This post can get long.
There are three forms. The general form, 0=ax+by+c. The point slope form is (y1-y2)=m(x1-x2). The y intercept form if y=mx+b. The notation of the constants can be confusing and tends to mix things.
First, lets look at the slope intercept form.
y=mx+b. Note that this form makes it easiest to graph a line given the equation. To understand this form it is best to look at the simplest equation for a line, y=x or if you expand it y=1x+0. This means that the line crosses the y axis at (0,0). Also, you can tell by the 1, which is 1/1 that if you take any point on the line, (0,0) being the easiest, and want to find another point all you have to do is add 1 to the x and y coordinate, (0+1,0+1)=(1,1) is another point on the line.
If i want to turn the y intercept into the other forms all i need to do is go to the general form, 0=ax+by+c.
y=mx+b and subtract y from both sides and you get 0=mx-y+b.
To find the point slope form it takes an extra step. Take the y intercept form and you need to find a point on the line. The simplest point to get is to take when x=0 then y=b, the point being (0,b). Another simple point is to take y=0 and solve for x, which gives you the point (-b/m,0). Then just plug in the point into the point slope form, (y-b)=m(x-0), and you can see how easy it is to solve for y and to get y=mx+b.
The point intercept form is good for finding the equation of a line given two points. The slope m=(y2-y1)/(x2-x1)=(y1-y2)/(x1-x2). Keep in mind that it does not matter if you do y2-y1 or y1-y2 as long as you follow the same order on the x side. Once you have the slope m you take one of the points and you get (y-y1)=m(x-x1). To change it to another form is easy.
Solve for y to get the slope intercept form, y=mx +(-mx1+y1), where (-mx1+y1) is a number and the y intercept. To get to the general form is obvious from this step.
Lastly, the general form, 0=ax+by+c. This is probably the best form when you get into advanced mathematics, but it can take some experience to master. You may not make a lot of use for this form in graphing. But if you do have this equation you can simply find any point on the line, just let y=0 and solve for x, or let x=0 and solve for y.
If you let x=0 then 0=by+c. If you let y=0 when 0=ax+c. That gives you a point on the line. Then you can find another point on the line by letting x=x1 and y=y1 instead of x=0 and y=0. But pick your x and y carefully so you can get a nice solution. You want something which cancels out a and b in a nice fashion; another easy point is x=a, y=b, but not always.
I could say more and there are plenty of other tricks about lines, but they are hard to explain in text on a forum.
