Problem 30 - College Algebra and Calculus: An Applied Approach

Question

Find an equation of the line that passes through the point and has the indicated slope. Then sketch the line. Point \(\quad\) Slope \((-8,3)\) \(m=-\frac{1}{2}\)

Step-by-Step Solution

Answer

The equation of the line is \(y = -\frac{1}{2}x - 1\).

Step 1: Apply the Point-Slope Formula

The point-slope formula for the equation of a line is given by \(y - y_1 = m*(x - x_1)\) where \((x_1, y_1)\) are the coordinates of a point on the line and \(m\) is the slope of the line. Substituting the given values, we have \(y - 3 = -\frac{1}{2} *(x + 8)\).

Step 2: Simplify to the Slope-Intercept Form

Rearrange the previous equation to obtain the slope-intercept form \(y = mx + b\). We distribute the slope, get rid of the parentheses and solve for \(y\), we get \(y = -\frac{1}{2}x - 1\).

Step 3: Sketch the line

Using this equation, plot the y-intercept at \(-1\) on the y-axis. Then from that point, use the slope \(-\frac{1}{2}\) to find other points. This means from the y-intercept, move 1 unit down (because the slope is negative) and 2 units to the right (because the denominator of the slope fraction is 2). Then, draw a line through these points.

Key Concepts

Point-Slope Formula

The point-slope formula is a straightforward method to find the equation of a line when you know a point on the line and its slope. This formula is helpful because it doesn't require the y-intercept to determine the equation.To use this formula, remember:

• The formula itself is: \(y - y_1 = m(x - x_1)\)
• \((x_1, y_1)\) are the coordinates of the given point.
• \(m\) is the slope of the line.

By plugging in our known values, we convert this into a specific equation for our situation. For example, if a point is \((-8, 3)\) and the slope is \(-\frac{1}{2}\), you insert these into the formula as follows: \(y - 3 = -\frac{1}{2}(x + 8)\). This represents the entire equation of our line in terms of the variables \(x\) and \(y\). As simple as this approach is, it sets the foundation for transforming the equation into other forms.

Slope-Intercept Form

After finding the equation using the point-slope formula, it's common to rearrange it into the slope-intercept form. This form gives more intuitive information about the line, like where it crosses the y-axis and highlights the slope again.The slope-intercept form is written as:

• \(y = mx + b\)
• \(m\) is still the slope.
• \(b\) is the y-intercept, meaning it's where the line intersects the y-axis.

To convert from point-slope to slope-intercept form, distribute the slope through the parentheses and solve for \(y\). For our example, \(y - 3 = -\frac{1}{2}(x + 8)\), distributing gives \(y - 3 = -\frac{1}{2}x - 4\). By shifting terms, we simplify to \(y = -\frac{1}{2}x - 1\), making it clear that the slope is \(-\frac{1}{2}\) and the y-intercept is \(-1\). Once you have this form, drawing or interpreting the graph becomes straightforward.

Linear Equations

Linear equations are foundational elements of algebra, representing straight lines when graphed on a coordinate plane. They express the relationship between two variables, usually using the standard forms we've discussed: point-slope and slope-intercept.Here's what makes a linear equation stand out:

• Its graph is always a straight line.
• It can be manipulated into various forms for different uses, like calculating slopes or intercepts.
• Common in real-life applications to represent constant relationships where one variable changes at a consistent rate relative to another.

For example, looking at the equation \(y = -\frac{1}{2}x - 1\), we interpret this as a line that descends since the slope \(-\frac{1}{2}\) is negative. Using these equations effectively allows prediction and exploration of relationships in algebra, filling crucial roles both in simple problems, like finding equation forms, and larger mathematical modeling tasks.