Is there another way of graphing a line $y = mx + c$ rather than letting $m = 0, 1$? I was graphing the rule $y = 2x - 3$. Since this rule is of the form $y = mx + c$, we have substituted $m = 2$ and $c = -3$ when $x = 0$. This means the gradient (slope, steepness, etc) of the line $y$ we want to graph will be $2$ since $m$ denotes the gradient. We can mark our first co-ordinate on the graph by knowing that $x = 0$ and $c = -3$, forming the ordered pair $(0, -3)$ which is a point on our graph. Since every point is of the form $(x, y)$ then $c$ is equal to the $y$-intercept (the point where the line crosses the $y$ axis).
Now I have a question. My teacher asked me to let $x = 1$ after I set it equal to $0$, and according to the rule, $y = 2\cdot 1 - 3 = 2 - 3 = -1$ so we form another point $P(1, -1)$ where we write $P$ in front of the ordered pair to make it clear that it is a point on the graph. Since the rule is of the form $y = mx + c$, we are forming a linear graph where our line is always straight (of course since the measured gradient of the line at each of its points must be the same; a constant gradient). So from here, we can just draw a straight line through the two points we have plotted on the graph and Bob's your uncle.
However, I figured out something interesting. We could re-write $y = 2x -3$ as $y + 1 = 2(x - 1)$ and here we have constants $1, -1$ which are exactly the $x$ and $y$ coordinates of the point on our line for which $m = 1$. Is this a coincidence?
I am a little eager and impatient, and I do not want to wait for my next maths class to discuss this.
Thank you in advance.

Edit:
I have found my answer here $\longrightarrow$ How to Graph Equations $-$ Linear, Quadratic, Cubic, Radical, &amp; Rational Functions
But thanks for the help anyway :)
Once you have recognized that the graph will be a straight line, you just need two different points on it to actually draw it with a ruler.

For $y=mx+c$, setting $x=0$ and $x=1$  is a particularly  convenient way to get those two points (because those two numbers are both trivially easy to multiply by $m$, so it takes only a single addition to find all the numbers you need). But you don't have to do it that way if you have a different way to choose two points on the line you'd rather use.

If, instead, you have the equation on the form $y-b=m(x-a) $ like you rewrote it to, it makes perfect sense to let $(a, b) $ be one of your two points.
to graph the line $$y=mx+c$$ we will need two different points:
if $$x=0$$ we get $$y=c$$ this is the first point:
$$P_1(0,c)$$
second: $$y=0$$ we get $$x=-\frac{c}{m}$$ we get
$$P_2\left(\frac{-c}{m},0\right)$$