Problem 34 - College Algebra

Question

For the following exercises, given each set of information, find a linear equation satisfying the conditions, if possible. Passes through (-2,8) and (4,6)

Step-by-Step Solution

Answer

The linear equation is \(y = -\frac{1}{3}x + \frac{22}{3}\).

Step 1: Identify the Components

To find the linear equation, we need to identify the slope and the y-intercept. We start by identifying the two points given: \((-2, 8)\) and \((4, 6)\).

Step 2: Calculate the Slope

Use the slope formula, \(m = \frac{y_2 - y_1}{x_2 - x_1}\), where \((x_1, y_1) = (-2, 8)\) and \((x_2, y_2) = (4, 6)\). Substituting these values gives:\[m = \frac{6 - 8}{4 - (-2)} = \frac{-2}{6} = -\frac{1}{3}\]

Step 3: Use Point-Slope Form

With the slope \(m = -\frac{1}{3}\) and a point, use the point-slope form equation, \(y - y_1 = m(x - x_1)\) with the point \((-2, 8)\). This gives:\[y - 8 = -\frac{1}{3}(x + 2)\]

Step 4: Simplify to Slope-Intercept Form

Distribute the slope and simplify to get the equation in slope-intercept form, \(y = mx + b\):\[y - 8 = -\frac{1}{3}x - \frac{2}{3}\]Add 8 to both sides to isolate \(y\):\[y = -\frac{1}{3}x + \frac{24}{3} - \frac{2}{3}\]So, \[y = -\frac{1}{3}x + \frac{22}{3}\].

Step 5: Verify the Equation (Optional)

To ensure the equation is correct, check if both points satisfy it:\(-2, 8\) and \(4, 6\).For \((x, y) = (-2, 8)\):\[y = -\frac{1}{3}(-2) + \frac{22}{3} = \frac{2}{3} + \frac{22}{3} = \frac{24}{3} = 8\]For \((x, y) = (4, 6)\):\[y = -\frac{1}{3}(4) + \frac{22}{3} = -\frac{4}{3} + \frac{22}{3} = \frac{18}{3} = 6\] Both points satisfy the equation.

Key Concepts

Point-Slope Form

The point-slope form of a linear equation is a very useful format when you know a point on the line and the slope. This form is expressed as \( y - y_1 = m(x - x_1) \), where \((x_1, y_1)\) is a point on the line and \(m\) is the slope of that line.

• Start by identifying a specific point \((x_1, y_1)\) that the line passes through. In the given problem, you could choose either of your provided points, such as \((-2, 8)\).
• The slope \(m\) has already been calculated as \(-\frac{1}{3}\), allowing us to substitute these values into the equation structure.

This makes the point-slope form very handy for creating linear equations directly from basic information. It's a bridge from data to equation, keeping things manageable. When starting with point-slope form, the algebraic handling is both flexible and direct, guiding you through to later forms like slope-intercept when you choose to simplify further.

Slope-Intercept Form

Slope-intercept form is one of the most familiar forms of linear equations, making it extremely easy to work with once you have it simplified. This form is useful because it tells you directly both the slope and the intercept with the y-axis. Its equation is written as \( y = mx + b \), where \(m\) is the slope and \(b\) is the y-intercept.

• This form emphasizes how changes in \(x\) affect the value of \(y\), since the slope defines this rate of change.
• The y-intercept \(b\) is the point where the line crosses the y-axis, a critical feature for graphing.

Converting from point-slope to slope-intercept involves simplifying the equation, as shown in the step-by-step solution. After distributing the slope through the terms, you isolate \(y\) to make the equation easy to use and interpret at a glance. This conversion makes your equation more straightforward for solving new problems or discussing linear relationships.

Slope Calculation

Understanding how to calculate the slope is fundamental to working with linear equations effectively. The slope (often represented as \(m\)) of a line is calculated using the formula \( m = \frac{y_2 - y_1}{x_2 - x_1} \), which measures how steep the line is.

• The values \((x_1, y_1)\) and \((x_2, y_2)\) are points that lie on the line.
• A positive slope means the line inclines upwards as you move along, while a negative slope shows it is declining.
• A slope of zero implies a perfectly horizontal line.

In our example, plugging in the points \((-2, 8)\) and \((4, 6)\) into the slope formula showed us how the y-values decrease as the x-values increase, resulting in a negative slope of \(-\frac{1}{3}\). This calculation is critical, as it gives the line's exact inclination, deciding the entire equation form. With this foundational grasp of the slope, you are better equipped to craft accurate equations and understand the behavior of linear graphs.