Problem 33 - College Algebra with Corequisite Support

Question

For the following exercises, given each set of information, find a linear equation satisfying the conditions, if possible. Passes through (-1,4) and (5,2)

Step-by-Step Solution

Answer

The linear equation is \(y = -\frac{1}{3}x + \frac{11}{3}\).

Step 1: Determine the Slope

To find the slope (\(m\)), we use the formula \[m = \frac{y_2 - y_1}{x_2 - x_1}\]where \((x_1, y_1) = (-1, 4)\) and \((x_2, y_2) = (5, 2)\).Substituting these points into the formula, we get:\[m = \frac{2 - 4}{5 - (-1)} = \frac{-2}{6} = -\frac{1}{3}\]

Step 2: Use Point-Slope Form

The point-slope form of a line equation is given by:\[y - y_1 = m(x - x_1)\]where \((x_1, y_1)\) is either one of the points on the line.Let's use the point \((-1, 4)\) and the slope \(m = -\frac{1}{3}\).Plug these values into the formula:\[y - 4 = -\frac{1}{3}(x + 1)\]

Step 3: Simplify to Slope-Intercept Form

Now, we simplify the equation from Step 2 to the slope-intercept form \(y = mx + b\):\[y - 4 = -\frac{1}{3}x - \frac{1}{3}\]Add 4 to both sides to isolate \(y\):\[y = -\frac{1}{3}x + \frac{11}{3}\]

Key Concepts

Slope

The concept of slope is crucial in understanding linear equations. The slope, often represented by the letter \(m\), measures how steep a line is, or the rate at which \(y\) changes with respect to \(x\). To calculate the slope between two points \((x_1, y_1)\) and \((x_2, y_2)\), you use the formula:

• \[m = \frac{y_2 - y_1}{x_2 - x_1}\]

This formula represents the "rise" over the "run," or how much \(y\) changes for each change in \(x\).For example, if the points given are \((-1,4)\) and \((5,2)\), substitute these values into the formula to get:

• \[m = \frac{2 - 4}{5 - (-1)} = \frac{-2}{6} = -\frac{1}{3}\]

This tells us that for every 3 units we move to the right along the x-axis, the line falls 1 unit down. The slope is negative, indicating the line is descending as we move from left to right.

Point-Slope Form

Once the slope is known, we can use the point-slope form of a linear equation. This form is especially useful when you know a point on the line and its slope. The point-slope form is expressed as:

• \[y - y_1 = m(x - x_1)\]

Here, \((x_1, y_1)\) is a specific point on the line, and \(m\) is the slope.Choose one of the given points, like \((-1, 4)\), and use the slope calculated earlier, which is \(-\frac{1}{3}\). Substituting these into the formula gives:

• \[y - 4 = -\frac{1}{3}(x + 1)\]

This equation shows the relationship between \(x\) and \(y\) for any point lying on the line. It is a flexible way to describe a line before transitioning to other forms.

Slope-Intercept Form

After using the point-slope form, it often makes sense to convert the equation into the slope-intercept form for simplicity and clarity. The slope-intercept form is written as:

• \[y = mx + b\]

Here, \(m\) remains the slope, and \(b\) is the y-intercept, the point where the line crosses the y-axis.Starting with the earlier point-slope equation \(y - 4 = -\frac{1}{3}(x + 1)\), we aim to express \(y\) explicitly:

• First, expand: \[y - 4 = -\frac{1}{3}x - \frac{1}{3}\]
• Then, add 4 to both sides to solve for \(y\): \[y = -\frac{1}{3}x + \frac{11}{3}\]

The result, \(y = -\frac{1}{3}x + \frac{11}{3}\), expresses the line in slope-intercept form. It provides an easy way to identify the slope \(-\frac{1}{3}\) and y-intercept \(\frac{11}{3}\), making it straightforward to graph or understand the line's behavior.