y$ in $Ax+By = C$ ($B \neq 0$) you get slope-intercept form. In other words, $$By = C-Ax$$ $$y = \frac{C}{B}-\frac{Ax}{B}$$ so that $m = -A/B$ and $b = C/B$ in $y = mx+b$.
As Gonzalo points out, if $B = 0$ then we get $Ax = C$ or $x = C/A$ which is a line parallel to the $y$-axis through the point $(C/A,0)$.
Using The Pearson Complete Guide For Aieee 2/e
By Khattar as a point of reference for the items below.
You want to convert
$$\tag 1 x-2y-3=0$$
into normal (perpendicular) form.
We have:
$$Ax + By = -C \rightarrow (1)x + (-2)y = -(-3)$$
This means $C \lt 0$, so we divide both sides of $(1)$ by:
$$\sqrt{A^2 + B^2} = \sqrt{(1)^2 + (-2)^2} = \sqrt{5}$$
This yields:
$$\dfrac{1}{\sqrt{5}} x - \dfrac{2}{\sqrt{5}} y = -\dfrac{-3}{\sqrt{5}}$$
In normal form, this is:
$$\cos(\alpha)~x + \sin(\alpha)y = p \rightarrow \dfrac{1}{\sqrt{5}} x - \dfrac{2}{\sqrt{5}} y = -\dfrac{-3}{\sqrt{5}}$$
Note that this has a positive $x$ coordinate, and a negative $y$ coordinate, which puts us in the $4^{th}$ quadrant.
So, we have:
$$\cos(\alpha) = \dfrac{1}{\sqrt{5}}, \sin(-\alpha) = -\dfrac{2}{\sqrt{5}} , p = -\dfrac{-3}{\sqrt{5}} = \dfrac{3}{\sqrt{5}}$$
This gives us:
$$\alpha \approx 1.10714871779409 ~\mbox{radians}~ \approx 63.434948822922 {}^{\circ}$$
Since we are in the $4^{th}$ quadrant, we can write this angle as:
$$\alpha \approx 2 \pi - 1.10714871779409 ~\mbox{radians}~ \approx 5.176036589385497 ~\mbox{radians}~ \approx 296.565 {}^{\circ}$$
As another reference point, see $10.3.4$ (including examples) at Straight Lines.
Best Answer
$b$ represents the y-intercept of the line: where the line crosses the y-axis.
It can be found by setting $x = 0$ (the line crosses the y-axis when and only when $x = 0.$)
An example of such a line, $$y = 3x + \underbrace{7}_{\large b}$$ crosses the y-axis at the point $(0, 7)$.
When $x = 0$, you see that $y = 7 = b$.