The problem asks for the equation of a line that passes through the origin and is perpendicular to the line $8x + y = 4$. Let's solve this step-by-step:

Step 1: Find the slope of the given line $8x + y = 4$

To find the slope, we first rewrite the equation in slope-intercept form $y = mx + b$, where $m$ is the slope.

8x + y = 4 \quad \text{(Subtract 8x from both sides)} $y = -8x + 4$ So, the slope $m_1$ of the given line is $-8$.

Step 2: Find the slope of the perpendicular line

The slope of a line perpendicular to another is the negative reciprocal of the original slope. The negative reciprocal of $-8$ is:

$m_2 = \frac{1}{8}$

Step 3: Use the point-slope form to find the equation of the new line

Since the new line passes through the origin (0, 0), we can use the slope-intercept form $y = mx + b$ with $m = \frac{1}{8}$ and $b = 0$ (since it passes through the origin).

$y = \frac{1}{8}x$

Thus, the equation of the line is:

$y = \frac{1}{8}x$

Final Answer:

$y = \frac{1}{8}x$

Would you like further details on any step, or do you have any questions?

Here are 5 related questions to expand on this topic:

1. What is the geometric significance of the negative reciprocal when finding perpendicular slopes?
2. How do we determine the y-intercept for lines that do not pass through the origin?
3. What would the equation look like if the new line passed through a different point, such as (2, 3)?
4. How can you derive the perpendicular slope using vector dot products?
5. How can this method be extended to three-dimensional coordinate geometry?

Tip: Always rewrite equations in slope-intercept form when identifying slopes to ensure clarity.