Problem 56 - Precalculus Mathematics for Calculus

Question

Sketch the triangle with the given vertices, and use a determinant to find its area. $$(1,0),(3,5),(-2,2)$$

Step-by-Step Solution

Answer

The area of the triangle is 9.5 square units.

Step 1: Sketch the Triangle

First, make a rough sketch of the triangle by plotting the given vertices on a coordinate plane. Plot the points at \((1,0)\), \((3,5)\), and \((-2,2)\). Connect them to form a triangle.

Step 2: Setup the Determinant Formula for Area

The area \(A\) of a triangle with vertices \((x_1, y_1), (x_2, y_2), (x_3, y_3)\) can be calculated using the determinant:\[A = \frac{1}{2} \left| x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2) \right|\]

Step 3: Substitute the Vertices into the Formula

Substitute the coordinates of the given vertices: - \((x_1, y_1) = (1,0)\)- \((x_2, y_2) = (3,5)\)- \((x_3, y_3) = (-2,2)\)The formula becomes:\[A = \frac{1}{2} \left| 1(5-2) + 3(2-0) + (-2)(0-5) \right|\]

Step 4: Calculate the Determinant Expression

Calculate inside the absolute value:\[1(5-2) = 1 \times 3 = 3\]\[3(2-0) = 3 \times 2 = 6\]\[(-2)(0-5) = -2 \times -5 = 10\]Combine these:\[3 + 6 + 10 = 19\]

Step 5: Compute the Final Area

Substitute the result back into the formula:\[A = \frac{1}{2} \left| 19 \right| = \frac{19}{2} = 9.5\]Thus, the area of the triangle is 9.5 square units.

Key Concepts

Coordinate Geometry

Coordinate Geometry is a fascinating branch of mathematics that allows us to explore the shapes and sizes of geometric figures through the use of a coordinate system. In the case of a triangle, we use the Cartesian plane to represent points as coordinates. This helps us visualize the placement and structure of geometric shapes and perform calculations on them.

Given a set of points, such as \(1,0\), \(3,5\), and \(-2,2\), we can plot these on the plane to form a triangle. Each point is described by an \(x, y\) coordinate, with the first number representing the horizontal position (x-axis) and the second the vertical position (y-axis).

• **Plotting points**: On a graph, identify the x and y coordinates and mark the points accordingly.
• **Drawing lines**: Connect the points sequentially to draw the sides of the triangle.

By using coordinate geometry, we've laid the groundwork for applying more advanced techniques to analyze the triangle's properties.

Determinants

Determinants are a mathematical tool that can transform many complex mathematical operations into simpler computation. Specifically, a determinant is a special number calculated from a square matrix. This concept is crucial in solving systems of linear equations, finding area and volume of certain geometric shapes, and more.

When relating determinants to a triangle on a coordinate plane, we use a specific determinant formula to find the triangle's area, when given its vertices.

• The general formula employed is:
  \[ A = \frac{1}{2} \left| x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2) \right| \]
• Here, \(x_1, y_1, x_2, y_2, x_3, \) and \(y_3\) represent the coordinates of the triangle's vertices.

Through determinants, one can efficiently calculate the area without manually measuring or calculating the lengths of the sides. This method is particularly useful for triangles situated at various angles and positions.

Triangle Area Calculation

Calculating the area of a triangle using its vertex coordinates is an essential application of both coordinate geometry and determinants. Instead of relying on the traditional base-times-height method, using coordinates allows for a straightforward approach to finding the area.

Consider the vertices \(1,0\), \(3,5\), and \(-2,2\). By plugging these into the determinant formula mentioned before, we calculate each term inside the determinant separately: \(1(5-2) = 3\), \(3(2-0) = 6\), and \((-2)(0-5) = \text{-10}\).

• **Combine the terms**: Add the results: \(3 + 6 + 10 = 19\).
• **Divide by two**: Take \(\frac{1}{2}\) of this total.
• **Absolute value**: Calculate the absolute value to ensure the area is positive, as area can't be negative.

Ultimately, the triangle's area is determined as \(9.5\) square units, revealing a dynamic interplay between geometry and algebra to achieve a precise result.