The problem asks for the equation of a line that is perpendicular to the line $y = 2x + 1$ and passes through the point $(-8, 4)$.

Step-by-step solution:

1. Identify the slope of the given line:
   The equation of the given line is $y = 2x + 1$. Here, the slope (m) is 2.

2. Find the slope of the perpendicular line:
   The slopes of two perpendicular lines are negative reciprocals of each other. So, the slope of the perpendicular line will be $m = -\frac{1}{2}$.

3. Use the point-slope form of the equation of a line:
   The point-slope form is $y - y_1 = m(x - x_1)$, where $(x_1, y_1)$ is a point on the line and $m$ is the slope.
   Substituting $m = -\frac{1}{2}$, $x_1 = -8$, and $y_1 = 4$: $y - 4 = -\frac{1}{2}(x + 8)$

4. Simplify the equation: Distribute the slope: $y - 4 = -\frac{1}{2}x - 4$ Add 4 to both sides: $y = -\frac{1}{2}x$

So, the equation of the line is $y = -\frac{1}{2}x$.

Answer:

The correct option is B: $y = -\frac{1}{2}x$.

Would you like more details on any of the steps, or have any questions?

Here are 5 related questions to practice:

1. How do you find the slope of a line from its equation in standard form?
2. What is the equation of a line parallel to $y = 3x + 5$ passing through $(2, -1)$?
3. How do you derive the equation of a line given two points?
4. Can two lines with the same slope be perpendicular?
5. What is the point-slope form of the equation of a line?

Tip: For perpendicular lines, always remember to take the negative reciprocal of the slope!