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.