We are tasked to solve for the values of $n$ and $q$ given the problem.

Problem Breakdown:

• Points $A(-3, 6)$, $B(n, q)$, and $C(-3, -4)$ are collinear.
• The ratio of $AB:AC$ is $2:5$.

Key Idea:

Using the concept of section formula, if three points are collinear and the ratio of their segments is known, the coordinates of the point dividing the segment can be found. The section formula for a point dividing a line in the ratio $m:n$ is: $(x, y) = \left( \frac{m x_2 + n x_1}{m+n}, \frac{m y_2 + n y_1}{m+n} \right).$

Step 1: Use the section formula

Here, $B(n, q)$ divides the segment $AC$ in the ratio $2:5$, so: $n = \frac{2(-3) + 5(-3)}{2+5}, \quad q = \frac{2(-4) + 5(6)}{2+5}.$

Step 2: Simplify

For $n$: $n = \frac{-6 - 15}{7} = \frac{-21}{7} = -3.$

For $q$: $q = \frac{-8 + 30}{7} = \frac{22}{7}.$

Final Answer:

$n = -3, \quad q = \frac{22}{7}.$

Let me know if you have questions or need further clarification!

Relative Questions:

1. How do we determine if points are collinear using the slope method?
2. Can you derive the section formula for dividing a line segment?
3. How is the ratio of division used in 3D coordinates?
4. What happens if $AB:AC$ is given as 1:1?
5. How can you verify the values of $n$ and $q$ are correct?

Tip: Always double-check your substitution when applying formulas to ensure precision in calculations.