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

Finding the area of a triangle is a fundamental concept in geometry, and this article delves into a specific scenario: calculating the area of a triangle bounded by the y-axis, a given line, and a perpendicular line passing through the origin. This problem combines concepts of linear equations, perpendicular lines, and geometric area calculations. Let's break down the process step-by-step.

1. Understanding the Problem: Triangle Bounded by Lines

At its core, this problem asks us to determine the area of a triangle formed by three lines. The first line is the y-axis, a vertical line defined by the equation x = 0. The second line is a given linear function, f(x) = 7 - (1/3)x. The third line is a line perpendicular to f(x) that passes through the origin (0,0). To find the area, we'll need to determine the vertices of the triangle, which are the points where these lines intersect. Understanding the slope and y-intercept of the given line is crucial, as it will help us find the equation of the perpendicular line. The problem beautifully integrates algebraic concepts with geometric applications, making it a valuable exercise in mathematical reasoning. Visualizing the problem by sketching the lines can greatly aid in understanding the triangle's formation and the relationships between the lines. This initial understanding is the key to unlocking the solution, allowing us to systematically approach the problem and find the area with confidence. Remember, geometry is as much about visualization as it is about calculation. So, take a moment to picture the triangle and its boundaries before diving into the equations.

2. Finding the Perpendicular Line: Slope and Equation

To determine the area of the triangle, we first need to find the equation of the line perpendicular to f(x) = 7 - (1/3)x and passing through the origin. The given line, f(x), has a slope of -1/3. A fundamental property of perpendicular lines is that their slopes are negative reciprocals of each other. Therefore, the slope of the line perpendicular to f(x) will be the negative reciprocal of -1/3, which is 3. This negative reciprocal relationship is the cornerstone of finding the perpendicular line. Since this perpendicular line passes through the origin (0,0), we can use the point-slope form of a linear equation, y - y1 = m(x - x1), where m is the slope and (x1, y1) is the point. Plugging in the slope (m = 3) and the origin (0,0), we get y - 0 = 3(x - 0), which simplifies to y = 3x. This is the equation of the line perpendicular to f(x) and passing through the origin. Now we have the equations for two of the triangle's sides: f(x) = 7 - (1/3)x and y = 3x. The next step involves finding the intersection points of these lines to determine the vertices of our triangle. Understanding the relationship between slopes and perpendicularity is essential for solving a wide range of geometry problems, and this step provides a practical application of that concept.

3. Determining the Vertices: Intersection Points

To accurately calculate the area of the triangle, we need to identify the coordinates of its vertices. These vertices are the points where the lines intersect. We already know one vertex: the origin (0,0), which is the intersection of the y-axis and the perpendicular line y = 3x. To find the other vertices, we need to determine where the line f(x) = 7 - (1/3)x intersects the y-axis and the line y = 3x. First, let's find the intersection of f(x) with the y-axis. The y-axis is defined by the equation x = 0. Substituting x = 0 into f(x) gives us f(0) = 7 - (1/3)(0) = 7. So, the intersection point is (0,7). Next, we find the intersection of f(x) and the line y = 3x. To do this, we set the two equations equal to each other: 7 - (1/3)x = 3x. Solving for x, we get 7 = 3x + (1/3)x, which simplifies to 7 = (10/3)x. Multiplying both sides by 3/10, we find x = 21/10. Now, substitute this value of x into y = 3x to find the corresponding y-coordinate: y = 3(21/10) = 63/10. Thus, the third vertex is (21/10, 63/10). We now have all three vertices of the triangle: (0,0), (0,7), and (21/10, 63/10). With these vertex coordinates in hand, we can proceed to calculate the area of the triangle.

4. Calculating the Area: Base and Height

Now that we have determined the vertices of the triangle – (0,0), (0,7), and (21/10, 63/10) – we can proceed to calculate its area. The most straightforward method for this triangle is to use the formula: Area = (1/2) * base * height. We can choose the segment along the y-axis between (0,0) and (0,7) as the base of the triangle. The length of this base is the difference in the y-coordinates, which is 7 - 0 = 7 units. The height of the triangle is the perpendicular distance from the third vertex (21/10, 63/10) to the y-axis. This distance is simply the x-coordinate of the vertex, which is 21/10 units. Therefore, the base of the triangle is 7 and the height is 21/10. Plugging these values into the area formula, we get: Area = (1/2) * 7 * (21/10) = (1/2) * (147/10) = 147/20. So, the area of the triangle is 147/20 square units, or 7.35 square units. This calculation demonstrates the practical application of the area formula and highlights the importance of accurately determining the base and height relative to each other. Understanding how to choose the base and height is key to simplifying area calculations in various geometric problems.

5. Final Answer: Area of the Triangle

Having meticulously worked through each step, we have successfully calculated the area of the triangle bounded by the y-axis, the line f(x) = 7 - (1/3)x, and the line perpendicular to f(x) that passes through the origin. We found the vertices of the triangle by determining the intersection points of these lines, and then we applied the formula for the area of a triangle: Area = (1/2) * base * height. Our calculations revealed the area to be 147/20 square units, which is equivalent to 7.35 square units. This final answer represents the culmination of our efforts, showcasing the power of combining algebraic and geometric concepts to solve a challenging problem. The process involved understanding the properties of linear equations, perpendicular lines, and geometric figures. The result, 147/20 square units, is not just a numerical answer; it's a testament to the application of mathematical principles to a real-world geometric scenario. This exercise underscores the importance of a systematic approach to problem-solving, where each step builds upon the previous one to reach a logical and accurate conclusion. The final area, 7.35 square units, provides a concrete measure of the space enclosed by the three lines, completing our exploration of this geometric problem.