I know the equation, $y = mx + b$ where $m$ is slope and $b$ is $y$-intercept, is a straight line. But I know also that $Ax + By = C$ is a straight line equation, but how does it represent a straight line?
[Math] How is $Ax + By = C$ the equation of a straight line
algebra-precalculus
Related Solutions
$b$ 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 $y = 7 = b$.
In space $\mathbb{R}^2$ with Euclidean metric, one of the possible definitions of a straight line is that
A line is an equivalence class of ordered triplet $(a,b,c)$ of reals $a,b,c$ such that at least one of $a,b$ is nonzero, and where equivalence relation is given by $(a,b,c)\sim(a',b',c')$ if and only if there is real $\lambda\neq 0$ such that $(a,b,c)=\lambda(a',b',c')$
You can define a line on $\mathbb{R}^2$ via locus of equation $ax+by+c=0$. Then one can show that there is a one to one correspondence given by definition above: important thing is that equations $ax+by+c=0$ and $\lambda(ax+by+c)=0$ represent the same line if $\lambda\neq 0$, so really you need to think of $(a,b,c)$ and $\lambda(a,b,c)$ as the same thing, which is expressed by "equivalence classes of ~". Also note that the condition saying at least one of $a,b$ being nonzero prevents us having $0\cdot x+0\cdot y +c=0$, which represents whole space if $c=0$ and emptyset otherwise (and both are clearly not lines)
You accept this as a definition of line and show its usual expected properties. If $b\neq 0$, then we can have usual slope-form of line via $y=-\frac{a}{b} x -\frac{c}{b}$ dividing. If $a=0$ we cannot do this and equation reduces to $y=-c/b$, a horizontal line. Similarly if $b=0$ the equation represents vertical lines (of "infinite" slope)
From the slope form, let $m=-a/b$, $-c/b=c'$. Then $y=mx+c'$
and at any point $(a,b)$ on the line, $$\frac{f(a+\Delta x)-f(a)}{(a+\Delta x) - a}=\frac{m(a+\Delta x)-ma}{\Delta x}=\frac{m\Delta x}{\Delta x}=m$$
So sending $\Delta x\to 0$ yields $dy/dx=m$
Best Answer
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 $(C/A,0)$.