A line having a slope of 3/4 passes through the point (−8,4).Write the equation of this line in slope-intercept form.

Let the required line be y = mx+c in the slope intercept
form, where m is the slope and c is the y intercept.

We
determine m and c by the using given conditions.

m = 3/4,
given.

So y = (3/4)x+c
(1).

The line at (1) has the point (-8, 4) on
it.

Therefore  4 =
(3/4)*(-8)+c...(2)

(1)-(2) gives: y-4 =
(3/4)(x+8).

=> y -4 = (3/4)x +
6

We  multiply 4  to get integral
coefficients.

4(y-4) =
3x+24

We
rearrange.

3x-4y+24+16=
0

3x-4y+40 = 0 is the required
line.