Problem 42 - Precalculus with Limits

Question

In Exercises 39-44, use a determinant to determine whether the points are collinear. \((0, \frac{1}{2})\), \((2, -1)\), \((-4, \frac{7}{2})\)

Step-by-Step Solution

Answer

The points \((0, \frac{1}{2})\), \((2, -1)\), and \((-4, \frac{7}{2})\) are collinear.

Step 1: Creating the matrix

To start, a 3x3 matrix is created using the coordinates of the given points:\n\[ \left| \begin{array}{ccc} 1 & 0 & \frac{1}{2} \ 1 & 2 & -1 \ 1 & -4 & \frac{7}{2} \end{array} \right| \].

Step 2: Calculate the determinant

Then, calculate the determinant of the matrix. The determinant of a 3x3 matrix can be calculated using minors and cofactors.\nFor the given matrix, it is calculated as such: \[1*(2*\frac{7}{2} - -1*(-4)) - 2*(1*\frac{7}{2} - -1*1) + \frac{1}{2}*(1*-4 - 2*1)\] which simplifies to \[1*(7 - -4) - 2*(\frac{7}{2} - -1) + \frac{1}{2}*(-4 - 2)\].

Step 3: Simplification and final conclusion

Simplify it to get 11 - 2*(\frac{5}{2}) - 3 = 0.\nThe determinant of the matrix is zero, which means that the given points are collinear.

Key Concepts

Collinearity

Collinearity is a concept in geometry that describes the condition where three or more points lie on a single straight line. Understanding this idea can be essential in various mathematical tasks, such as in coordinate geometry.
To put it formally, if points are collinear, there is no distinct line that you can draw connecting any two of these points that would not pass through the other point(s) in the group.
So, how can we determine if three points (such as the ones given in the exercise:

• Point 1: \(0, \frac{1}{2}\)
• Point 2: \(2, -1\)
• Point 3: \(-4, \frac{7}{2}\)

) are collinear?
A practical method is to use determinants, as they can systematically establish collinearity by checking if the determinant of the matrix formed by these points is zero.

Matrix

In mathematics, a matrix is a rectangular array of numbers or other mathematical objects, arranged in rows and columns. Matrices are fundamental in many areas of math and are key players in solving linear equations, transformations, and more.
For our exercise, we construct a matrix from given points' coordinates. Each row in the matrix represents the coordinates of a point and an additional column of ones.
This configuration, where the last column consists of ones, helps when computing determinants to check for collinearity.

• For example, with points \(\left(0, \frac{1}{2}\right), (2, -1), \left(-4, \frac{7}{2}\right)\), the matrix looks like this:

\[\begin{bmatrix}1 & 0 & \frac{1}{2} \ 1 & 2 & -1 \ 1 & -4 & \frac{7}{2} \\end{bmatrix}\]
This representation helps us calculate the determinant easily, which facilitates finding whether the points are collinear or not.

3x3 Matrix

A 3x3 matrix consists of 3 rows and 3 columns, and it is a common form used in various calculations, particularly in linear algebra for problems involving three dimensions.
In the context of our problem, a 3x3 matrix aids in calculating the determinant to assess collinearity.
The determinant of a 3x3 matrix \(\begin{vmatrix} a & b & c \ d & e & f \ g & h & i \end{vmatrix}\) is calculated using the formula:

• \(a(ei - fh) - b(di - fg) + c(dh - eg)\)

Applying this method to the matrix derived from the points, we compute the determinant as shown in the solution steps. This determinant plays a crucial role: if it equals zero, the points are collinear.
This approach not only provides a systematic way to check collinearity but also demonstrates the power and applicability of matrices in solving geometric problems.