Problem 44 - Thomas Calculus

Question

Find the areas of the triangles whose vertices are given in Exercises \(41-47 .\) $$ A(-6,0), \quad B(10,-5), \quad C(-2,4) $$

Step-by-Step Solution

Answer

The area of the triangle is 42 square units.

Step 1: Write the Formula for the Area of a Triangle

The area of a triangle given its vertices \( (x_1, y_1), (x_2, y_2), (x_3, y_3) \) is calculated using the formula: \[ \text{Area} = \frac{1}{2} \left| x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2) \right|.\] We'll use this formula to find the area of the triangle with vertices \(A(-6,0)\), \(B(10,-5)\), and \(C(-2,4)\).

Step 2: Substitute the Given Vertices

Substitute the coordinates \(A(x_1, y_1) = (-6, 0)\), \(B(x_2, y_2) = (10, -5)\), and \(C(x_3, y_3) = (-2, 4)\) into the area formula.\[ \text{Area} = \frac{1}{2} | -6(-5 - 4) + 10(4 - 0) + (-2)(0 + 5) |\]

Step 3: Simplify the Expression

Calculate each term within the absolute value:- First Term: \(-6(-5-4) = -6(-9) = 54\)- Second Term: \(10(4 - 0) = 10 \times 4 = 40\)- Third Term: \((-2)(0 + 5) = -2 \times 5 = -10\)Now, substitute these values:\[ \text{Area} = \frac{1}{2} | 54 + 40 - 10 |\]

Step 4: Evaluate the Determinant

Add the calculated values inside the absolute value:\[ 54 + 40 - 10 = 84 \]Then, substitute back into the equation:\[ \text{Area} = \frac{1}{2} | 84 |\]

Step 5: Calculate the Final Area

Divide by 2 to get the area:\[ \text{Area} = \frac{1}{2} \times 84 = 42\]

Step 6: Conclusion

The area of the triangle with vertices \(A(-6,0)\), \(B(10,-5)\), and \(C(-2,4)\) is \textbf{42} square units.

Key Concepts

Area Formula for Triangles

In geometry, there are various ways to compute the area of a triangle. One effective method utilizes the coordinates of the vertices. The formula for the area when vertices are given as \((x_1, y_1), (x_2, y_2), (x_3, y_3)\) is: \[ \text{Area} = \frac{1}{2} \left| x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2) \right|. \]This formula stems from breaking down the triangle into simpler geometric parts. By calculating this determinant-like expression, we can find the signed area, simplifying the process greatly. This method considers both the arrangement and distance between the points, which is crucial when dealing with coordinates. It’s a powerful tool for students and professions involving geometrical computation.

Coordinate Geometry

Understanding coordinate geometry is essential when working with triangles on the XY plane. Here, each point on the plane is defined by a coordinate pair \((x, y)\), which helps in determining geometrical properties:

• Distance between two points.
• Slope of a line segment.
• Relative positions of various shapes.

In our example, we had vertices \[A(-6,0), \; B(10,-5), \; C(-2,4).\] These coordinates allow us to perform calculations easily and directly. Coordinate geometry offers tools not only for calculating areas but also for understanding spatial relationships in a practical way. Embracing these concepts can enable better analytical skills in mathematics and its applications.

Determinants in Geometry

Determinants are powerful mathematical tools often used in algebra, calculus, and geometry. In the context of geometry, particularly in calculating triangle areas, determinants help simplify complex expressions into manageable form.With the formula \[ \frac{1}{2} \left| x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2) \right|, \] parts of the calculation mimic determinants:

• Swapping between addition and subtraction of terms.
• Handling coordinates as matrix-like structures.
• Enabling compact expressions for outcome like the triangle area.

This approach is convenient and efficient. It also connects linear algebra concepts with geometric interpretations, making it an integral part of advanced geometry studies. Knowing this relationship broadens one's mathematical understanding, allowing for deeper insight into both fields.