What is the equation of the line using the y=mx+b? This is how far I've gotten so far: y-4=2/3(x-1). m=2/3 , (1,-4)

You can just use algebra to rearrange the terms into the y=mx+b form.

y - 4 = 2/3(x - 1)

                  ^ distribute   

-> y - 4 = 2x/3 -2/3

       +4             +4= 12/3

-> y = 2x/3 + 10/3

Take it from here

y - (-4)  = (2/3)(x - 1)   simplify

y + 4  = (2/3)x - 2/3      subtract 4 from both sides

y = (2/3)x - 2/3 - 4

y = (2/3)x - (4 + 2/3)

y = (2/3)x - 14/3

@CPhill... where did the y-(-4) come from tho....

Let the point  be  (1, -4)  = (x1, y1)

And in the point-slope form we have :

y - y1  = m ( x - x1)    substituting, we have

y - (-4)  = (2/3) (x - 1)      

y + 4  = (2/3) (x - 1)     and so forth

Another way to chack that my answer is correct is to sub 1 into the final equation......so we have

y = (2/3)(1)  - 14/3

y = 2/3 - 14/3

y = -12/3   = -4      which   gives us the point (1, -4)