Problem 53 - College Algebra with Modeling and Visualization

Question

Find an equation of the line satisfying the conditions. Perpendicular to \(y=15,\) passing through \((4,-9)\)

Step-by-Step Solution

Answer

The equation of the line is \( x = 4 \).

Step 1: Identify the Slope of the Given Line

The equation of the line is given as \( y = 15 \). This is a horizontal line. The slope of a horizontal line is 0. We need to consider this when finding a line perpendicular to it.

Step 2: Determine the Slope of the Perpendicular Line

For a line to be perpendicular to a horizontal line, such as \( y = 15 \), it must be vertical. The slope of a vertical line is undefined. Instead of finding a slope, we realize the line will be in the form \( x = a \).

Step 3: Use the Given Point to Find the Equation

We know the perpendicular line must pass through the point \((4, -9)\). Since perpendicular to a horizontal line, it must be a vertical line at \( x = 4 \).

Step 4: Write the Equation of the Line

Using the information from previous steps, we write down the equation of the line as \( x = 4 \). This equation is a vertical line passing through the given point.

Key Concepts

Perpendicular Lines

Perpendicular lines are a fascinating concept in geometry, referring to two lines that intersect at a right angle, or 90 degrees. This relationship is key when discussing different line orientations on a graph. Understanding perpendicularity is crucial when working with linear equations, especially when tasked with finding the equation of a line that must be perpendicular to another.

• Perpendicular lines have slopes that are negative reciprocals of each other.
• If one line is vertical with an undefined slope, the perpendicular line is horizontal with a slope of zero, and vice versa.

When given the equation of a horizontal line, like the example from the exercise, determining the perpendicular is straightforward. A horizontal line has a slope of zero, so its perpendicular will be a vertical line. Vertical lines are represented by equations of the form \( x = \text{constant} \). They touch the x-axis and do not change in the y-direction.

Slope of a Line

The slope of a line is a crucial concept in understanding how lines behave on a graph. It measures the steepness and direction of the line and is calculated as the ratio of the vertical change (rise) to the horizontal change (run) between two points on the line.For a line represented by the equation \( y = mx + c \), the slope \( m \) determines how slanted the line is. Different lines have different slopes:

• Positive slope: Line rises from left to right.
• Negative slope: Line falls from left to right.
• Zero slope: Horizontal line – no rise, just run.
• Undefined slope: Vertical line – all rise, no run.

In the exercise, the line \( y = 15 \) is horizontal, indicating a slope of zero. A line perpendicular to this horizontal line will be vertical, which means it cannot have a numerical slope as its slope is undefined. Hence, perpendicular lines emerge into this expert discussion by altering the slope orientation.

Vertical and Horizontal Lines

Understanding vertical and horizontal lines is essential, especially when dealing with perpendicular and parallel concepts in coordinate geometry.**Horizontal Lines**- Horizontal lines run parallel to the x-axis.- They have a constant y-value, such as \( y = 15 \), and their slope is always zero.- These lines are perfectly flat with no rise regardless of the run.**Vertical Lines**- Vertical lines run parallel to the y-axis.- They have a constant x-value and an undefined slope.- The equation of a vertical line takes the form \( x = a \), where \( a \) is a constant.When tasked with finding a line perpendicular to a horizontal line, you instantly think about a vertical line. In the given exercise, determining that the line is perpendicular to \( y = 15 \), and passes through \( (4, -9) \), we can quickly establish its equation as \( x = 4 \). This indicates a vertical line positioned at \( x = 4 \) on the graph, lining up perfectly in perpendicular opposition to the horizontal component.