To solve this problem, we need to find the value of $k$ so that the line passing through points $(-3, k)$ and $(3, 6)$ is parallel to the line passing through points $(13, -1)$ and $(8, 1)$.

Steps to Solve:

1. Find the Slope of the Second Line (Reference Line): For two points $(x_1, y_1)$ and $(x_2, y_2)$, the slope $m$ of the line passing through them is: $m = \frac{y_2 - y_1}{x_2 - x_1}$ For the points $(13, -1)$ and $(8, 1)$: $m = \frac{1 - (-1)}{8 - 13} = \frac{2}{-5} = -\frac{2}{5}$ So, the slope of the line containing $(13, -1)$ and $(8, 1)$ is $-\frac{2}{5}$.

2. Set Up the Slope Equation for the First Line: Since the line containing $(-3, k)$ and $(3, 6)$ is parallel to the line containing $(13, -1)$ and $(8, 1)$, it must have the same slope, $-\frac{2}{5}$.

   The slope of the line passing through $(-3, k)$ and $(3, 6)$ is: $m = \frac{6 - k}{3 - (-3)} = \frac{6 - k}{6}$ Setting this equal to $-\frac{2}{5}$, we have: $\frac{6 - k}{6} = -\frac{2}{5}$

3. Solve for $k$: Cross-multiply to clear the fraction: $5(6 - k) = -2 \cdot 6$ $30 - 5k = -12$ Rearranging terms: $-5k = -12 - 30$ $-5k = -42$ Divide by $-5$: $k = \frac{42}{5} = 8.4$

Answer:

$k = 8.4$

Would you like further clarification or details on any step?

Here are five related questions to expand understanding:

1. What is the definition of a slope and how does it relate to parallel lines?
2. How would the approach change if the lines were required to be perpendicular instead of parallel?
3. How can the concept of slope be applied to analyze the steepness of real-world inclines?
4. What is the general equation of a line, and how can it be derived from two points?
5. How does changing one point on a line affect the slope?

Tip: For parallel lines, remember that their slopes must be identical, as parallelism implies no difference in their angle of inclination.