Question

Find the area of the triangle with vertices at \((-1,4)\), \((2,-2)\), and \((5,1)\) by integration.

Short answer

The area of the triangle is 11.5 square units.

Step-by-step solution

Determine the Equations of the Lines

First, identify the equations of the lines connecting each pair of points, forming the sides of the triangle. For example, find the slope \(m\) of the line between points \((-1,4)\) and \((2,-2)\):\[ m = \frac{-2 - 4}{2 - (-1)} = \frac{-6}{3} = -2. \]Using the point-slope form equation \(y - y_1 = m(x - x_1)\), with \( (x_1, y_1) = (-1, 4)\):\[ y - 4 = -2(x + 1), \text{ so } y = -2x - 2. \]Repeat this process to find the other two sides' equations: - Between \((2, -2)\) and \((5, 1)\): \(y = \frac{1}{3}x - \frac{8}{3}.\)- Between \((-1, 4)\) and \((5, 1)\): \(y = -\frac{1}{2}x + \frac{9}{2}.\)

Set Up Integration Limits

To use integration, consider the range of x-values over which the triangle extends. Observing the vertices, the triangle spans from \(x = -1\) to \(x = 5\). These will be the limits of integration.

Determine the Areas Using Integration

Integrate vertically from the line with the lower y-values to the line with the higher y-values. To do so, you need to identify which is the "upper" curve and which is the "lower" curve over each sub-interval of \([-1, 5]\).- From \(x = -1\) to \(x = 2\), "upper" is \(y = -2x - 2\) and "lower" is the x-axis (y = 0).- From \(x = 2\) to \(x = 5\), "upper" is \(y = \frac{1}{3}x - \frac{8}{3}\) and "lower" is \(y = -\frac{1}{2}x + \frac{9}{2}.\)Thus, calculate:- Integral \(\int_{-1}^{2} (-2x - 2) \, dx\) for the first region.- Integral \(\int_{2}^{5} \left( \frac{1}{3}x - \frac{8}{3} - \left(-\frac{1}{2}x + \frac{9}{2}\right) \right) \, dx\) for the second region.

Evaluate the Integrals

Perform the integration and evaluate:1. For \(\int_{-1}^{2} (-2x - 2) \, dx\): \[ = \left[-x^2 - 2x\right]_{-1}^{2} \] Evaluate: \[ = [-(2)^2 - 2(2)] - [(-(-1)^2) - 2(-1)] \] \[ = -4 - 4 - (-1 + 2) \] \[ = -8 + 1 \] \[ = -7. \] (Absolute necessitated since areas are non-negative) Actual area = 7.2. For \(\int_{2}^{5} \left( \frac{1}{3}x - \frac{8}{3} + \frac{1}{2}x - \frac{9}{2}\right) \, dx\): Combine first: \(\frac{1}{3}x + \frac{1}{2}x - (\frac{8}{3} + \frac{9}{2})\) Evaluate: \[ \int_{2}^{5} \left(\frac{5}{6}x - \frac{43}{6}\right) \, dx \] \[ = \left[\frac{5}{12}x^2 - \frac{43}{6}x\right]_{2}^{5} \] Calculate: \[ = \left(\frac{5}{12}(5)^2 - \frac{43}{6}(5)\right) - \left(\frac{5}{12}(2)^2 - \frac{43}{6}(2)\right) \] \[ = (\frac{125}{12} - \frac{215}{6}) - (\frac{20}{12} - \frac{86}{6}) \] Simplify and solve to get 4.5.

Sum the Areas

Add the absolute area values calculated:7 (from \(x = -1\) to \(x = 2\)) and 4.5 (from \(x = 2\) to \(x = 5\)).Thus, the total area of the triangle is \(7 + 4.5 = 11.5\).

Key concepts

Equations of Lines

Understanding how to find the equations of lines is crucial when working with geometric problems, like determining the area of a triangle. The first step is identifying the line's slope, which indicates the steepness and direction of the line. Slope (\(m\)) is found using the formula \(m = \frac{y_2 - y_1}{x_2 - x_1}\). For example, given the points \((-1,4)\) and \((2,-2)\), we find the slope as \(-2\).

Next, use the point-slope form to write the equation of the line: \(y - y_1 = m(x - x_1)\). This formula helps form a line's equation by substituting one of the points and the slope you calculated. Thus, with a point \((-1, 4)\), and a slope \(m = -2\), the equation becomes \(y = -2x - 2\).

Repeat this process for other line segments by using different pairs of vertices like \((2, -2)\) to \((5, 1)\) and \((-1, 4)\) to \((5, 1)\). Understanding line equations gives a geometric framework necessary for further calculations, including integration.

Integration Limits

When using integration to find an area, it's important to set up the correct integration limits, which define the range over which integration is performed. The integration limits represent the extent of the shape along the x-axis, which, for a triangle, is dictated by the x-values of its vertices.

In our triangle with vertices at \((-1, 4)\), \((2, -2)\), and \((5, 1)\), we observe that the x-coordinate stretches from \(x = -1\) to \(x = 5\). These become our integration limits. This indicates that the calculation over each section of the x-axis will consider areas generated along each curve determined by the line equations.

Choosing correct integration limits ensures that we cover the entire base of the triangle in our calculations, and don't omit or include unwanted sections. Properly applied, it confirms a holistic evaluation of the defined area beneath curves in calculus.

Evaluate Integrals

Evaluating integrals is the core operation to find the area between curves when integration is involved. When tasked with finding a geometric area's size, it necessitates finding areas between established functions, which in triangular problems often revolve around its sides' line equations.

The essential process involves:

• Identifying which line is above (upper curve) and which one is below (lower curve).
• Setting up integrals for each segment with these lines acting as the boundary.

As seen, from \(x = -1\) to \(x = 2\), the line \(y = -2x - 2\) serves as the upper curve compared to the x-axis. Here, we compute the integral \(\int_{-1}^{2} (-2x - 2) \, dx\). The integration process involves determining the antiderivative followed by evaluating it over the previously defined limits.

In the segment from \(x = 2\) to \(x = 5\), utilize the lines \(y = \frac{1}{3}x - \frac{8}{3}\) and \(y = -\frac{1}{2}x + \frac{9}{2}\), where one acts as upper and the other as lower within this specific context. Proper execution of integrations thus yields calculated areas, complemented further by evaluating differences in reconstructed antiderivatives along set limits.

Triangle Vertices Coordinates

The coordinates of a triangle's vertices form the building blocks of solving for its area, serving as the initial information needed to proceed further. Each vertex will have an x and a y component, providing a concrete point in a defined Cartesian plane.

For instance, with vertices at \((-1,4)\), \((2,-2)\), and \((5,1)\), you are provided foundational data required to establish line equations connecting each vertex pair. This positioning provides insight into the extent of the triangle on the plane, showing potential height of figures formed.

Knowing the vertices allows for precise determination of slopes and configuration of integration areas subsequently. Also, these coordinates help evaluate interlinking lines' directions, which can greatly aid in assessing orderliness, implying toward which lines should become upper or lower when performing integration. Essentially, interests in finding areas often start by evaluating and interpreting the coordinates, an indispensable preliminary step that steers all calculations to follow.