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The problem asks for the area of a triangle bounded by the y-axis, the line \( f(x) = 7 - \frac{3}{5}x \), and a line perpendicular to \( f(x) \) that passes through the origin.

### Step 1: Equation of the line perpendicular to \( f(x) \)
The slope of the line \( f(x) = 7 - \frac{3}{5}x \) is \( -\frac{3}{5} \). A line perpendicular to this would have a slope that is the negative reciprocal of \( -\frac{3}{5} \), which is \( \frac{5}{3} \).

Since this perpendicular line passes through the origin, its equation is:
\[
y = \frac{5}{3}x
\]

### Step 2: Finding the points of intersection
- **Intersection with the y-axis**: This occurs when \( x = 0 \). For the line \( f(x) = 7 - \frac{3}{5}x \), substituting \( x = 0 \):
  \[
  f(0) = 7
  \]
  So, the point of intersection with the y-axis is \( (0, 7) \).

- **Intersection of the two lines**: To find the point of intersection between \( f(x) = 7 - \frac{3}{5}x \) and \( y = \frac{5}{3}x \), set the two equations equal to each other:
  \[
  7 - \frac{3}{5}x = \frac{5}{3}x
  \]
  Multiply through by 15 to eliminate the fractions:
  \[
  15(7) - 15\left(\frac{3}{5}x\right) = 15\left(\frac{5}{3}x\right)
  \]
  \[
  105 - 9x = 25x
  \]
  Solving for \( x \):
  \[
  105 = 34x \quad \Rightarrow \quad x = \frac{105}{34}
  \]

  Substituting this value of \( x \) into \( y = \frac{5}{3}x \) to find \( y \):
  \[
  y = \frac{5}{3} \times \frac{105}{34} = \frac{525}{102} = \frac{175}{34}
  \]

  Thus, the point of intersection is \( \left( \frac{105}{34}, \frac{175}{34} \right) \).

### Step 3: Area of the triangle
The triangle is bounded by the y-axis, the line \( f(x) \), and the perpendicular line through the origin. The vertices of the triangle are:
- \( (0, 0) \) (the origin),
- \( (0, 7) \) (the y-intercept of \( f(x) \)),
- \( \left( \frac{105}{34}, \frac{175}{34} \right) \) (the point of intersection).

The area of a triangle with vertices at \( (x_1, y_1) \), \( (x_2, y_2) \), and \( (x_3, y_3) \) can be calculated using the formula:
\[
\text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|
\]
Substituting the points \( (0, 0) \), \( (0, 7) \), and \( \left( \frac{105}{34}, \frac{175}{34} \right) \):
\[
\text{Area} = \frac{1}{2} \left| 0(7 - \frac{175}{34}) + 0(\frac{175}{34} - 0) + \frac{105}{34}(0 - 7) \right|
\]
\[
\text{Area} = \frac{1}{2} \left| \frac{105}{34} \times (-7) \right| = \frac{1}{2} \times \frac{105 \times 7}{34} = \frac{1}{2} \times \frac{735}{34} = \frac{735}{68}
\]

Thus, the area of the triangle is \( \frac{735}{68} \), which simplifies to approximately **10.81 square units**.

Would you like more details or further clarification on any step? Here are some related questions for further exploration:

1. What is the geometric meaning of the negative reciprocal of a slope?
2. How do we find the intersection of two non-parallel lines algebraically?
3. What other methods can we use to find the area of a triangle given different types of information?
4. What happens to the area if we shift the lines to different y-intercepts?
5. Can this area formula for triangles be extended to quadrilaterals or other polygons?

**Tip**: Always cross-check the points of intersection before calculating the area, especially when dealing with algebraic manipulations.

Find the area of a triangle bounded by the y-axis, the line f(x) = 7 - (3/5)x, and the line perpendicular to f(x) that passes through the origin.

Learn how to calculate the area of a triangle bounded by the y-axis, the line f(x) = 7 - (3/5)x, and its perpendicular line passing through the origin. This problem covers key coordinate geometry concepts, including finding the intersection of lines and using triangle area formulas.

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