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### 8. Lines \( L_1 \) and \( L_2 \)

Given:
- \( L_1 \): \( ax - 3y = 27 \)
- \( L_2 \): \( y = \frac{2}{3}x + 4 \)

The lines are perpendicular.

#### (a) Find the value of \( a \).

To determine when two lines are perpendicular, the product of their slopes must equal \(-1\).

First, rewrite \( L_1 \) in slope-intercept form:
\[
ax - 3y = 27 \quad \Rightarrow \quad y = \frac{a}{3}x - 9
\]
Thus, the slope of \( L_1 \) is \( \frac{a}{3} \).

The equation for \( L_2 \) is already in slope-intercept form, and the slope of \( L_2 \) is \( \frac{2}{3} \).

Since the lines are perpendicular:
\[
\left(\frac{a}{3}\right) \times \left(\frac{2}{3}\right) = -1
\]
Solving this:
\[
\frac{2a}{9} = -1 \quad \Rightarrow \quad a = -\frac{9}{2}
\]

Thus, \( a = -\frac{9}{2} \).

#### (b) Determine the coordinates of the intersection of the lines.

Now that we know \( a = -\frac{9}{2} \), the equation of \( L_1 \) becomes:
\[
-\frac{9}{2}x - 3y = 27
\]
Let's solve for the intersection by substituting \( y = \frac{2}{3}x + 4 \) (from \( L_2 \)) into \( L_1 \)'s equation.

First, solve for \( y \):
\[
-\frac{9}{2}x - 3\left(\frac{2}{3}x + 4\right) = 27
\]
Simplifying:
\[
-\frac{9}{2}x - 2x - 12 = 27
\quad \Rightarrow \quad -\frac{13x}{2} - 12 = 27
\quad \Rightarrow \quad -\frac{13x}{2} = 39
\quad \Rightarrow \quad x = -6
\]

Now substitute \( x = -6 \) into the equation for \( L_2 \):
\[
y = \frac{2}{3}(-6) + 4 = -4 + 4 = 0
\]

So, the intersection point is \( (-6, 0) \).

---

### 9. Ski resort problem

You are given the coordinates of three stations:
- \( B = (0, 0, 0) \)
- \( P = (500, 400, 300) \)
- \( Q = (900, 600, 700) \)

#### (a) Determine the distance from \( B \) to \( P \).

The distance between two points in 3D space is given by:
\[
d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}
\]

For the distance from \( B(0, 0, 0) \) to \( P(500, 400, 300) \):
\[
d = \sqrt{(500 - 0)^2 + (400 - 0)^2 + (300 - 0)^2} = \sqrt{500^2 + 400^2 + 300^2}
\]
\[
d = \sqrt{250000 + 160000 + 90000} = \sqrt{500000} = 707.11 \, \text{meters}
\]

Thus, the distance from \( B \) to \( P \) is approximately \( 707.11 \, \text{m} \).

#### (b) Determine the total distance covered from \( B \) to \( Q \).

Now, we need to compute the distance from \( P(500, 400, 300) \) to \( Q(900, 600, 700) \):
\[
d = \sqrt{(900 - 500)^2 + (600 - 400)^2 + (700 - 300)^2} = \sqrt{400^2 + 200^2 + 400^2}
\]
\[
d = \sqrt{160000 + 40000 + 160000} = \sqrt{360000} = 600 \, \text{meters}
\]

Thus, the total distance from \( B \) to \( Q \) is:
\[
707.11 \, \text{m} + 600 \, \text{m} = 1307.11 \, \text{m}
\]

---

Would you like further details or clarifications? Here are some related questions to explore:

1. How do you verify if two lines are parallel instead of perpendicular?
2. What is the significance of the slope-intercept form in analyzing lines?
3. Can you calculate the midpoint of the ski lift path from \( B \) to \( Q \)?
4. How do you find the equation of a plane that passes through points \( B \), \( P \), and \( Q \)?
5. What are the real-world applications of distance calculations in 3D space?

**Tip:** When calculating distances in 3D space, always ensure you square the differences in each coordinate before adding them up.

Lines L1 and L2 are given by the equations L1: ax - 3y = 27 and L2: y = (2/3)x + 4. The two lines are perpendicular. Find the value of a. Hence, determine the coordinates of the intersection of the lines. A ski resort is designing two new ski lifts. One lift connects station B at the base of the mountain and station P at the top of a ski run. The other lift connects station P with station Q, which is at the top of another ski lift. The three stations are placed on a three-dimensional coordinate system, measured in meters, with B = (0, 0, 0), P = (500, 400, 300), and Q = (900, 600, 700). Determine the distance covered by a skier on the lift from the base to P. Determine the total distance covered by a skier on the ski lifts from the base of the mountain to Q.

Solve for the value of a when two lines L1 and L2 are perpendicular and determine their intersection point. Also, calculate the distance between points in 3D space for a ski lift system. This problem covers key algebraic and geometric concepts suitable for Grades 10-12.

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