Problem 41 - A Graphical Approach to College Algebra

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 equal 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 coordinates as given. $$P(2,5), Q(-1,3), R(4,0)$$

Step-by-Step Solution

Answer

The area is 9.5 square units.

Step 1: Write down the determinant formula

The area of the triangle with vertices at coordinates \((x_1, y_1), (x_2, y_2), (x_3, y_3)\) is given by the absolute value of \(D\). The determinant \(D\) is defined as \[D = \frac{1}{2} \times \begin{vmatrix} x_1 & y_1 & 1 \ x_2 & y_2 & 1 \ x_3 & y_3 & 1 \end{vmatrix}\].

Step 2: Insert the coordinates into the determinant

Substitute the given coordinates \(P(2, 5), Q(-1, 3), R(4, 0)\) into the determinant. This gives us: \[ D = \frac{1}{2} \times \begin{vmatrix} 2 & 5 & 1 \ -1 & 3 & 1 \ 4 & 0 & 1 \end{vmatrix} \].

Step 3: Calculate the determinant of the matrix

The determinant of a 3x3 matrix \(\begin{vmatrix}a & b & c \ d & e & f \ g & h & i \end{vmatrix}\) can be calculated using the formula \(a(ei - fh) - b(di - fg) + c(dh - eg)\). Substitute \(a=2, b=5, c=1, d=-1, e=3, f=1, g=4, h=0, i=1\) into this formula.

Step 4: Simplify the determinant

Apply the formula: \[ 2(3\times1 - 1\times0) - 5(-1\times1 - 4\times1) + 1(-1\times0 - 3\times4) \]. This simplifies to \[ 2(3) - 5(-1 - 4) + 1(-0 - 12) \].

Step 5: Calculate each term

Evaluate each expression separately: \[ 2 \times 3 = 6,\quad 5(1 + 4) = 5 \times 5 = 25,\quad 1 \times (-12) = -12 \].

Step 6: Find the sum of the expanded terms

Add the results of each of the terms together: \[ 6 + 25 - 12 = 19 \].

Step 7: Compute the final area of the triangle

The area of the triangle is the absolute value of half the determinant: \[ \frac{1}{2} \times |19| = \frac{19}{2} = 9.5 \]. The area is thus 9.5 square units.

Key Concepts

Area of a Triangle

Finding the area of a triangle using its vertices is a common problem in coordinate geometry. The traditional formula requires the base and height of the triangle, but with coordinate geometry, we can find the area using a determinant. This is particularly useful when you don’t have a physical representation of the triangle to measure from. When using the determinant method, we arrange the coordinates of the vertices into a 3x3 matrix. The matrix has the following form:

• The first column is the x-coordinates \( (x_1, x_2, x_3) \).
• The second column is the y-coordinates \( (y_1, y_2, y_3) \).
• The third column is filled with ones.

The determinant of this matrix can be calculated and used to determine the area of the triangle, where the area \( A \) is given by:\[ A = \frac{1}{2} \times |D| \]In our example, we found the area to be 9.5 square units, emphasizing the convenience and power of the determinant method in finding area directly from coordinates.

Matrix Mathematics

Matrix mathematics is an essential tool in algebra, allowing us to deal efficiently with coordinates and transform them through operations such as addition, multiplication, and calculating determinants. Here, we particularly focus on calculating the determinant of a 3x3 matrix, which helps us solve geometric problems like finding the area of a triangle.A 3x3 matrix looks like this:\[\begin{array}{ccc} a & b & c \ d & e & f \ g & h & i \\end{array}\]The determinant \( D \) of this matrix is calculated with the formula:\[ D = a(ei - fh) - b(di - fg) + c(dh - eg) \]For our triangle's coordinates \( P(2,5), Q(-1,3), R(4,0) \), the process involved substituting the coordinates to find \( D \) and subsequently using it to calculate the triangle's area. This formula not only assists in solving geometry problems but also has wider applications in computer graphics, physics, and engineering.

Coordinate Geometry

Coordinate geometry, also known as analytic geometry, involves using coordinate systems to solve geometric problems. By placing geometric figures within a coordinate plane, problems that might require extensive graphic interpretation and scale can instead be handled with algebraic calculations. The key power of coordinate geometry lies in its precision and the ability to manage complex figures algebraically. It translates geometry into numbers and equations. Therefore, with a clear set of formulas and principles:

• We can find distances between points.
• We can calculate slopes, midpoints, and more.
• Most importantly in our example, we can find the area of triangles using determinants.

In our specific case, coordinate geometry allowed us to assign values to the vertices of a triangle and use those positions to determine its area without physically measuring the triangle. This makes it a robust tool not only for theoretical applications but also for practical, real-world solving of geometric problems.