m is the slope
b is the y-intercept.
if you are given 2 points, you find the slope by y = (y2-y1) / (x2-x1)
any one of the points can be x1,y1.
the other of the points is then x2,y2.
you solve for the slope first.
once you find the slope, you take any one of the points and replace x with the x-coordinate of that point and y with the y coordinate of that point to solve for the y-intercept.
for example:
your points are (1,1) and (2,3)
assign (x1,y1) to (1,1)
assign (x2,y2) to 2,3)
your slope is (y2-y1) / (x2-x1) = (3-1) / (2-1) = 2 / 1 = 2
the equation of y = mx + b becomes y = 2 * x + b
now you take any one of the points and replace x with the x-coordinate of that point and y with the y-coordinate of that point and solve for b.
we'll use (1,1).
y = 2 * x + b becomes 1 = 2 * 1 + b
simplify to get 1 = 2 + b
subtract 2 from both sides of this equation to get:
1 - 2 = b
solve for b to get b = -1
replace b with -1 and your equation becomes y = 2 * x - 1
here's a graph of your equation.
there is another way to find the y-intercept once you know the slope.
that is to use the point slope form of the equation.
that form is y-y1 = m * (x-x1).
the slope is 2 and the point we are using is (1,1).
that means we assigned (1,1) to (x1,y1).
all you need is a point on the line.
any point on the line will do.
this equation becomes (y-1) = 2 * (x-1).
simplify to get y-1 = 2x - 2
add 1 to both sides of this equation to get y = 2x - 1.
that's the slope intercept form of the equation, once again.
whether you start with y = 2x + b and replace y with 1 and x with 1 gets you the same answer as when you use y-1 = 2 * (x - 1).
the second form is called the point slope form.
here's some references that might help.
http://www.regentsprep.org/regents/math/algebra/ac1/eqlines.htm
http://www.purplemath.com/modules/strtlneq.htm
http://www.purplemath.com/modules/strtlneq2.htm
http://www.regentsprep.org/regents/math/algebra/ac1/indexAC1.htm