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Chapter 8: Problem 59

three points, \(\left(x_{1}, y_{1}\right),\left(x_{2}, y_{2}\right),\) and \(\left(x_{3}, y_{3}\right),\) are collinear if and only if \(\left|\begin{array}{lll}x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1 \\ x_{3} & y_{3} & 1\end{array}\right|=0\) Apply determinants to determine whether the points, \((-2,-1),(1,5),\) and \((3,9),\) are collinear.

Short Answer

Expert verified
The points \((-2,-1), (1,5), (3,9)\) are collinear since the determinant is zero.

Step by step solution

01

Writing the Points in a Determinant

To determine if the points \((-2,-1), (1,5), (3,9)\) are collinear, form a 3x3 matrix using the given points:\[\begin{vmatrix} -2 & -1 & 1 \1 & 5 & 1 \3 & 9 & 1 \end{vmatrix}\] We need to compute this determinant and check if it equals zero.
02

Calculate the Determinant

Use the formula for a 3x3 determinant:\[det = a(ei-fh) - b(di-fg) + c(dh-eg)\]For the matrix:\[\begin{vmatrix} -2 & -1 & 1 \1 & 5 & 1 \3 & 9 & 1 \end{vmatrix}\]Substitute the values: \(a = -2, b = -1, c = 1, d = 1, e = 5, f = 1, g = 3, h = 9, i = 1\).Then calculate using:\[-2(5 \cdot 1 - 1 \cdot 9) - (-1)(1 \cdot 1 - 1 \cdot 3) + 1(1 \cdot 9 - 5 \cdot 3)\]
03

Simplifying the Determinant Expression

Calculate the value:\[det = -2(5 - 9) + 1(1 - 3) + 1(9 - 15)\]\[det = -2(-4) + 1(-2) + 1(-6)\]\[det = 8 - 2 - 6 = 0\]The value of the determinant is zero.
04

Conclusion on Collinearity

Since the determinant of the matrix is zero, it implies that the points \((-2,-1), (1,5), (3,9)\) are collinear. This means they lie on a straight line.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Determinants
Determinants play a crucial role in linear algebra, especially when it comes to determining specific properties of points or vectors in geometry. A determinant is a scalar value that can be computed from the elements of a square matrix. It provides a lot of information about the matrix, including whether a set of points is collinear or the matrix is invertible.

For a 2x2 matrix, the determinant is computed as \((ad - bc)\). For a 3x3 matrix, the calculation is a bit more complex. The formula is given as:
  • \((a(ei - fh) - b(di - fg) + c(dh - eg))\)
Where each letter corresponds to a specific element in the 3x3 matrix placed in row-major order. Calculating the determinant can seem daunting at first, but with practice, finding the value becomes straightforward.

The outcome of the determinant helps us understand properties such as collinearity; if the determinant is zero, it indicates that the points fall on a single straight line, making them collinear.
3x3 Matrix
A 3x3 matrix is an arrangement of numbers in three rows and three columns. Each element in the matrix refers to a number located at the intersection of a specific row and column.

In the context of the problem, the points given, \((-2, -1), (1, 5), (3, 9)\), are used to form a 3x3 matrix with the format:
  • First row from the first point: \((-2, -1, 1)\)
  • Second row from the second point: \((1, 5, 1)\)
  • Third row from the third point: \((3, 9, 1)\)
The column filled with '1's is essential. It ensures the computation is aligned with the rule for determining collinearity.

Mathematically, working with 3x3 matrices involves various operations, like adding, subtracting, multiplying, and finding determinants. Each operation has its role, but determinants are particularly significant when examining geometric properties.
Geometry
Geometry focuses on understanding shapes, sizes, and the properties of space. One critical concept in geometry is collinearity, which refers to points lying on a single straight line. This property can be verified using determinants.

Given three points in the plane, checking if they are collinear involves creating a 3x3 matrix and computing its determinant. If the result is zero, the points are collinear; this means they align perfectly in a geometric sense, without deviation from a straight path.

Understanding the concept of collinearity helps in various real-world applications, such as computer graphics, technical drawing, and modeling geometric congruence. Geometric principles, when coupled with linear algebra tools like determinants, allow us to solve complex problems with a neat, algebraic approach.

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Most popular questions from this chapter

In Exercises 85 and \(86,\) for the system of linear inequalities, assume \(a, b, c,\) and \(d\) are real numbers. $$\begin{array}{l}x \geq a \\\x < b \\\y > c \\\y \leq d\end{array}$$ What will the solution be if \(a>b\) and \(c>d ?\)

Explain the mistake that is made. Solve the system of equations using an augmented matrix. $$ \begin{array}{l} y-x+z=2 \\ x-2 z+y=-3 \\ x+y+z=6 \end{array} $$ Solution: Step 1: Write as an augmented matrix. \(\left[\begin{array}{rrr|r}1 & -1 & 1 & 2 \\ 1 & -2 & 1 & -3 \\ 1 & 1 & 1 & 6\end{array}\right]\) Step 2: Reduce the matrix using Gaussian elimination. \(\left[\begin{array}{rrr|r}1 & -1 & 1 & 2 \\ 0 & 1 & 0 & 5 \\ 0 & 0 & 0 & -6\end{array}\right]\) Step 3: Identify the solution. Row 3 is inconsistent, so there is no solution. This is incorrect. The correct answer is \(x=1, y=2, z=3\) What mistake was made?

Explain the mistake that is made. Perform the indicated row operations on the matrix. \(\left[\begin{array}{cccc}1 & -1 & 1 & 2 \\ 2 & -3 & 1 & 4 \\ 3 & 1 & 2 & -6\end{array}\right]\) a. \(R_{2}-2 R_{1} \rightarrow R_{2}\) b. \(R_{3}-3 R_{1} \rightarrow R_{3}\) Solution: a. \(\left[\begin{array}{rrr|r}1 & -1 & 1 & 2 \\ 0 & -3 & 1 & 4 \\ 3 & 1 & 2 & -6\end{array}\right]\) b. \(\left[\begin{array}{rrr|r}1 & -1 & 1 & 2 \\ 2 & -3 & 1 & 4 \\ 0 & 1 & 2 & -6\end{array}\right]\) This is incorrect. What mistake was made?

You are asked to model a set of three points with a quadratic function \(y=a x^{2}+b x+c\) and determine the quadratic function. a. Set up a system of equations, use a graphing utility or graphing calculator to solve the system by entering the coefficients of the augmented matrix. b. Use the graphing calculator commands \([\text { STAT }]\) QuadReg to model the data using a quadratic function. Round your answers to two decimal places. $$(-9,20),(2,-18),(11,16)$$

In calculus, when solving systems of linear differential equations with initial conditions, the solution of a system of linear equations is required. solve each system of equations. $$\begin{aligned} 2 c_{1}+2 c_{2}+2 c_{3} &=0 \\ 2 c_{1} &-2 c_{3}=2 \\ c_{1}-c_{2}+c_{3} &=6 \end{aligned}$$
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