The problem involves two straight lines, $L_1: hx + ky + 24 = 0$ and $L_2: 3x - 4y - 15h = 0$, which are perpendicular to each other. Additionally, the $y$-intercept of $L_1$ is given as $-8$, and we need to find the $x$-intercept of $L_2$.

Step-by-step solution:

1. $L_1: hx + ky + 24 = 0$

We are told that the $y$-intercept of $L_1$ is $-8$. To find the $y$-intercept, we set $x = 0$ in the equation of the line:

$h(0) + ky + 24 = 0 \quad \Rightarrow \quad ky = -24 \quad \Rightarrow \quad y = \frac{-24}{k}.$

Since the $y$-intercept is given as $-8$, we set $y = -8$:

$-8 = \frac{-24}{k} \quad \Rightarrow \quad k = 3.$

2. $L_2: 3x - 4y - 15h = 0$

The second equation is given as $L_2: 3x - 4y - 15h = 0$. The lines $L_1$ and $L_2$ are perpendicular, so their slopes must satisfy the condition that the product of their slopes is $-1$.

• The slope of $L_1: hx + ky + 24 = 0$ is $\frac{-h}{k} = \frac{-h}{3}$.
• The slope of $L_2: 3x - 4y - 15h = 0$ is $\frac{3}{4}$.

For the lines to be perpendicular:

$\left(\frac{-h}{3}\right) \times \left(\frac{3}{4}\right) = -1 \quad \Rightarrow \quad \frac{-h}{4} = -1 \quad \Rightarrow \quad h = 4.$

3. Find the $x$-intercept of $L_2$

To find the $x$-intercept of $L_2$, we set $y = 0$ in the equation $3x - 4y - 15h = 0$:

$3x - 15h = 0 \quad \Rightarrow \quad 3x = 15h \quad \Rightarrow \quad x = \frac{15h}{3} = 5h.$

Since $h = 4$:

$x = 5 \times 4 = 20.$

Thus, the $x$-intercept of $L_2$ is $\boxed{20}$.

Summary:

• The answer is D) 20.

Would you like further details or explanations? Here are some related questions for deeper understanding:

1. How do you find the slope of a line from its general equation?
2. Why do perpendicular lines have slopes whose product is $-1$?
3. Can you explain the significance of $x$-intercepts and $y$-intercepts in a graph?
4. How would the solution change if the lines were parallel instead of perpendicular?
5. What happens to the $x$-intercept if we change the value of $h$?

Tip: When two lines are perpendicular, their slopes are negative reciprocals, which means multiplying their slopes will always give $-1$.