Problem 70 - College Algebra and Calculus: An Applied Approach

Question

The equations of two lines are given. Determine if lines \(L_{1}\) and \(L_{2}\) are parallel, perpendicular, or neither. \(L_{1}: y=\frac{3}{4} x+1 ; L_{2}: y=-\frac{4}{3} x+3\)

Step-by-Step Solution

Answer

The two lines \(L_{1}\) and \(L_{2}\) are perpendicular

Step 1: Identify the slopes of the two lines

The lines \(L_{1}\) and \(L_{2}\) are given in the slope-intercept form \(y = mx + b\) where \(m\) is the slope. Therefore, the slope of \(L_{1}\) is \(\frac{3}{4}\) and the slope of \(L_{2}\) is -\(\frac{4}{3}\).

Step 2: Check if the lines are parallel

Parallel lines have the same slope. Here, the slope of \(L_{1}\) is \(\frac{3}{4}\) while the slope of \(L_{2}\) is -\(\frac{4}{3}\). Since the slopes are not equal, the lines \(L_{1}\) and \(L_{2}\) are not parallel.

Step 3: Check if the lines are perpendicular

Perpendicular lines have slopes that are negative reciprocals of each other. So, if \(L_{1}\) is perpendicular to \(L_{2}\), the slope of \(L_{1}\) should be the negative reciprocal of the slope of \(L_{2}\). The negative reciprocal of -\(\frac{4}{3}\) is \(\frac{3}{4}\). Since the slope of \(L_{1}\) is \(\frac{3}{4}\), the lines \(L_{1}\) and \(L_{2}\) are perpendicular.

Key Concepts

Slope-Intercept Form

Analyzing the relationships between lines often starts with understanding the slope-intercept form of a linear equation, which is expressed as \( y = mx + b \). This form provides us with two critical pieces of information: the slope \( m \) and the y-intercept \( b \). The slope represents the rate at which the line ascends or descends, indicating the steepness of the line. A positive slope means the line goes upward from left to right, while a negative slope indicates the line descends. The y-intercept \( b \) is the point where the line crosses the y-axis.

When given an equation like \( y = \frac{3}{4}x + 1 \), we can quickly identify that the slope is \( \frac{3}{4} \) and the y-intercept is 1. This initial recognition is the foundational step in determining the relationships between lines, as seen in exercises where one must compare the slopes of two lines to understand their interaction.

Parallel and Perpendicular Lines

Understanding the relationship between lines includes identifying whether they are parallel or perpendicular to each other.

Parallel Lines

Two lines are considered parallel if they have the same slope and will never intersect. For example, if two lines in the slope-intercept form \( y = mx + b \) have the identical value of \( m \) but different intercepts \( b \) , they are parallel.

h4>Perpendicular LinesPerpendicular lines, on the other hand, intersect at a 90-degree angle, and their slopes have a specific relationship - they are negative reciprocals of each other. This means that if one line has a slope of \( a \), the other must have a slope of \( -\frac{1}{a} \). This is a crucial aspect in determining the perpendicularity of two given lines.

To determine if the lines in our exercise, \( L_{1} \) and \( L_{2} \), are parallel or perpendicular, we must compare their slopes. Since their slopes are not equal and are negative reciprocals, this confirms that \( L_{1} \) and \( L_{2} \) are perpendicular.

Negative Reciprocals

The concept of negative reciprocals is essential when dealing with perpendicular lines. A negative reciprocal is taken by inverting a number and changing its sign. For instance, the negative reciprocal of \( \frac{3}{4} \) is \( -\frac{4}{3} \) and vice versa. These relationships are used to determine if two lines are perpendicular.

For example, if we want to find the line perpendicular to \( y = \frac{3}{4}x + 1 \), we would look for a line with a slope that is the negative reciprocal of \( \frac{3}{4} \), which would be \( -\frac{4}{3} \). As shown in the exercise, the line \( L_{2} \), with the equation \( y = -\frac{4}{3}x + 3 \), has this exact slope, confirming that \( L_{2} \) is perpendicular to \( L_{1} \). The concept of negative reciprocals not only helps to confirm perpendicularity but is also a powerful tool in creating equations of lines that are required to have this specific relationship.