let the equation bel :

y- y1 = m(x-x1)

Given the slope m = 2/3

==> y-y1 = (2/3)(x-x1)

Let (x1,y1) be any point of the line.

We know that the y-intercept = -8

Then the point (0,-8) is on the line:

==> y-(-8) = (2/3)(x-0)

==> y+ 8 = (2/3)x

**==> y= (2/3)x - 8 **

The equation of the line with slope m and y intercept c is y = mx+c. The line is inclined to x axis with a slope of arctanm and intersects y axis at c units from the origin.

Given m = 2/3 and c = -8.

So substituting the values of m and c in y = mx+c, we get:

y = (2/3)x+(-8) .....(1)

y = (2/3)x - 8. is the required equation in slope intercept form. We can write this in ax+by+k = 0 form also.

(1)*3 gives:

3y = 2x-8*3

3y = 2x-24.

Subtract 3y and rearrrange:

2x-3y-24 = 0 which is in the standard ax+by+k = 0 form.