Problem 54 - College Algebra

Question

Determinants are used to write an equation of a line passing. through two points. An equation of the line passing through the distinct points \(\left(x_{1}, y_{1}\right)\) and \(\left(x_{2}, y_{2}\right)\) is given by $$\left|\begin{array}{lll}x & y & 1 \\\x_{1} & y_{1} & 1 \\\x_{2} & y_{2} & 1\end{array}\right|=0$$ Use the determinant to write an equation of the line passing through \((-1,3)\) and \((2,4) .\) Then expand the determinant, expressing the line's equation in slope-intercept form.

Step-by-Step Solution

Answer

The equation of the line through points (-1,3) and (2,4) using the determinant approach is \(y =\frac{1}{3}x +\frac{10}{3}\)

Step 1: Form the Determinant

Form the determinant using the two points (-1,3) and (2,4), it will look like this: \\[ \left|\begin{array}{ccc} x & y & 1 \\ -1 & 3 & 1 \\ 2 & 4 & 1 \end{array}\right| = 0 \]

Step 2: Expand the Determinant

Expanding the determinant using the Rule of Sarrus: \\ \( [x (3*1 - 4*1) - y (-1*1 - 2*1) + 1*(-1*4 - 2*3)] = 0 \) \\ Simplifying, we get \(x - 3y - 10 = 0 \)

Step 3: Rewrite in Slope-Intercept Form

Rearrange the equation to get it into slope-intercept form, which is \(y = mx + c\), where \(m\) represents the slope and \(c\) represents the y-intercept. \\ The equation becomes \(y =\frac{1}{3}x +\frac{10}{3}\)

Key Concepts

Equation of a line

When you want to find the equation of a line that goes through two points, there's a special tool you can use: determinants. A determinant is like a special box that helps us mathematically check certain properties. We borrow this concept from linear algebra to help us find lines that connect points.
To find the line equation going through two points \((x_1, y_1)\) and \((x_2, y_2)\), we use this determinant setup:
\[\left|\begin{array}{ccc} x & y & 1 \ x_{1} & y_{1} & 1 \ x_{2} & y_{2} & 1 \end{array}\right| = 0\]
This equation arises from the need to ensure that both points satisfy the linear equation we are about to discover. Solve this determinant equation by expanding it, and you get the equation of the line.

Slope-intercept form

A line on a graph can be described by a simple formula called the slope-intercept form:
\(y = mx + c\).
This formula is great because it easily tells us two things:

• \(m\), the slope, which shows how steep the line is.
• \(c\), the y-intercept, or where the line crosses the y-axis.

To convert an equation into slope-intercept form:

• Rearrange the terms so \(y\) is by itself on one side of the equation.
• The result will be an easy-to-read formula showing the line's characteristics.

In the example we have, the equation \(x - 3y - 10 = 0\) can be changed to
\(y = \frac{1}{3}x + \frac{10}{3}\), showing a slope \(m = \frac{1}{3}\) and intercept \(c = \frac{10}{3}\).

Rule of Sarrus

The Rule of Sarrus is a quick shortcut to calculate the determinant for a 3x3 matrix. This rule helps us to find determinants efficiently without getting lost in complex arithmetic.
To use this rule, write the first two columns of the 3x3 matrix again to the right of the matrix.

• For the diagonal (\(\searrow\)) going downwards to the right, multiply the numbers along these lines and add them up.
• For the crossed diagonal (\(warrow\)) going upwards to the right, multiply those numbers and subtract the sum from the previous result.

In the example for the determinant:\[\left|\begin{array}{ccc} x & y & 1 \ -1 & 3 & 1 \ 2 & 4 & 1 \end{array}\right|\]The Rule of Sarrus leads us to:\[x(3 \times 1 - 4 \times 1) - y(-1 \times 1 - 2 \times 1) + 1(-1 \times 4 - 2 \times 3)\]Which simplifies to the linear equation of the line. It's a straightforward yet powerful concept!

Expanding determinants

Expanding determinants means carefully "opening up" these boxes (our determinants) to find the equation inside.
The process involves picking a row or column and multiplying each element by a minor (determinant of the remaining elements after removing the current element's row and column), then adding and subtracting these results.

• You start by selecting a row or column, often the first.
• Apply the cofactor expansion, multiplying each element by its determinant and a sign factor based on its position.

In the given task, expanding means calculating:
\[x(3 \times 1 - 4 \times 1) - y(-1 \times 1 - 2 \times 1) + 1(-1 \times 4 - 2 \times 3)\]
Once you break down the pieces by expanding, you can simplify to form an equation that represents the line connecting your points. This process turns out an essential algebraic manipulation that clarifies not only the geometry but also helps in transitioning to other forms like the slope-intercept form.