Find the general equation of the line that passes through the point P(1;-4) and is parallel to the line L:3x+y-8=0

Given line \( L: 3x + y - 8 = 0 \)

1. The slope-intercept form of the line \( L \) is found by solving for \( y \):
y = -3x + 8

2. From the slope-intercept form, the slope \( m \) of the line \( L \) is:
m = -3

3. Since parallel lines have the same slope, the new line passing through point \( P(1, -4) \) also has a slope of \( -3 \). Using the point-slope form of a line equation:
y - y_1 = m(x - x_1)
where \( (x_1, y_1) = (1, -4) \) and \( m = -3 \),
y + 4 = -3(x - 1)

4. Solve for \( y \) to get it in slope-intercept form:
y + 4 = -3x + 3
y = -3x + 3 - 4
y = -3x - 1

5. Convert back to the standard form of the equation:
3x + y + 1 = 0

So, the general equation is:

3x + y + 1 = 0