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.

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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.