Problem 58 - Precalculus Mathematics for Calculus

Question

Sketch the triangle with the given vertices, and use a determinant to find its area. $$(-2,5),(7,2),(3,-4)$$

Step-by-Step Solution

Answer

The area of the triangle is 33 square units.

Step 1: List the Vertices

The vertices of the triangle are given as \((-2, 5)\), \((7, 2)\), and \((3, -4)\). These points will be used to calculate the area.

Step 2: Write the Formula for the Area

The area \(A\) of a triangle with vertices \((x_1, y_1)\), \((x_2, y_2)\), and \((x_3, y_3)\) can be found 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 Values into the Formula

Substitute \((x_1, y_1) = (-2, 5)\), \((x_2, y_2) = (7, 2)\), and \((x_3, y_3) = (3, -4)\) into the determinant formula:\[ A = \frac{1}{2} \left| -2(2 - (-4)) + 7((-4) - 5) + 3(5 - 2) \right| \].

Step 4: Calculate the Determinant

Calculate each term in the expression:- For \(-2(2 - (-4)) = -2 \times 6 = -12\).- For \(7((-4) - 5) = 7 \times (-9) = -63\).- For \(3(5 - 2) = 3 \times 3 = 9\).Combine these: \[-12 - 63 + 9 = -66.\]

Step 5: Calculate the Area

The area is half of the absolute value of the result. Thus, calculate:\[ A = \frac{1}{2} \left| -66 \right| = \frac{1}{2} \times 66 = 33. \]

Step 6: Double Check Your Calculations

Review each step for accuracy, confirming that each mathematical operation has been correctly completed, especially the signs and calculations of terms in the determinant.

Key Concepts

Determinant Formula

The determinant formula is a powerful tool in mathematics, particularly useful in calculating the area of a triangle when its vertices coordinates are given. To find the area of a triangle using the determinant, you can use the formula:

• For a triangle with points \( (x_1, y_1) \), \( (x_2, y_2) \), and \( (x_3, y_3) \), the area \( A \) is:\[ A = \frac{1}{2} \left| x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2) \right|. \]
• This formula essentially calculates the determinant of a matrix formed by the vertices of the triangle, projected onto the coordinate plane.
• The absolute value ensures that the area is always positive, which is essential since an area cannot be negative.

This approach is especially beneficial because it does not require the length of the sides, only the coordinates. It's efficient and avoids the more lengthy process of using more complex geometry formulas.

Coordinates of Vertices

To determine the area of a triangle using the determinant formula, the essential data required are the coordinates of the vertices. The coordinates are precise points on the coordinate plane that define the corners of the triangle.

• In our example, the vertices of the triangle are \( (-2, 5), (7, 2), (3, -4) \).
• These points are substituted directly into the determinant formula.For instance: \[(-2) \](for \(x_1\), and \[5\](for \(y_1\) and so forth.
• By understanding these coordinates, you can effectively map out the geometry of the triangle on a graph, offering a tangible visual representation important for comprehension.

Having a clear understanding of the coordinates makes it easier to manage any potential errors in calculation, as manual plotting can serve as a double-check method.

Triangle Geometry

The use of determinants in finding the area of a triangle is grounded in the principles of triangle geometry, where understanding the fundamental properties of triangles becomes important.

• A triangle is composed of three sides, three angles, and three vertices each crucial to geometric calculations.
• By using determinants, this approach exploits the relationship between these vertices in a two-dimensional plane to find the area.
• This understanding brings efficiency by reducing the need for direct measurement of sides or angles, which could be more time-consuming or prone to error.

Furthermore, triangle geometry offers insights into congruence, similarity, and the Pythagorean theorem, all of which have relationships tied to the spatial distribution of the triangle's coordinates.
In summary, this problem-solving piece not only provides an efficient calculation method but also enriches the comprehension of geometrical shapes and their properties.