Problem 49 - A Graphical Approach to College Algebra

Question

Find the equation of the line satisfying the given conditions, giving it in slope-intercept form if possible. Through \((-5,8),\) parallel to \(y=-0.2 x+6\)

Step-by-Step Solution

Answer

The equation of the line is \(y = -0.2x + 7\) in slope-intercept form.

Step 1: Identify the Slope from the Given Line

First, we need to identify the slope of the line given by the equation \(y = -0.2x + 6\). In the slope-intercept form \(y = mx + b\), \(m\) represents the slope. For this line, \(m = -0.2\).

Step 2: Use the Slope for the New Line

Since the new line is parallel to the given line, it shares the same slope. Therefore, the slope \(m\) of the new line is also \(-0.2\).

Step 3: Use the Point-Slope Form

We will use the point-slope form of a line equation, given by \(y - y_1 = m(x - x_1)\), where \((x_1, y_1)\) is the given point \((-5, 8)\) and \(m = -0.2\). Substitute these into the equation: \[ y - 8 = -0.2(x + 5) \]

Step 4: Simplify the Equation to Slope-Intercept Form

Expand and simplify the equation from the previous step to get it into slope-intercept form \(y = mx + b\): First, distribute the \(-0.2\): \[ y - 8 = -0.2x - 1 \]Next, solve for \(y\) by adding 8 to both sides:\[ y = -0.2x - 1 + 8 \]Simplify the equation:\[ y = -0.2x + 7 \]

Step 5: Verify the Equation

Ensure that the equation meets the conditions: The line passes through \((-5, 8)\) and has the same slope as the given line. Since we've used the point \((-5, 8)\) during the substitution in the point-slope form, the equation is indeed valid.

Key Concepts

Slope-Intercept Form

The slope-intercept form is one of the most common ways to represent the equation of a straight line in algebra. It is written as \( y = mx + b \), where:

• \( m \) represents the slope of the line, which tells us how steep the line is or the rate of change.
• \( b \) is the y-intercept, which is the point where the line crosses the y-axis.

This form is particularly useful because it provides a clear picture of both the slope and y-intercept at a glance, making it easier to graph the line on a coordinate plane. It defines the line's direction and position with just these two constants, allowing quick adjustments or predictions about the line's behavior. Whenever you have the slope and intercept, you can construct the entire line!For instance, in our problem, transforming the equation into \( y = -0.2x + 7 \) helps us understand that the line falls left to right at a steepness of \(-0.2\) and crosses the y-axis at (0,7). This makes graphing and understanding the line's behavior straightforward.

Parallel Lines

Parallel lines are special because they never meet each other. They run side by side, maintaining the same distance from each other at all points, like railroad tracks. In algebra, for two lines to be parallel, they must have the same slope.

• If two lines have the same slope, \( m_1 = m_2 \), they are parallel.
• This means their steepness is identical.

For the exercise, given that the line is parallel to \( y = -0.2x + 6 \), it must have the same slope of \(-0.2\). This condition ensures the new line won't touch or intersect the original line, maintaining a consistent distance throughout its length. To check if two lines are parallel, compare their slopes: if they are identical, the lines are indeed parallel.

Point-Slope Form

The point-slope form is a versatile way to express the equation of a line whenever 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 given point on the line.
• \( m \) is the slope of the line.

The main benefit of this form is that it directly incorporates a known point, simplifying the process of finding the exact line equation when a specific point is involved. It is also a stepping stone to converting to other forms, like slope-intercept form, for further simplification or graphing. In our exercise, using point-slope form \( y - 8 = -0.2(x + 5) \) allowed us to easily incorporate the point \((-5, 8)\) and the slope \(-0.2\), directly leading to a streamlined way to solve and eventually convert to slope-intercept form. This approach highlights how adaptable the point-slope form can be when connecting known values for precise equation setup.