Problem 41 - 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) $$P(2,5), Q(-1,3), R(4,0)$$

Step-by-Step Solution

Answer

The area of the triangle is \(\frac{19}{2}\).

Step 1: Set Up the Determinant Formula

The formula for the area of a triangle given its vertices \((x_1, y_1), (x_2, y_2), (x_3, y_3)\) is \(D = \frac{1}{2} \left| \begin{vmatrix} x_1 & y_1 & 1 \ x_2 & y_2 & 1 \ x_3 & y_3 & 1 \end{vmatrix} \right|\). The vertices for triangle PQR are \(P(2,5), Q(-1,3), R(4,0)\). Plug these into the formula.

Step 2: Set Up the 3x3 Matrix

Substitute the coordinates into the matrix: \[ \begin{vmatrix} 2 & 5 & 1 \ -1 & 3 & 1 \ 4 & 0 & 1 \end{vmatrix} \]

Step 3: Compute the Determinant

Calculate the determinant using the formula for a 3x3 matrix:\[ \begin{vmatrix} 2 & 5 & 1 \ -1 & 3 & 1 \ 4 & 0 & 1 \end{vmatrix} = 2(3 \cdot 1 - 1 \cdot 0) - 5(-1 \cdot 1 - 1 \cdot 4) + 1(-1 \cdot 0 - 3 \cdot 4) \]= \[ 2(3) + 5(-5) + 1(-12) \] = 6 + 25 - 12.

Step 4: Simplify the Result

Compute the simplified determinant: 6 + 25 - 12 = 19. Now, since the area \(|D|\) is \(\frac{1}{2} \text{ of the absolute value of the determinant},\)

Step 5: Calculate the Area

The area of the triangle \(\triangle PQR\) is \(\frac{1}{2} \times |19| = \frac{19}{2}\).

Step 6: Conclusion: State the Final Area

The area of the triangle formed by the points \(P(2,5), Q(-1,3), R(4,0)\) is \(\frac{19}{2}\).

Key Concepts

Area of a Triangle

The area of a triangle can be calculated using the determinant of a matrix when the coordinates of its vertices are known. For a triangle with vertices at \((x_{1}, y_{1}), (x_{2}, y_{2}), (x_{3}, y_{3})\), the formula to find the area is:\[\text{Area} = \frac{1}{2} \left| D \right|,\]where \(D\) is the determinant of the 3x3 matrix formed by these coordinates. This formula leverages the geometric properties of determinants, allowing us to easily find the area without directly calculating through geometric means such as base and height. Using the determinant method can be particularly helpful when dealing with vertices in a coordinate plane, as it simplifies complex calculations into a straightforward operation.

Vertex Coordinates

Understanding vertex coordinates is essential in calculating the area of a triangle using the determinant method. Each vertex of a triangle in the coordinate plane can be represented by an ordered pair \((x, y)\). These coordinates are crucial inputs that determine the shape and dimensions of the triangle. To illustrate, consider a triangle with vertices at points \(P(2,5), Q(-1,3), R(4,0)\). Each point consists of an \(x\)-coordinate and a \(y\)-coordinate:

• \(x_1 = 2, y_1 = 5 \) for point P
• \(x_2 = -1, y_2 = 3 \) for point Q
• \(x_3 = 4, y_3 = 0 \) for point R

These coordinates are then placed into the matrix for calculating the determinant, which is a crucial step in finding the area of the triangle using the determinant approach.

Matrix Calculation

Matrix calculation involves finding the determinant of a 3x3 matrix that is structured from the triangle's vertex coordinates. Once you have the coordinates of the triangle vertices, you can set up the matrix: \[\begin{vmatrix} x_1 & y_1 & 1 \x_2 & y_2 & 1 \x_3 & y_3 & 1 \end{vmatrix}\]For our example with points \(P(2,5), Q(-1,3), R(4,0)\):\[\begin{vmatrix} 2 & 5 & 1 \-1 & 3 & 1 \4 & 0 & 1 \end{vmatrix}\]To calculate the determinant of this matrix, perform the following operations:

• Multiply the elements along the main diagonal from top-left to bottom-right and subtract the product of the other diagonal.
• Simplify your result by carrying out the standard operations for calculating a determinant.

Performing the calculation for the given matrix:\[= 2(3 \cdot 1 - 1 \cdot 0) - 5(-1 \cdot 1 - 4) + 1(-1 \cdot 0 - 3 \cdot 4)\]Simplifying further results in 19. Finally, to find the area of the triangle, compute \[\frac{1}{2} \times |19|.\] This matrix method streamlines the computation for any triangle when its vertices are known. This makes working with triangles in coordinate geometry much more efficient!