Problem 31 - Precalculus Mathematics for Calculus

Question

Find an equation of the line that satisfies the given conditions. Through \((1,-6) ; \quad\) parallel to the line \(x+2 y=6\)

Step-by-Step Solution

Answer

The line's equation is \(y = -\frac{1}{2}x - \frac{11}{2}\).

Step 1: Identify the Slope of the Given Line

First, rewrite the equation of the given line to identify its slope. The line is given as \(x + 2y = 6\). Solve for \(y\) to put it in slope-intercept form \(y = mx + b\), where \(m\) is the slope. Start by isolating \(y\) on one side: \(2y = -x + 6\), then divide by 2: \(y = -\frac{1}{2}x + 3\). The slope \(m\) of the given line is \(-\frac{1}{2}\).

Step 2: Apply the Concept of Parallel Lines

Lines that are parallel have the same slope. Thus, the line we are trying to find will also have the slope \(m = -\frac{1}{2}\).

Step 3: Use the Point-Slope Form Equation

With a point \((x_1, y_1)\) given as \((1, -6)\) and a slope \(m = -\frac{1}{2}\), use the point-slope form of the equation of a line, which is \(y - y_1 = m(x - x_1)\). Substitute \(x_1 = 1\), \(y_1 = -6\), and \(m = -\frac{1}{2}\) into this equation: \(y + 6 = -\frac{1}{2}(x - 1)\).

Step 4: Solve for y to find the Equation

Simplify the equation from Step 3 to the slope-intercept form \(y = mx + b\). Start by distributing the slope on the right: \(y + 6 = -\frac{1}{2}x + \frac{1}{2}\). Subtract 6 from both sides to solve for \(y\): \(y = -\frac{1}{2}x + \frac{1}{2} - 6\), which simplifies to \(y = -\frac{1}{2}x - \frac{11}{2}\). Thus, the equation of the line is \(y = -\frac{1}{2}x - \frac{11}{2}\).

Key Concepts

Slope

The slope of a line is a fundamental concept in understanding linear equations and line characteristics. It tells us how steep the line is and the direction in which it rises or falls. We represent slope by the letter \( m \).
The formula for calculating the slope between two points \((x_1, y_1)\) and \((x_2, y_2)\) on a line is:

• \( m = \frac{y_2 - y_1}{x_2 - x_1} \)

This formula shows the change in \( y \) values (the rise) over the change in \( x \) values (the run).
For the line \( y = -\frac{1}{2}x + 3 \), the slope is \( -\frac{1}{2} \), which indicates that for every unit the line moves to the right (positive direction), it falls down by half a unit. Understanding the slope gives us insight into how the line behaves graphically.
When learning about slopes, remember:

• A positive slope means the line rises from left to right.
• A negative slope means the line falls from left to right.
• A zero slope indicates a horizontal line.
• An undefined slope refers to a vertical line.

Parallel Lines

Parallel lines are lines in the same plane that never meet. They have the same direction and never intersect, regardless of how far they are extended. An essential characteristic of parallel lines is that they have the same slope.
When asked to find a line parallel to another, you simply need to ensure the slopes match. In our original exercise, the line \( x + 2y = 6 \) can be rewritten in slope-intercept form as \( y = -\frac{1}{2}x + 3 \).
This equation tells us that the slope is \( -\frac{1}{2} \).
Since the new line must be parallel to it, its slope will also be \( -\frac{1}{2} \).
This is a key property we utilize when working with parallel line equations. Ensuring that both lines have the same slope guarantees that they will be parallel.

Point-Slope Form

The point-slope form of a line is a way to write the equation of a line when you know a point on the line and its slope. This form is useful for constructing the equation quickly without much algebra. The formula is:

• \( y - y_1 = m(x - x_1) \)

Here, \((x_1, y_1)\) represents a specific point on the line, and \( m \) is the slope of the line.
Using point-slope form helps one quickly build an equation when a problem provides a point and the need for parallelism (matching slope).
In our original problem, we used point \((1, -6)\) and slope \( -\frac{1}{2} \).
Plugging these into the formula:

• \( y - (-6) = -\frac{1}{2}(x - 1) \)

We simplify it to build a complete line equation, which provides clarity on how the line moves through the particular point with the given slope. Understanding the point-slope form is vital for quickly modeling lines and making calculations simpler for many geometry and algebra problems.