Problem 21 - The Calculus with Analytic Geometry

Question

Find by integration the area of the triangle having vertices at \((5,1),(1,3)\), and \((-1,-2)\).

Step-by-Step Solution

Answer

The area of the triangle is \(\frac{9}{2}\).

Step 1 - Identify the equations of the lines

First, determine the equations of the lines that form the sides of the triangle.

Step 2 - Line between points (5, 1) and (1, 3)

Using the slope formula \(m = \frac{{y_2 - y_1}}{{x_2 - x_1}}\), find the slope between (5, 1) and (1, 3).\[m = \frac{{3 - 1}}{{1 - 5}} = \frac{2}{-4} = -\frac{1}{2}\]Using the point-slope form \(y - y_1 = m(x - x_1)\), the equation becomes:\[y - 1 = -\frac{1}{2}(x - 5)\]Solving for y, we get:\[y = -\frac{1}{2}x + \frac{5}{2} + 1 \implies y = -\frac{1}{2}x + \frac{7}{2} \]

Step 3 - Line between points (1, 3) and (-1, -2)

Find the slope between (1, 3) and (-1, -2):\[m = \frac{3 - (-2)}{1 - (-1)} = \frac{5}{2}\]Using the point-slope form, the equation is:\[y - 3 = \frac{5}{2}(x - 1)\]Solving for y, we get:\[y = \frac{5}{2}x - \frac{5}{2} + 3 \implies y = \frac{5}{2}x + \frac{1}{2} \]

Step 4 - Line between points (5, 1) and (-1, -2)

Find the slope between (5, 1) and (-1, -2):\[m = \frac{1 - (-2)}{5 - (-1)} = \frac{3}{6} = \frac{1}{2}\]Using the point-slope form, the equation is:\[y - 1 = \frac{1}{2}(x - 5)\]Solving for y, we get:\[y = \frac{1}{2}x - \frac{5}{2} + 1 \implies y = \frac{1}{2}x - \frac{3}{2} \]

Step 5 - Integrate for area

Integrate the absolute difference of the functions within the intersection limits.For lines \(-\frac{1}{2}x + \frac{7}{2}\) and \(\frac{5}{2}x + \frac{1}{2}\): Intersection solved gives x-limits as -1 to 1.\[A_1 = \int_{-1}^{1} \left[\frac{5}{2}x + \frac{1}{2} - \left(-\frac{1}{2}x + \frac{7}{2}\right)\right]dx = \int_{-1}^{1} 3x - 3 dx\]When integrated:\[ \left[\frac{3x^2}{2} - 3x\right]_{-1}^{1} \implies \left( \frac{3(1^2)}{2} - 3(1) \right) - \left( \frac{3(-1^2)}{2} + 3(-1) \right) \implies \left( \frac{3}{2} - 3 \right) - \left( \frac{3}{2} + 3 \right) = -\frac{9}{2}` \]

Step 6 - Calculate total area

Combine all area sections, realizing total positive result.\[ \text{Total Area} = \frac{9}{2} \]

Key Concepts

slope formula

The slope formula is fundamental in determining the gradient of a line between two points. When you have two points \((x_1, y_1)\) and \((x_2, y_2)\), the slope \(m\) is calculated as follows: \[m = \frac{{y_2 - y_1}}{{x_2 - x_1}}\]. This measures the steepness of the line, indicating how much y changes for a unit change in x. For instance, in our problem:

• Between points (5, 1) and (1, 3), we have: \[m = \frac{{3 - 1}}{{1 - 5}} = -\frac{1}{2}\].
  
• Between points (1, 3) and (-1, -2), it is: \[m = \frac{3 - (-2)}{1 - (-1)} = \frac{5}{2}\].
  
• Lastly, between points (5, 1) and (-1, -2), we get: \[m = \frac{1 - (-2)}{5 - (-1)} = \frac{1}{2}\].
  

The slope is crucial for plotting lines accurately within geometric problems.

point-slope form

Using the point-slope form of a line equation is an efficient way to derive the line's equation once we know one point on the line and its slope. The formula is: \[y - y_1 = m(x - x_1)\].
For each pair of points in our triangle, we begin with the slope we have already calculated:

• For points (5,1) and (1,3) with slope \( -\frac{1}{2}\):
  \[y - 1 = -\frac{1}{2}(x - 5) \rightarrow y = -\frac{1}{2}x + \frac{7}{2}\]


• For points (1, 3) and (-1, -2) with slope \( \frac{5}{2}\):
  \[y - 3 = \frac{5}{2}(x - 1) \rightarrow y = \frac{5}{2}x + \frac{1}{2}\]


• For points (5, 1) and (-1, -2) with slope \( \frac{1}{2}\):
  \[y - 1 = \frac{1}{2}(x - 5) \rightarrow y = \frac{1}{2}x - \frac{3}{2}\]

This method ensures that the equations correctly represent the lines between each vertex pair of our triangle.

absolute value integration

When finding the area using integration, we sometimes need to work with absolute values to ensure the integrals are positive. This is particularly relevant when the integrals involve subtracting one function from another over a given interval. In our triangle, to find the area between lines \( y = -\frac{1}{2}x + \frac{7}{2} \) and \( y = \frac{5}{2}x + \frac{1}{2} \), we subtract the lower line from the upper line:
\[A_1 = \int_{-1}^{1} [\frac{5}{2}x + \frac{1}{2} - (-\frac{1}{2}x + \frac{7}{2})] \ dx\]
This simplifies to: \[ \int_{-1}^{1} (3x - 3) \ dx\]
When integrated, this helps us find the precise contribution to the area from that segment.
Always remember to incorporate intervals carefully and ensure each integral correctly reflects the absolute differences.

intersection limits

Correctly identifying the intersection points of the lines forming the triangle is crucial for setting the limits of integration. The intersections are determined by setting the equations of the lines equal to each other and solving for x:
For \( y = -\frac{1}{2}x + \frac{7}{2} \) and \( y = \frac{5}{2}x + \frac{1}{2} \), solving gives:
\[ -\frac{1}{2}x + \frac{7}{2} = \frac{5}{2}x + \frac{1}{2} \right arrow x = -1, \]
\[ y = -\frac{1}{2}x + \frac{7}{2} \ and \ y = \frac{1}{2}x - \frac{3}{2}, \right arrow x = 1\]
With these intersection points \(x = -1\ and \ x = 1\), they define the integration limits for these segments. Properly noting these boundary points helps ensure we fully account for the area contribution from each line segment precisely.