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We are tasked with finding the area of a triangle bounded by the \( y \)-axis, the line \( f(x) = 9 - \frac{2}{5}x \), and a line that is perpendicular to \( f(x) \) and passes through the origin.

### Step 1: Understanding the geometry
1. The line \( f(x) = 9 - \frac{2}{5}x \) has a slope of \( -\frac{2}{5} \).
2. The slope of a line perpendicular to this will be the negative reciprocal, which is \( \frac{5}{2} \).
3. Since the perpendicular line passes through the origin, its equation is \( y = \frac{5}{2}x \).

### Step 2: Finding the points of intersection
1. The triangle is formed by the lines and the \( y \)-axis, so we need the points where these lines intersect:
   - The \( y \)-axis (at \( x = 0 \)) for both lines.
   - The intersection of \( f(x) = 9 - \frac{2}{5}x \) and \( y = \frac{5}{2}x \).

#### Intersection with the \( y \)-axis:
- For \( f(x) = 9 - \frac{2}{5}x \), when \( x = 0 \), \( y = 9 \).
- For \( y = \frac{5}{2}x \), when \( x = 0 \), \( y = 0 \).

Thus, one vertex of the triangle is at \( (0, 0) \), and another is at \( (0, 9) \).

#### Intersection of the two lines:
Set \( f(x) = y \), so:
\[
9 - \frac{2}{5}x = \frac{5}{2}x
\]
Multiply the entire equation by 10 to eliminate fractions:
\[
90 - 4x = 25x
\]
Solving for \( x \):
\[
90 = 29x \quad \Rightarrow \quad x = \frac{90}{29}
\]
Substitute \( x = \frac{90}{29} \) into \( y = \frac{5}{2}x \) to find \( y \):
\[
y = \frac{5}{2} \times \frac{90}{29} = \frac{450}{58} = \frac{225}{29}
\]

Thus, the intersection point is \( \left( \frac{90}{29}, \frac{225}{29} \right) \).

### Step 3: Calculate the area of the triangle
The triangle is bounded by the following points:
- \( (0, 0) \) (the origin),
- \( (0, 9) \) (on the \( y \)-axis),
- \( \left( \frac{90}{29}, \frac{225}{29} \right) \) (the intersection of the two lines).

We can use the formula for the area of a triangle with vertices \( (x_1, y_1) \), \( (x_2, y_2) \), and \( (x_3, y_3) \):
\[
\text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|
\]
Substitute \( (x_1, y_1) = (0, 0) \), \( (x_2, y_2) = (0, 9) \), and \( (x_3, y_3) = \left( \frac{90}{29}, \frac{225}{29} \right) \):
\[
\text{Area} = \frac{1}{2} \left| 0(9 - \frac{225}{29}) + 0\left(\frac{225}{29} - 0\right) + \frac{90}{29}(0 - 9) \right|
\]
\[
= \frac{1}{2} \left| \frac{90}{29} \times (-9) \right|
\]
\[
= \frac{1}{2} \times \frac{810}{29} = \frac{405}{29}
\]

### Final Answer:
The area of the triangle is \( \frac{405}{29} \) square units.

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Do you have any questions, or would you like more details? 

Here are 5 related questions:
1. How do you determine the slope of a perpendicular line?
2. Why is the negative reciprocal used for perpendicular lines?
3. Can you explain how the area of a triangle is calculated from vertex coordinates?
4. What are other methods to find the area of a triangle in geometry?
5. How does changing the slopes of lines affect the shape of the triangle?

**Tip:** Always plot the triangle or geometric shape when solving area problems for better visualization!

Find the area of a triangle bounded by the y-axis, the line f(x) = 9 - 2/5x, and the line perpendicular to f(x) that passes through the origin.

This problem helps find the area of a triangle formed by the y-axis, the line f(x) = 9 - 2/5x, and a line perpendicular to it through the origin. Concepts include slopes of perpendicular lines, finding points of intersection, and using the formula for the area of a triangle.

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