Problem 43
Question
Triangle area Find a formula for the area of the triangle in the \(x y\) -plane with vertices at \((0,0),\left(a_{1}, a_{2}\right),\) and \(\left(b_{1}, b_{2}\right) .\) Explain your work.
Step-by-Step Solution
Verified Answer
The area is \( \frac{1}{2} |a_1b_2 - b_1a_2| \).
1Step 1: Understanding the Triangle in the Coordinate Plane
We are given a triangle with vertices at \((0, 0)\), \((a_1, a_2)\), and \((b_1, b_2)\). The task is to find the formula for the area of this triangle. We can visualize the vertices as points on the coordinate plane.
2Step 2: Use the Determinant Formula for Area
To find the area of a triangle given its vertices in the coordinate plane, we use the formula derived from the determinant of a matrix containing the vertices. The area \(A\) is given by:\[A = \frac{1}{2} \left| a_1(b_2 - 0) + b_1(0 - a_2) + 0(a_2 - b_2) \right|\]This simplifies to: \[A = \frac{1}{2} \left| a_1b_2 - b_1a_2 \right|\]
3Step 3: Apply the Determinant Formula
By substituting the coordinates into the determinant-based formula, we compute:\[A = \frac{1}{2} \left| a_1b_2 - b_1a_2 \right|\]This gives us the area of the triangle defined by the given vertices in terms of \(a_1, a_2, b_1,\) and \(b_2\).
Key Concepts
Determinant FormulaCoordinate GeometryTriangle VerticesArea Calculation
Determinant Formula
The determinant formula is a powerful tool in mathematics, especially in linear algebra, used for various applications including finding areas of polygons. When computing the area of a triangle with vertices in a coordinate plane, we can use a matrix determinant to simplify the calculation. Here, the determinant helps express the area of the triangle without having to draw it or perform complex geometric operations.
The formula for the area using the determinant is:
The formula for the area using the determinant is:
- Take the coordinates of the vertices:
- (0, 0),
- (a_1, a_2),
- (b_1, b_2).
- Substitute the coordinates into the area formula:
- Use: \[ A = \frac{1}{2} \left| a_1b_2 - b_1a_2 \right| \]
Coordinate Geometry
Coordinate geometry, also known as analytic geometry, involves the use of coordinates to represent geometric shapes. This approach enables mathematicians and students alike to tackle geometric problems using algebra. By plotting points on a plane using their coordinates, geometric figures such as triangles can be analyzed with precision and ease.
In the case of our triangle:
In the case of our triangle:
- We consider its vertices
- (0, 0),
- (a_1, a_2),
- (b_1, b_2).
- These points are plotted on the xy-plane, forming the vertices of the triangle.
- Using these coordinates within the coordinate system allows us to apply algebraic formulas to find measurements, like the area.
Triangle Vertices
Vertices are the backbone of any polygon, marking the precise points where two sides of the shape meet. In a triangle, these are considered as its corners, anchoring the shape in space. Understanding the importance of vertices helps in visualizing and calculating characteristics like area and perimeter.
In our example, the triangle's vertices are given as:
In our example, the triangle's vertices are given as:
- (0, 0): This is the origin, simplifying calculations as one point is at an axis intercept.
- (a_1, a_2): This point represents a vertex spaced somewhere within the plane, described by its coordinates.
- (b_1, b_2): Similarly, this vertex complements the other two, completing the triangle.
Area Calculation
Calculating the area of a triangle using its vertices in the coordinate plane provides a reliable and quick method. The determinant formula, as discussed earlier, helps in getting the area without complex geometric construction. Here’s how the calculation unfolds:
- Identify the triangle's vertices as coordinates.
- Plug these into the determinant-based area formula.
- Use the formula: \[ A = \frac{1}{2} \left| a_1b_2 - b_1a_2 \right| \]
- This formula simplifies to give the absolute area, ensuring no negative values are considered, since the area is always a positive quantity.
Other exercises in this chapter
Problem 42
Find the centers and radii of the spheres in Exercises \(41-44\) $$ \left(x+\frac{1}{2}\right)^{2}+\left(y+\frac{1}{2}\right)^{2}+\left(z+\frac{1}{2}\right)^{2}
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Sketch the surfaces in Exercises \(13-76\) $$ y^{2}-x^{2}=z $$
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Work along a line Find the work done by a force \(\mathbf{F}=5 \mathbf{i}\) (magnitude 5 \(\mathrm{N}\) ) in moving an object along the line from the origin to
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In Exercises 39–44, find the distance from the point to the plane. $$ (0,-1,0), \quad 2 x+y+2 z=4 $$
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