Problem 9 - College Algebra and Calculus: An Applied Approach

Question

Use a determinant to find the area of the triangle with the given vertices. $$ (-3,5),(2,6),(3,-5) $$

Step-by-Step Solution

Answer

The area of the triangle with vertices (-3,5), (2,6), and (3,-5) is 28 square units.

Step 1: Write Down the Formula

The formula to calculate the area of a triangle using the determinant method is given by: \[ \text{Area} = \frac{1}{2} |x_{1}(y_{2} - y_{3}) + x_{2}(y_{3} - y_{1}) + x_{3}(y_{1} - y_{2})| \] where (x1,y1), (x2,y2), and (x3,y3) are the coordinates of the vertices of the triangle.

Step 2: Substitute The Values into the Formula

Substitute the given values into the formula: \[ \text{Area} = \frac{1}{2} |-3(6 - (-5)) + 2(-5 - 5) + 3(5 - 6)| \]

Step 3: Calculate the Area

Simplify the expression inside the absolute value sign and then multiply by \(\frac{1}{2}\) to find the area: \[ \text{Area}= \frac{1}{2} |-3(11) + 2(-10) + 3(-1)| = \frac{1}{2} |-33 - 20 - 3| = \frac{1}{2} |-56|= 28 \]

Key Concepts

Area of a Triangle

Understanding the area of a triangle is crucial for many mathematical applications. The area is essentially the space contained within the three sides of the triangle. Knowing how to calculate this area helps in fields like geometry, engineering, and even art.
For a basic triangle, calculating the area could involve using simpler methods, such as the base times height divided by two. But when you're dealing with triangles on a coordinate plane, where the vertices are given by their coordinates, a different method can be effectively used.
This is where the determinant method comes in, especially useful for triangles on a graph. The area is expressed with a mathematical formula allowing you to plug in the triangle's coordinates directly.

Formula for Area

The formula for finding the area of a triangle using the determinant method is: \\[ \\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| \\] This may seem complicated at first glance, but it's quite straightforward once you understand the components.
Each part of the formula involves selecting coordinates from the triangle's vertices:

• \(x_1, y_1\) is the first vertex,
• \(x_2, y_2\) is the second vertex,
• \(x_3, y_3\) is the third vertex.

The calculation uses these points to form cross products, which are then summed and divided by two, providing the absolute value of that result—ensuring that area remains positive.

Triangle Vertices

In the context of coordinate geometry, each vertex of a triangle is a point with specific coordinates. In our problem, the vertices of the triangle are given as

• \((-3,5)\)
• \((2,6)\)
• \((3,-5)\)

Each pair represents an \((x, y)\) coordinate on the Cartesian plane. These vertices are essential because they define the triangle's shape and size. Using these coordinates, you can determine the slope of each side, the perimeter, and, most relevant to our discussion, the area using the determinant method.
Making sure you have the correct coordinates is the first step in calculating the area accurately, as any mistake here can lead to incorrect results.

Determinant Method

The determinant method provides an efficient way to calculate the area of a triangle in a coordinate plane. This method involves using the determinant of a matrix constructed from the coordinates of the triangle's vertices.
Here's how it works:

• Arrange the vertices in a 3x2 matrix, such that each row corresponds to one vertex.
• The formula replaces the typical matrix-related tasks with operations specifically designed for area calculation.

This method embraces the idea of linear algebra applied to geometry. It allows you to tackle complex geometric problems with straightforward calculations, employing simple arithmetic like addition, subtraction, and absolute values to find exact areas.
Moreover, it provides a visual and analytical understanding of how the position of the vertices impacts the overall space that the triangle occupies on the plane. The determinant method is particularly useful when dealing with expansive sets of vertices, as each calculation step builds upon familiar algebraic and geometric concepts.