We are tasked with finding the equation of a line that passes through the point $A(3, -4)$ and is parallel to the line $y = -x + 8$.

Steps:

1. Identify the slope: The given line equation $y = -x + 8$ is in slope-intercept form $y = mx + b$, where $m$ is the slope.

   • The slope of the given line is $m = -1$.
   • Since the required line is parallel to this line, its slope is also $-1$.

2. Use point-slope form: The point-slope form of a line is: $y - y_1 = m(x - x_1)$ Where $(x_1, y_1)$ is a point on the line, and $m$ is the slope.

   • Here, $(x_1, y_1) = (3, -4)$ and $m = -1$.

   Substituting these values: $y - (-4) = -1(x - 3)$ Simplifying: $y + 4 = -1(x - 3)$ Distribute the $-1$: $y + 4 = -x + 3$ Subtract 4 from both sides: $y = -x - 1$

Thus, the equation of the line is: $y = -x - 1$

Would you like more details or have any questions about this? Here are some follow-up problems to consider:

1. Find the equation of the line parallel to $y = 2x + 3$ and passing through $(1, 2)$.
2. Write the equation of a line perpendicular to $y = -x + 8$ and passing through the point $(3, -4)$.
3. Determine the equation of the line passing through $(2, -1)$ and parallel to $y = \frac{1}{2}x + 7$.
4. Find the slope of the line that passes through the points $(3, -4)$ and $(0, 8)$.
5. Convert the line $y = -x + 8$ to standard form.

Tip: Parallel lines always have the same slope, while perpendicular lines have slopes that are negative reciprocals of each other.