Problem 42 - A Graphical Approach to Precalculus with Limits

Question

A triangle with vertices at \(\left(x_{1}, y_{1}\right),\left(x_{2}, y_{2}\right)\) and \(\left(x_{3}, y_{3}\right),\) as shown in the figure, has area cqual to the absolute value of \(D,\) where $$D=\frac{1}{2} \operatorname{det}\left[\begin{array}{lll}x_{1} & y_{1} & 1 \\\x_{2} & y_{2} & 1 \\\x_{3} & y_{3} & 1\end{array}\right]$$ Use \(D\) to find the area of each triangle with coondinates as given. (GRAPH CAN'T COPY) $$P(2,-2), Q(0,0), R(-3,-4)$$

Step-by-Step Solution

Answer

The area of the triangle is 1 square unit.

Step 1: Identify the Points

Identify the coordinates of the vertices of the triangle. The given points are \(P(2, -2)\), \(Q(0, 0)\), and \(R(-3, -4)\).

Step 2: Set Up the Determinant Matrix

Place the coordinates of the points into a 3x3 matrix:\[\begin{bmatrix}2 & -2 & 1 \0 & 0 & 1 \-3 & -4 & 1\end{bmatrix}\]

Step 3: Calculate the Determinant of the Matrix

Use the rule of Sarrus or cofactor expansion to find the determinant of the matrix:\[D = 2(0 \cdot 1 - 1 \cdot (-4)) - (-2)(0 \cdot 1 - 1 \cdot (-3)) + 1(0 \cdot (-3) - 1 \cdot 0)\]

Step 4: Simplify the Expression

Calculate each term:- First term: \(2 \times 4 = 8\)- Second term: \(+2 \times 3 = 6\)- Third term: \(0 \times 0 = 0\)Thus, the determinant \(D = 8 - 6 + 0 = 2\).

Step 5: Calculate the Absolute Value of the Determinant

The area of the triangle is given by the formula:\[\text{Area} = \frac{1}{2} \times |D|\]Substitute \(D = 2\) into the equation:\[\text{Area} = \frac{1}{2} \times |2| = \frac{1}{2} \times 2 = 1\]

Key Concepts

Area of a Triangle

To find the area of a triangle using its vertex coordinates, we rely on a special formula derived from the concept of determinants in matrix algebra. If a triangle's vertices are \((x_1, y_1), (x_2, y_2), (x_3, y_3)\), we can calculate its area using the determinant of a 3x3 matrix where each row corresponds to a vertex, and the third column is filled with ones. This matrix method offers a systematic way of determining the area without direct measurement of sides or angles.

The formula is:\[\text{Area} = \frac{1}{2} \left| \begin{vmatrix} x_1 & y_1 & 1 \ x_2 & y_2 & 1 \ x_3 & y_3 & 1 \end{vmatrix} \right|\]This equation highlights the connection between coordinate geometry and linear algebra, showing how geometric properties can be distilled into matrix forms.

The requirement for the absolute value ensures that the area remains a positive quantity, reflecting consistent geometric understanding.

Coordinate Geometry

Coordinate geometry allows us to work with geometric figures using algebraic techniques. This approach facilitates the computation of distances, angles, and areas on a coordinate plane. We represent each point by its coordinates, which can be modified and manipulated algebraically, offering a bridge between abstract math and physical space.

In the context of finding the area of a triangle, you'll first identify the vertices' coordinates and then utilize algebraic expressions, such as determinants, for calculations. By transforming geometric shapes into algebraic equations, it becomes easier to apply various mathematical methods to solve problems.

• Vertices of the triangle translate into points on a plane.
• Formulas based on coordinates yield precise results like area.
• Algebraic summation and multiplication replace geometric measurement.

Matrix Algebra

Matrix algebra simplifies handling computations in coordinate geometry, especially when it comes to determinants. A determinant is a special value computed from a square matrix. In the context of this exercise, we use determinants to find the area of triangles on a plane.

The process involves setting up a 3x3 matrix with the vertices' coordinates and a column of ones. The reason behind using a matrix is its ability to consolidate information and perform multiple operations efficiently. With a matrix, we're applying a set of linear transformations, and the determinant helps in calculating the result of these transformations.

Here's why matrix algebra is useful:

• Efficiently packages many calculations into one operation.
• Can be used to solve various geometric problems compactly.
• Helps visualize transformations and changes in geometric space.