To find the value of $k$ such that the line through the points $(1-2k, -3)$ and $(2k-1, 2k+1)$ has a slope of 5, we can use the formula for the slope of a line given two points $(x_1, y_1)$ and $(x_2, y_2)$:

$\text{slope} = \frac{y_2 - y_1}{x_2 - x_1}$

Substituting the given points:

$\text{slope} = \frac{(2k+1) - (-3)}{(2k-1) - (1-2k)}$

Simplify the equation:

$\text{slope} = \frac{2k + 1 + 3}{2k - 1 - 1 + 2k}$

$\text{slope} = \frac{2k + 4}{4k - 2}$

We are given that the slope is 5, so:

$\frac{2k + 4}{4k - 2} = 5$

Now, solve for $k$:

1. Cross-multiply to eliminate the fraction:

$2k + 4 = 5(4k - 2)$

2. Expand and simplify:

$2k + 4 = 20k - 10$

3. Rearrange the equation to isolate $k$:

$4 + 10 = 20k - 2k$

$14 = 18k$

$k = \frac{14}{18} = \frac{7}{9}$

So, the value of $k$ is $\frac{7}{9}$.

Would you like more details or have any questions? Here are 5 related questions to explore:

1. How would the slope change if $k$ were doubled?
2. What is the equation of the line when $k = \frac{7}{9}$?
3. How do you find the midpoint of the line segment between the two points?
4. What happens to the slope if both points move up by the same amount?
5. How can you determine if the line is horizontal or vertical based on $k$?

Tip: Always verify your solution by substituting it back into the original equation to check for correctness.