Problem 60 - Precalculus: Mathematical for Calculus

Question

(a) If three points lie on a line, what is the area of the "triangle" that they determine? Use the answer to this question, together with the determinant formula for the area of a triangle, to explain why the points \(\left(a_{1}, b_{1}\right)\) \(\left(a_{2}, b_{2}\right),\) and \(\left(a_{3}, b_{3}\right)\) are collinear if and only if $$\left|\begin{array}{lll} a_{1} & b_{1} & 1 \\ a_{2} & b_{2} & 1 \\ a_{3} & b_{3} & 1 \end{array}\right|=0$$ (b) Use a determinant to check whether cach set of points is collinear, Graph them to verify your answer. (i) \((-6,4),(2,10),(6,13)\) (ii) \((-5,10),(2,6),(15,-2)\)

Step-by-Step Solution

Answer

Neither set of points is collinear as their determinants are non-zero.

Step 1: Understanding Collinear Points and Triangle Area

When three points lie on a straight line, they are said to be collinear. In such cases, they do not form a triangle, and hence, the area is zero. To determine whether the points are collinear using the determinant formula, we can leverage the fact that if the area is zero, then the points are collinear.

Step 2: Determinant Formula for Collinearity

The area of a triangle formed by points \((a_1, b_1), (a_2, b_2), (a_3, b_3)\) can be calculated using the determinant: \[\text{Area} = \frac{1}{2} \left| \begin{array}{ccc}a_1 & b_1 & 1 \a_2 & b_2 & 1 \a_3 & b_3 & 1 \\end{array} \right|\]. The points are collinear if this determinant equals zero. Thus, if \(|A| = 0\), the points lie on a single line.

Step 3: Calculate Determinant for Given Points (i)

For points \((-6,4), (2,10), (6,13)\): Calculate the determinant: \[\left| \begin{array}{ccc}-6 & 4 & 1 \2 & 10 & 1 \6 & 13 & 1\end{array} \right| = -6\times(10 - 13) - 4\times(2 - 6) + 1\times(2\times13 - 10\times6)\]Calculate the expression: \[= -6(-3) - 4(-4) + 1(-32 + 20)\]\[= 18 + 16 - 12 = 22\]Since the determinant is not zero, the points are not collinear.

Step 4: Calculate Determinant for Given Points (ii)

For points \((-5,10), (2,6), (15,-2)\): Calculate the determinant: \[\left| \begin{array}{ccc}-5 & 10 & 1 \2 & 6 & 1 \15 & -2 & 1 \end{array} \right| = -5(6 + 2) - 10(2 - 15) + 1(2(-2) - 6(15))\]Calculate the expression: \[= -5(8) - 10(-13) + 1(-4 - 90)\]\[= -40 + 130 - 94 = -4\]Since the determinant is not zero, the points are not collinear.

Step 5: Conclusion: Graphical Verification

Graph both sets of points to visually verify collinearity. In this exercise, we find that neither -6,4; 2,10; 6,13 nor -5,10; 2,6; 15,-2 are collinear as their respective determinants are not zero, which is confirmed by a graphical check.

Key Concepts

Determinant Formula

The determinant formula is a powerful tool used in geometry to determine the area of a triangle from the vertices' coordinates. It also helps verify if points are collinear. This concept applies to points \( (a_1, b_1), (a_2, b_2), (a_3, b_3) \).

When these points form a triangle, the area can be computed using the determinant: \[ \text{Area} = \frac{1}{2} \left| \begin{array}{ccc} a_1 & b_1 & 1 \ a_2 & b_2 & 1 \ a_3 & b_3 & 1 \end{array} \right| \]. This determinant formula tells us that if the calculated value is zero, the area is zero. Thus, points are collinear as they lie on the same line and do not form a triangle.

This method is efficient for quickly checking collinearity, which otherwise could take more complex algebraic solutions. It relies intricately on linear algebra concepts and properties of matrices, which bring a precise approach to geometrical problems.

Area of a Triangle

In geometry, determining the area of a triangle is a fundamental operation. The area helps in understanding various relations of points in the plane.

To calculate the area of a triangle with vertices at \( (a_1, b_1), (a_2, b_2), (a_3, b_3) \), the determinant formula is quite straightforward: \[ \text{Area} = \frac{1}{2} \left| \begin{array}{ccc} a_1 & b_1 & 1 \ a_2 & b_2 & 1 \ a_3 & b_3 & 1 \end{array} \right| \].

This formula derives from the concept that for a triangle formed by three points, the area is half the absolute value of the determinant of a matrix formed by those points.

• Any three non-collinear points form a definite triangle with a specific area.
• Collinear points result in zero area, confirming that a triangle cannot be formed.

The determinant approach to finding the triangle's area is deeply linked to algebraic structures, cementing the connection between geometry and algebra.

Matrices in Geometry

Matrices are indispensable in various geometric computations. Their structure makes them suited for representing multidimensional data such as coordinate points.

In the context of collinear points, a matrix helps visualize problem data and apply algebraic operations like determinant calculations. For example, to determine whether three points, \( (a_1, b_1), (a_2, b_2), (a_3, b_3) \), are collinear, you form a specific matrix: \[ \begin{array}{ccc} a_1 & b_1 & 1 \ a_2 & b_2 & 1 \ a_3 & b_3 & 1 \end{array} \].

This matrix approach simplifies computational processes for areas and collinearity:

• Providing an algebraic method for spatial relations.
• Enabling use of linear algebra techniques, such as matrix manipulation and determinant evaluation.

Understanding matrices in geometry allows for solving complex spatial problems and provides insights into spatial relations and properties.