Problem 44 - 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) $$P(-1,0), Q(-3,5), R(5,2)$$

Step-by-Step Solution

Answer

The area of the triangle is 17 square units.

Step 1: Write the determinant formula

To find the area of the triangle, we need to compute the determinant of a matrix formed by the triangle's vertices. The formula for the area \(D\) of the triangle is: \[D = \frac{1}{2} \left| \begin{vmatrix} x_1 & y_1 & 1 \ x_2 & y_2 & 1 \ x_3 & y_3 & 1 \end{vmatrix} \right|\] In this problem, the vertices of the triangle are \(P(-1,0), Q(-3,5), R(5,2)\).

Step 2: Substitute the coordinates into the matrix

Substitute the coordinates of the points into the determinant formula: \[ \begin{vmatrix} -1 & 0 & 1 \ -3 & 5 & 1 \ 5 & 2 & 1 \end{vmatrix} \]

Step 3: Calculate the determinant

Expand the determinant using cofactor expansion:\[ \begin{vmatrix} -1 & 0 & 1 \ -3 & 5 & 1 \ 5 & 2 & 1 \end{vmatrix} = -1 \cdot (5 \cdot 1 - 1 \cdot 2) - 0 \cdot (-3 \cdot 1 - 5 \cdot 5) + 1 \cdot (-3 \cdot 2 - 5 \cdot 5) \]Simplify the expression:\[ = -1 \cdot (5 - 2) + 0 \cdot 0 + 1 \cdot (-6 - 25) \] \[ = -1 \cdot 3 + 0 \cdot 0 + 1 \cdot (-31) \] \[ = -3 - 31 = -34 \]

Step 4: Calculate the area

The formula for the area of the triangle is the absolute value of half of the determinant:\[ \text{Area} = \frac{1}{2} \cdot |D| = \frac{1}{2} \cdot |-34| = \frac{1}{2} \cdot 34 \] \[ \text{Area} = 17 \]

Key Concepts

Triangle Area

The area of a triangle can be determined using a powerful tool in mathematics called the determinant. In traditional geometry, you might find the area by sketching the figure or using lengths and angles.But with coordinate geometry, you use the coordinates of the triangle's vertices.

By placing the triangle's vertex coordinates \((x_1, y_1), (x_2, y_2), (x_3, y_3)\) into a specific formula, we can find the area using determinants. Here's the formula:

• 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 formula comes from a matrix algebra concept. It calculates the "signed area" of the triangle, then takes the absolute value to give the true geometric area.The "half" factor in the formula ensures the result matches traditional calculations for triangle area.

Matrix Algebra

In the context of determining a triangle's area, matrix algebra plays an integral role. A matrix is essentially an array of numbers arranged into rows and columns. In this exercise, the numbers in the matrix are the coordinates of the triangle's vertices, with an additional column comprised solely of ones for simplicity.

• The determinant of a matrix is a special number calculated from this array. It's a scalar that gives a lot of information, especially in linear transformations, such as the area.
• To compute the determinant of a \(3 \times 3\) matrix, we use cofactor expansion. This involves multiplying each element in a row by the determinant of a smaller matrix obtained by removing that row and column.

For our triangle, \[\begin{vmatrix}-1 & 0 & 1 \ -3 & 5 & 1 \ 5 & 2 & 1 \end{vmatrix}\]serves as the matrix whose determinant gives us the area information. After calculations, we determine a "signed" value which reflects the geometry's orientation. Taking its absolute value ensures we're only considering the size.

Coordinate Geometry

Coordinate geometry is a melding of algebra and geometry using coordinate points to solve geometric problems. This exercise employs coordinate geometry to find areas efficiently by considering vertices of geometric shapes. In contrast to classical geometry, which often relies on intricate constructions, calculations in coordinate geometry leverage formulas to arrive at the same results.

• In our context, the vertices of triangle \(P(-1, 0), Q(-3, 5), R(5, 2)\) point to a space on the coordinate plane.
• This plane helps to visualize and compute the spatial relationships through algebraic means.

The use of determinants derived from matrices of these vertex points showcases one of the significant advantages of coordinate geometry. By converting the spatial problem of area into an algebraic calculation, we sidestep the need for drawing models or using geometric measurements directly.This combination makes it both a powerful and efficient method in mathematical analysis.