Problem 59 - Precalculus: Mathematics for Calculus

Question

Sketch the triangle with the given vertices, and use a determinant to find its area. $$(-1,3),(2,9),(5,-6)$$

Step-by-Step Solution

Answer

The triangle's area is 31.5 square units.

Step 1: Understand the Determinant Formula for Area of Triangle

The area of a triangle with vertices at \((x_1, y_1), (x_2, y_2), (x_3, y_3)\) is given by the determinant formula:\[\text{Area} = \frac{1}{2} \left| x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2) \right|\].

Step 2: Assign Points to Variables

Assign the provided vertices to the formula:\((x_1, y_1) = (-1, 3), (x_2, y_2) = (2, 9), (x_3, y_3) = (5, -6)\).

Step 3: Substitute the Coordinates into the Formula

Substitute the coordinates into the formula:\[\text{Area} = \frac{1}{2} \left| -1(9 + 6) + 2(-6 - 3) + 5(3 - 9) \right|\].

Step 4: Simplify the Expression Inside the Determinant Absolute Value

Calculate each term individually:\(-1 \times 15 = -15\), \(2 \times -9 = -18\), and \(5 \times -6 = -30\). Put these together: \(-15 - 18 - 30 = -63\).

Step 5: Calculate the Absolute Value and Complete the Formula

Take the absolute value: \(|-63| = 63\). Then compute the area:\[\text{Area} = \frac{1}{2} \times 63 = 31.5\].

Step 6: Sketch the Triangle

Plot the points \((-1, 3), (2, 9), (5, -6)\)on a coordinate plane and connect them to form a triangle. This visualizes the triangle for better understanding and verification.

Key Concepts

Coordinate Geometry

Coordinate geometry, also known as analytic geometry, allows us to represent geometric shapes in a numerical way using coordinates. In this exercise, we have been given three points: \((-1, 3), (2, 9), (5, -6)\). These points can be plotted on a coordinate plane, which is a two-dimensional surface defined by an x-axis and a y-axis.
When you plot these points and connect them, they form a triangle. The clarity offered by coordinate geometry makes it easy to visualize geometric problems and solutions. This becomes especially useful when trying to determine geometric properties such as the area of shapes. Placing points precisely and being able to clearly understand and sketch their relationship on a coordinate plane is one of the fundamental skills in coordinate geometry.
Understanding how to find distances and areas using these points paves the way to solving more complex problems. It builds a bridge to further mathematical concepts where geometry meets algebra, and allows for precise calculations of various properties, rather than relying on visual estimations.

Precalculus

Precalculus acts as a preparatory stage for calculus, reinforcing skills and concepts necessary for advanced mathematics. One of the essential skills it covers is the understanding of geometric properties using algebraic methods, such as calculating the area of a triangle. By leveraging algebraic equations and concepts, precalculus helps to solidify students' grasp of mathematical reasoning.
In this exercise, we are using a formula that integrates algebra into geometry. The process of substituting vertex coordinates into the determinant formula and simplifying them is based on algebraic manipulation—a core component of precalculus. This practice in calculation and understanding complex equations readies students for calculus, where advanced and abstract mathematical ideas are explored.
Precalculus emphasizes connections between different mathematical areas, preparing students to tackle calculus concepts like limits, derivatives, and integrals. These connections are evident as we explore geometric aspects using algebraic tools, thereby setting a robust foundation for future learning and application.

Determinant Formula

The determinant formula is a powerful tool for finding the area of a triangle when its vertices are given. It is an algebraic method that simplifies calculations, especially in coordinate geometry. The formula:\[\text{Area} = \frac{1}{2} \left| x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2) \right|\]helps to calculate the area of a triangle defined by vertices \((x_1, y_1), (x_2, y_2), (x_3, y_3)\). This formula is efficient because it transforms the geometric problem into an algebraic calculation, thereby avoiding complex trigonometric methods or cumbersome manual processes.
Here, substituting the coordinates for the given vertices into the formula simplifies to calculating expressions like \(-1(9 + 6)\), \(2(-6 - 3)\), and \(5(3 - 9)\), breaking the task into individual manageable steps. After calculating and simplifying these expressions, we sum them up, and take the absolute value before multiplying by \(\frac{1}{2}\), ensuring the area is always a positive value.
This method is not only straightforward but also widely applicable in various mathematical and real-world scenarios, making it an invaluable part of precalculus and more advanced mathematical coursework.