Area Of Triangle Bounded By Y-Axis, A Line, And A Perpendicular Line

Finding the area of geometric shapes formed by lines and axes is a fundamental concept in coordinate geometry. In this article, we will delve into a problem that involves finding the area of a triangle bounded by the y-axis, a given line, and a line perpendicular to it passing through the origin. This problem combines concepts of linear equations, perpendicular lines, and area calculation, providing a comprehensive exercise in geometric problem-solving. Let's explore the step-by-step approach to solving this problem.

Problem Statement

Our objective is to find the area of a triangle that is enclosed by the following:

1. The y-axis
2. The line defined by the equation $f(x) = 3 - \frac{1}{6}x$
3. A line that is perpendicular to $f(x)$ and passes through the origin

Step 1: Understanding the Given Line

The given line is represented by the equation $f(x) = 3 - \frac{1}{6}x$. This is a linear equation in slope-intercept form, which can be rewritten as $y = -\frac{1}{6}x + 3$. From this, we can identify the slope and y-intercept of the line.

• Slope: The slope of the line is $m_1 = -\frac{1}{6}$. The slope indicates the steepness and direction of the line. A negative slope means the line slopes downward from left to right.
• Y-intercept: The y-intercept is the point where the line crosses the y-axis. In this equation, the y-intercept is 3. This means the line intersects the y-axis at the point (0, 3).

To visualize this line, we can plot a few points. We already know the y-intercept (0, 3). To find another point, we can set $x$ to a convenient value, say $x = 6$, which gives us $y = -\frac{1}{6}(6) + 3 = -1 + 3 = 2$. So, another point on the line is (6, 2). Plotting these points and drawing a line through them gives us a visual representation of the line $f(x)$.

Step 2: Finding the Perpendicular Line

Now, we need to find the equation of the line that is perpendicular to $f(x)$ and passes through the origin. The key concept here is the relationship between the slopes of perpendicular lines.

• Perpendicular Slopes: If two lines are perpendicular, the product of their slopes is -1. Let the slope of the perpendicular line be $m_2$. Then, $m_1 imes m_2 = -1$.

We know $m_1 = -\frac{1}{6}$, so we can find $m_2$:

$(- \frac{1}{6}) imes m_2 = -1$

$m_2 = 6$

So, the slope of the line perpendicular to $f(x)$ is 6. Since this line passes through the origin (0, 0), we can use the point-slope form of a linear equation to find its equation. The point-slope form is given by:

$y - y_1 = m(x - x_1)$

where $(x_1, y_1)$ is a point on the line and $m$ is the slope. Plugging in the origin (0, 0) and the slope $m_2 = 6$, we get:

$y - 0 = 6(x - 0)$

$y = 6x$

Thus, the equation of the line perpendicular to $f(x)$ and passing through the origin is $y = 6x$.

Step 3: Determining the Vertices of the Triangle

The triangle is formed by the intersection of three lines: the y-axis, $f(x)$, and the perpendicular line. To find the area of the triangle, we first need to determine the coordinates of its vertices. The vertices are the points where these lines intersect.

1. Intersection of $f(x)$ and the y-axis: The y-axis is represented by the equation $x = 0$. To find the intersection point, we substitute $x = 0$ into the equation $f(x) = 3 - \frac{1}{6}x$:

   $y = 3 - \frac{1}{6}(0) = 3$

   So, the first vertex is (0, 3).

2. Intersection of the perpendicular line and the y-axis: The perpendicular line is $y = 6x$. To find its intersection with the y-axis ($x = 0$), we substitute $x = 0$ into the equation:

   $y = 6(0) = 0$

   So, the second vertex is (0, 0), which is the origin.

3. Intersection of $f(x)$ and the perpendicular line: To find the intersection of $f(x) = 3 - \frac{1}{6}x$ and $y = 6x$, we set the expressions for $y$ equal to each other:

   $6x = 3 - \frac{1}{6}x$

   Multiply both sides by 6 to eliminate the fraction:

   $36x = 18 - x$

   Add $x$ to both sides:

   $37x = 18$

   Divide by 37:

   $x = \frac{18}{37}$

   Now, substitute this value of $x$ into $y = 6x$ to find the $y$-coordinate:

   $y = 6(\frac{18}{37}) = \frac{108}{37}$

   So, the third vertex is $(\frac{18}{37}, \frac{108}{37})$.

Therefore, the vertices of the triangle are (0, 0), (0, 3), and $(\frac{18}{37}, \frac{108}{37})$.

Step 4: Calculating the Area of the Triangle

Now that we have the coordinates of the vertices, we can calculate the area of the triangle. Since one of the vertices is at the origin and another lies on the y-axis, we can use a simplified method to calculate the area. The vertices are:

• A = (0, 0)
• B = (0, 3)
• C = $(\frac{18}{37}, \frac{108}{37})$

The length of the base of the triangle, which lies along the y-axis, is the distance between (0, 0) and (0, 3), which is 3 units. The height of the triangle is the x-coordinate of the third vertex, which is $\frac{18}{37}$.

The area of a triangle is given by the formula:

$Area = \frac{1}{2} imes base imes height$

Plugging in the values, we get:

$Area = \frac{1}{2} imes 3 imes \frac{18}{37}$

$Area = \frac{1}{2} imes \frac{54}{37}$

$Area = \frac{27}{37}$

Therefore, the area of the triangle bounded by the y-axis, the line $f(x) = 3 - \frac{1}{6}x$, and the line perpendicular to $f(x)$ that passes through the origin is $\frac{27}{37}$ square units.

Conclusion

In this article, we successfully calculated the area of a triangle formed by the y-axis and two lines: a given line and a line perpendicular to it passing through the origin. We began by understanding the equation of the given line and finding the equation of the perpendicular line. We then determined the vertices of the triangle by finding the points of intersection of these lines. Finally, we used the coordinates of the vertices to calculate the area of the triangle using the formula for the area of a triangle. This problem demonstrates the application of coordinate geometry principles in solving geometric problems, and it highlights the importance of understanding linear equations, slopes, and perpendicular lines.