Find the area of a triangle bounded by the y-axis, the line 
f(x) = 5 − 
4
7
x,
 and the line perpendicular to f(x) that passes through the origin. (Round your answer to two decimal places.)

To find the area of the triangle bounded by the y-axis, the line \( f(x) = 5 - \frac{4}{7}x \), and a line perpendicular to \( f(x) \) that passes through the origin, we can follow these steps:

### 1. **Find the Slope of \( f(x) \):**
   The line \( f(x) = 5 - \frac{4}{7}x \) is in slope-intercept form, \( y = mx + b \), where \( m \) is the slope and \( b \) is the y-intercept.

   \[
   \text{Slope of } f(x), m = -\frac{4}{7}
   \]

### 2. **Find the Slope of the Perpendicular Line:**
   The slope of a line perpendicular to another line is the negative reciprocal of the original line's slope.

   \[
   \text{Slope of the perpendicular line} = -\frac{1}{\left(-\frac{4}{7}\right)} = \frac{7}{4}
   \]

### 3. **Equation of the Perpendicular Line Passing Through the Origin:**
   The line passes through the origin, so its y-intercept is \( 0 \). The equation of the line with slope \( \frac{7}{4} \) is:

   \[
   y = \frac{7}{4}x
   \]

### 4. **Find the Intersection of the Two Lines:**
   To find the intersection point of \( f(x) = 5 - \frac{4}{7}x \) and \( y = \frac{7}{4}x \), set the equations equal to each other:

   \[
   5 - \frac{4}{7}x = \frac{7}{4}x
   \]

   Multiply every term by 28 (the least common multiple of 4 and 7) to clear the fractions:

   \[
   140 - 16x = 49x
   \]

   Combine like terms:

   \[
   140 = 65x
   \]

   Solve for \( x \):

   \[
   x = \frac{140}{65} = \frac{28}{13}
   \]

   Substitute \( x = \frac{28}{13} \) into \( y = \frac{7}{4}x \) to find the y-coordinate:

   \[
   y = \frac{7}{4} \cdot \frac{28}{13} = \frac{196}{52} = \frac{98}{26} = \frac{49}{13}
   \]

   The intersection point is \( \left(\frac{28}{13}, \frac{49}{13}\right) \).

### 5. **Find the Triangle's Area:**
   The vertices of the triangle are:
   - The origin: \( (0, 0) \)
   - The y-intercept of \( f(x) \): \( (0, 5) \)
   - The intersection point: \( \left(\frac{28}{13}, \frac{49}{13}\right) \)

   Use the area formula for a triangle with vertices at \( (x_1, y_1) \), \( (x_2, y_2) \), and \( (x_3, y_3) \):

   \[
   \text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|
   \]

   Substituting the points:

   \[
   \text{Area} = \frac{1}{2} \left| 0 \cdot (5 - \frac{49}{13}) + 0 \cdot (\frac{49}{13} - 0) + \frac{28}{13} \cdot (0 - 5) \right|
   \]

   \[
   \text{Area} = \frac{1}{2} \left| \frac{28}{13} \cdot (-5) \right|
   \]

   \[
   \text{Area} = \frac{1}{2} \left| \frac{-140}{13} \right|
   \]

   \[
   \text{Area} = \frac{1}{2} \cdot \frac{140}{13}
   \]

   \[
   \text{Area} = \frac{70}{13} \approx 5.38
   \]

**Therefore, the area of the triangle is approximately \( 5.38 \) square units.**

---

Would you like more details or have any questions?

Here are five related questions to explore further:

1. How would the area change if the y-intercept of \( f(x) \) were different?
2. What is the equation of a line that is parallel to \( f(x) \) and passes through a specific point?
3. How can you find the area of a triangle using a determinant formula?
4. How does the slope of a line affect its perpendicular line?
5. What are the properties of lines that make them perpendicular?

**Tip:** Remember, the slopes of perpendicular lines are negative reciprocals of each other.

Find the area of a triangle bounded by the y-axis, the line f(x) = 5 − 4/7x, and the line perpendicular to f(x) that passes through the origin. (Round your answer to two decimal places.)

Learn how to find the area of a triangle formed by the y-axis, the line f(x) = 5 − 4/7x, and its perpendicular line through the origin. This problem covers key concepts in coordinate geometry and is suitable for students in Grades 9-11.

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