Problem 5 - Elementary Algebra

Question

Find the equation of the line that contains the given point and has the given slope. Express equations in the form \(A x+B y=C\), where \(A, B\), and \(C\) are integers. (Objective 1a) $$(-4,8), m=-\frac{1}{3}$$

Step-by-Step Solution

Answer

The line equation is \( x + 3y = 20 \).

Step 1: Use Point-Slope Form

Begin by using the point-slope form of a line equation: \( y - y_1 = m(x - x_1) \)where - \((x_1, y_1)\) is the given point \((-4, 8)\), - \(m\) is the slope \(-\frac{1}{3}\). Substitute the given values to get the equation:\[ y - 8 = -\frac{1}{3}(x + 4) \]

Step 2: Simplify the Equation

Distribute the slope on the right-hand side of the equation:\[ y - 8 = -\frac{1}{3}x - \frac{4}{3} \]Add 8 to both sides to move the constant term on the left:\[ y = -\frac{1}{3}x - \frac{4}{3} + 8 \]

Step 3: Convert to Slope-Intercept Form

Convert the constant 8 to a fraction with a denominator of 3:\[ y = -\frac{1}{3}x - \frac{4}{3} + \frac{24}{3} \]Combine the fractions on the right:\[ y = -\frac{1}{3}x + \frac{20}{3} \]

Step 4: Convert to Standard Form

To convert to standard form \(Ax + By = C\), clear the fractions by multiplying every term by 3:\[ 3y = -x + 20 \]Rearrange to get all terms on one side:\[ x + 3y = 20 \]

Key Concepts

Point-Slope Form

The point-slope form is a linear equation format that is very useful for describing a line with a known slope and a point through which the line passes. The basic form is given by: \( y - y_1 = m(x - x_1) \). Here, \( (x_1, y_1) \) is the specific point on the line, and \( m \) represents the slope of the line.
In the problem given, you need to plug in the values of the slope, \( m = -\frac{1}{3} \), and the point \( (-4, 8) \) into the equation. This yields \( y - 8 = -\frac{1}{3}(x + 4) \). By using this form, finding the equation of a line becomes much simpler, especially when you have the slope and a point directly available.

• Benefits: Quickly forms an equation with given slope and point.
• Easy to derive into other forms like slope-intercept or standard form.

By simply inserting the given values, you create a base equation that can be further manipulated to fit other useful forms.

Slope-Intercept Form

The slope-intercept form is one of the most popular ways to describe a line. It is expressed as \( y = mx + b \), where \( m \) is the slope, and \( b \) is the y-intercept—the point where the line crosses the y-axis. Switching from point-slope form to slope-intercept form involves a bit of simplification.
In the step-by-step solution, the equation \( y - 8 = -\frac{1}{3}x - \frac{4}{3} \) is simplified by moving terms to get \( y = -\frac{1}{3}x + \frac{20}{3} \). This rearrangement confirms the slope is \(-\frac{1}{3}\) and reveals that \( b = \frac{20}{3} \).

• Benefits: Provides clear visual clues about the slope and position of the line on a graph.
• Necessary for graphing lines and evaluating linearity.

Reducing to this form is important, especially when interpreting graphs or comparing line equations.

Standard Form

The standard form for a linear equation is \( Ax + By = C \), where \( A \), \( B \), and \( C \) are integers, making it a versatile format. This form is particularly useful when solving systems of equations or needing integer coefficients.
In transitioning from slope-intercept form to the standard form, we take \( y = -\frac{1}{3}x + \frac{20}{3} \) and clear the fractions by multiplying through by 3, resulting in \( 3y = -x + 20 \). By rearranging to \( x + 3y = 20 \), we achieve the desired integer-coefficient format.

• Benefits: Compatible with various algebraic operations.
• Helps in quickly comparing and contrasting different linear equations.

This form helps when arranging and solving real-world problems where integer solutions are essential, eliminating fractions in the process.