Problem 27 - Algebra and Trigonometry

Question

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

Step-by-Step Solution

Answer

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

Step 1: Understand the Equation of a Line

The standard form of a linear equation is given by \(Ax + By = C\). The given line equation is \(x + 2y = 6\). To find the slope, rewrite it in the slope-intercept form \(y = mx + b\).

Step 2: Convert to Slope-Intercept Form

Rewrite \(x + 2y = 6\) in slope-intercept form by solving for \(y\). Start by moving \(x\) to the right side: \(2y = -x + 6\). Divide by 2: \(y = -\frac{1}{2}x + 3\). The slope \(m\) of the line is \(-\frac{1}{2}\).

Step 3: Determine the Slope of the Parallel Line

Lines that are parallel have the same slope. The slope of the line parallel to the given line is also \(-\frac{1}{2}\).

Step 4: Use the Point-Slope Form to Find the Equation

Use the point-slope form of a line equation: \(y - y_1 = m(x - x_1)\), where \(m\) is the slope and \((x_1, y_1)\) is the point through which the line passes. Substitute \(m = -\frac{1}{2}\), \(x_1 = 1\), and \(y_1 = -6\) into the formula.

Step 5: Simplify the Equation

Substitute into the equation: \(y - (-6) = -\frac{1}{2}(x - 1)\). This simplifies to \(y + 6 = -\frac{1}{2}x + \frac{1}{2}\). Solve for \(y\) by subtracting 6 from both sides: \(y = -\frac{1}{2}x - \frac{11}{2}\). This is the equation of the line.

Step 6: Write the Final Equation

The final equation of the line, through the point \((1, -6)\) and parallel to \(x + 2y = 6\), is \(y = -\frac{1}{2}x - \frac{11}{2}\).

Key Concepts

Slope-Intercept Form

The slope-intercept form is a way of writing a linear equation. It is expressed as \( y = mx + b \), where \( m \) represents the slope of the line and \( b \) is the y-intercept. The y-intercept is the point where the line crosses the y-axis. To identify the slope and intercept from an equation, it's helpful to first convert any line to this form.
To rewrite an equation in slope-intercept form, isolate \( y \) on one side of the equation. Take, for example, the equation \( x + 2y = 6 \). We start by moving \( x \) to the other side, simplifying to \( 2y = -x + 6 \). Next, divide every term by 2, resulting in \( y = -\frac{1}{2}x + 3 \). The slope (\( m \)) is \(-\frac{1}{2}\), and the y-intercept (\( b \)) is 3.
Understanding this form makes it easier to quickly discern how a line behaves and intersects the axes. Knowing how to switch an equation into this form is crucial for solving parallel and perpendicular line problems.

Point-Slope Form

Point-slope form is particularly useful when you know the slope of a line and one point on it. It is written as \( y - y_1 = m(x - x_1) \), where \( m \) is the slope and \((x_1, y_1)\) is a given point on the line.
This form comes in handy when constructing an equation for a line parallel or perpendicular to another line. For a line that passes through \((1, -6)\) with a slope of \(-\frac{1}{2}\), you substitute the values into the point-slope formula: \( y - (-6) = -\frac{1}{2}(x - 1) \). Simplifying this yields \( y + 6 = -\frac{1}{2}x + \frac{1}{2} \), which upon further simplification gives \( y = -\frac{1}{2}x - \frac{11}{2} \).
The point-slope form is an efficient way to quickly identify and write the equation of a line when given a slope and point, streamlining the problem-solving process.

Parallel Lines

When two lines are parallel, they never intersect and have the same slope. This is an important property in geometry and algebra since it can be used to establish relationships between multiple lines.
Given the original line equation \( x + 2y = 6 \), after converting it to the slope-intercept form \( y = -\frac{1}{2}x + 3 \), it becomes evident that the slope is \(-\frac{1}{2}\). A line parallel to this one will therefore also have a slope of \(-\frac{1}{2}\).
To demonstrate how parallel lines work practically, consider needing to find a line parallel to the given one, passing through a specific point, such as \((1, -6)\). Using the point-slope form and knowing the slope remains consistent, you can write the equation as \( y = -\frac{1}{2}x - \frac{11}{2} \). Parallel lines' consistent slopes offer a straightforward approach to solving problems related to line equations.