Use the given conditions to write an equation for the line in point-slope form and general form. Passing through (-4,8) and parallel to the line whose equation is 3x-5y-6=0

Step 1 ：Given a line passing through the point (-4,8) and parallel to the line whose equation is 3x-5y-6=0.

Step 2 ：The slope of a line in the form ax + by + c = 0 is -a/b. So, the slope of the line 3x - 5y - 6 = 0 is -3/5.

Step 3 ：Since parallel lines have the same slope, the slope of the line passing through (-4,8) is also -3/5.

Step 4 ：The point-slope form of a line is y - y1 = m(x - x1), where m is the slope and (x1, y1) is a point on the line. Substituting the given point and the calculated slope, we get the equation of the line in point-slope form as $y-8=-0.6\ast (x+4)$.

Step 5 ：The general form of a line is Ax + By + C = 0. Rearranging the point-slope form, we get the general form of the line as $0.6\ast x+y-5.6=0$.

Step 6 ：Final Answer: The equation of the line in point-slope form is $\boxed{{\displaystyle y-8=-0.6\ast (x+4)}}$ and in general form is $\boxed{{\displaystyle 0.6\ast x+y-5.6=0}}$.