How do you write the equation of a line in slope intercept form

when you have an inclination angle
and the line that it goes through
for example
the angle is 45 degrees and the line contains the point (-2,4)

Well the tangent of 45 degrees would be rise over run so
m = tan(45)
m = 1

y = mx + b
We have a sample point so we can solve for b
4 = 1*(-2) + b
b = 6

y = x + 6

Well, since you're working with degrees, I'll assume you have a background in trigonometry to do this. The tangent of any angle (45º in this case) is opposite over adjacent, which sounds a lot like "rise over run," which is the definition of your slope, "m."

So, by taking tan 45º, you get m=1

Then, use the equation

y-y.=m(x-x.) (the period means x-naught and y-naught, which are the values you put in)

so,

y-(4) = (1) (x- (-2))
y-4 = x+2
y = x + 6

there you go!

First you need it in the form y = mx + c where m is the slope or gradient and c is the intercept.
So you need to find m and c as you already have an x and y, (-2,4).
In this case when the slope is at an angle of 45 degrees, m = tan(45) = 1.
Plugging the values of m, x, and y into the general equation of the line gives,
4 = 1*(-2) + c
Rearranging to find c gives,
c = 4+2 = 6
So the equation of the line for this is,
y = x + 6

The slope of the line is m = tan 45º = 1. Now we use equation point-slope of a line

y - y1 = m(x-x1)

Substituting we get the equation of the line

y - 4 = 1(x - (-2))

or y = x + 6.