Problem 39
Question
Find the areas of the triangles whose vertices are given in Exercises \(39-42\) $$ A(0,0), \quad B(-2,3), \quad C(3,1) $$
Step-by-Step Solution
Verified Answer
The area of the triangle is 5.5 square units.
1Step 1: Understand the formula for the area of a triangle
To find the area of a triangle with vertices at \((x_1, y_1), (x_2, y_2), (x_3, y_3)\), we use the formula: \[ \text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \] This formula derives from the determinant of a matrix representing the coordinates of the vertices.
2Step 2: Identify coordinates and plug them into the formula
Given the vertices \( A(0,0), B(-2,3), C(3,1) \), identify the coordinates: \(x_1 = 0, y_1 = 0, x_2 = -2, y_2 = 3, x_3 = 3, y_3 = 1\). Plug these into the area formula: \[ \frac{1}{2} \left| 0(3-1) + (-2)(1-0) + 3(0-3) \right| \]
3Step 3: Simplify the expression inside the absolute value
Calculate each term separately: 1. \(0 \times 2 = 0\), 2. \(-2 \times 1 = -2\), 3. \(3 \times -3 = -9\). Add these values together: \[0 + (-2) + (-9) = -11\] The expression inside the absolute value becomes \(-11\).
4Step 4: Calculate the absolute value and area
Take the absolute value of \(-11\), which is \(11\). Substitute back into the area formula: \[ \frac{1}{2} \times 11 = 5.5 \] Thus, the area of the triangle is \(5.5\) square units.
Key Concepts
Determinant FormulaCoordinate GeometryAbsolute ValueTriangles in Mathematics
Determinant Formula
The determinant formula for finding the area of a triangle using its vertex coordinates is quite intriguing. This method is efficient and leverages the formal determinant from linear algebra. The formula helps in calculating the area of the triangle formed by points
- \( (x_1, y_1) \)
- \( (x_2, y_2) \)
- \( (x_3, y_3) \)
Coordinate Geometry
Coordinate Geometry, or analytic geometry, is the study of geometry using a coordinate system. Here, points are expressed as ordered pairs on a plane. This approach is perfect for solving spatial problems with algebra. When dealing with triangles, each vertex is defined by a pair of coordinates and arranged as \( (x, y) \). The beauty of coordinate geometry is in transforming geometric questions into algebraic equations using these coordinates.The determination of area, distance, and midpoint between points becomes straightforward with coordinate geometry. In our problem, each point such as
- \( A(0,0) \)
- \( B(-2,3) \)
- \( C(3,1) \)
Absolute Value
In mathematics, the absolute value signifies the non-negative value of a number without regard to its sign. Denoted by vertical bars around the number or expression, such as \( |-11| = 11 \), this concept preserves magnitude while ignoring direction.In the context of area calculation, absolute value ensures that the computed area is always a positive measure. This is crucial since a negative area doesn't make sense geometrically. After calculating the determinant value
- \( -11 \)
Triangles in Mathematics
Triangles are fundamental shapes in mathematics, significantly utilized in various applications from basic geometry to advanced calculus. Defined by three vertices and three sides, triangles are the simplest polygon type but hold immense applicability.
Understanding triangles involve examining properties like angles, side-length ratios, and special points (e.g., centroid, orthocenter), which reveal a great deal about geometry and trigonometry principles.
In our exercise, calculating the area of a triangle using vertex coordinates highlights the intersection of triangles with coordinate geometry and algebra. It's not just about visualizing a shape on paper; it's about understanding how triangles function within mathematical systems.
This base understanding of triangle properties invites further exploration into similar mathematical disciplines, making them essential in engineering, physics, and computer graphics.
Other exercises in this chapter
Problem 38
In Exercises \(35-38\) , find a. the direction of \(P_{1} P_{2}\) and b. the midpoint of line segment \(P_{1} P_{2}\) . $$ P_{1}(0,0,0) \quad P_{2}(2,-2,-2) $$
View solution Problem 39
Sketch the surfaces in Exercises \(13-76\) $$ z^{2}-x^{2}-y^{2}=1 $$
View solution Problem 39
In Exercises 39–44, find the distance from the point to the plane. $$ (2,-3,4), \quad x+2 y+2 z=13 $$
View solution Problem 39
In Exercises \(35-40,\) find the distance between points \(P_{1}\) and \(P_{2}\) $$ P_{1}(0,0,0), \quad P_{2}(2,-2,-2) $$
View solution