Problem 72 - Algebra for College Students

Question

How would you help a friend determine the equation of the line that is perpendicular to \(x-5 y=7\) and contains the point \((5,4)\) ?

Step-by-Step Solution

Answer

The perpendicular line's equation is \( y = -5x + 29 \).

Step 1: Find the Slope of the Given Line

The given line equation is \( x - 5y = 7 \). We need to rewrite this in the slope-intercept form \( y = mx + b \). Start by isolating \( y \):\[ x - 5y = 7 \ -5y = -x + 7 \ y = \frac{1}{5}x - \frac{7}{5}\]The slope \( m \) of the given line is \( \frac{1}{5} \).

Step 2: Determine the Slope of the Perpendicular Line

A line that is perpendicular to another line will have a slope that is the negative reciprocal of the original line's slope. The negative reciprocal of \( \frac{1}{5} \) is \( -5 \). Thus, the slope of the perpendicular line is \( -5 \).

Step 3: Use Point-Slope Form

The equation of a line can be expressed using the point-slope form: \( y - y_1 = m(x - x_1) \), where \( m \) is the slope and \((x_1, y_1)\) is the given point. Substitute \( m = -5 \), \( x_1 = 5 \), and \( y_1 = 4 \) into the formula:\[ y - 4 = -5(x - 5)\]

Step 4: Simplify the Equation

Expand and simplify to get the equation in slope-intercept form:\[ y - 4 = -5x + 25 \ y = -5x + 29\]

Step 5: Write the Final Equation

The equation of the line perpendicular to the given line and passing through the point \((5, 4)\) is \( y = -5x + 29 \).

Key Concepts

Slope-Intercept Form

Understanding the slope-intercept form is essential when dealing with linear equations. This form is expressed as \( y = mx + b \), where:

• \( y \) is the dependent variable (usually on the vertical axis).
• \( m \) represents the slope of the line, indicating its steepness and direction.
• \( x \) is the independent variable (usually on the horizontal axis).
• \( b \) is the y-intercept, the point where the line crosses the y-axis.

In the context of our exercise, we began with the equation \( x - 5y = 7 \). This needed to be rearranged into the slope-intercept form before we could find the slope. By isolating \( y \) to become \( y = \frac{1}{5}x - \frac{7}{5} \), it clearly shows that the slope \( m \) of the original line is \( \frac{1}{5} \) and indicates a line that rises slowly as we move from left to right. Obtaining this form is a crucial first step in understanding linear relationships and comparing slopes.

Slope of a Line

The slope of a line is a measure of its steepness and direction. It is calculated as the ratio of the vertical change (rise) to the horizontal change (run) between two points on the line. In other words, it is how much \( y \) changes for a unit change in \( x \).
When we have an equation such as \( y = \frac{1}{5}x - \frac{7}{5} \), as derived in our exercise, the slope of \( \frac{1}{5} \) tells us that for every 5 units increase in \( x \), \( y \) will increase by 1 unit. This results in a gentle upward slant.
For perpendicular lines, the key idea is the concept of negative reciprocals. If two lines are perpendicular, their slopes are negative reciprocals of each other. That means if line A has a slope of \( \frac{1}{5} \), line B, which is perpendicular, must have a slope of \( -5 \). This change indicates that the perpendicular line tilts steeply downward as opposed to the original line's gentle upward rise. It’s a foundational concept in geometry, crucial for understanding how different lines relate in a plane.

Point-Slope Form

The point-slope form is incredibly useful when you need to formulate the equation of a line, and all you have is a single point on the line and its slope. It is written as:\[ y - y_1 = m(x - x_1) \]

• \( m \) represents the slope of the line.
• \( (x_1, y_1) \) is a specific point on the line.
• \( x \) and \( y \) are the variables.

In our example, once we identified the slope of the perpendicular line as \( -5 \), we used the point-slope form to find the specific equation. By plugging in the point \((5, 4)\) and slope \( -5 \) into the formula, we derived the equation:\[ y - 4 = -5(x - 5) \]Simplifying this equation, we moved to the slope-intercept form to make it more accessible. This process demonstrates how point-slope form can act as a bridge from understanding specific characteristics of a line to visualizing it fully with an equation.