Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
Related questions
Question
Find distance between each pair of parallel lines.
![## Finding the Distance Between Each Pair of Parallel Lines
### Given Equations:
1. \( y = 2x + 1 \)
2. \( y = 2x - 4 \)
### Parameters:
- Slope (\( m \)) = 2
- \( b = 1 \)
- Perpendicular slope (\( \perp m \)) = \(-\frac{1}{2}\)
- Equation of perpendicular line: \( y = -\frac{1}{2}x + 1 \)
### Graph:
A coordinate grid is provided but no specific points are plotted. The grid can be used for graphing the lines and finding their intersection.
### Instructions:
Set the perpendicular line equation equal to one of the given parallel line equations and solve the system:
1. Solve:
\[
\left(-\frac{1}{2}x + 1\right) = (2x - 4)
\]
2. Simplify:
\[
-\frac{1}{2}x + 1 = 2x - 4
\]
3. Isolate x:
\[
-\frac{1}{2}x + 1 + 1 = 2x - 4 + 1
\]
\[
-\frac{1}{2}x + 2 = 2x - 3
\]
4. Multiply throughout to clear fractions and solve for \( x \):
\[
-x + 4 = 4x - 4
\]
\[
8 = 6x
\]
\[
x = \frac{8}{6} = \frac{4}{3}
\]
5. Substitute \( x \) into one equation to find \( y \).
6. Calculate the perpendicular distance to determine the distance between the points.
### Conclusion:
This exercise demonstrates finding the distance between parallel lines using their equations and calculating the perpendicular distance between points of intersection. The method ensures understanding of algebraic manipulations and geometric relationships among parallel lines.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F12a1ba4a-9eea-484c-989f-6e04f409a1c7%2F12cbc80d-c503-427a-8f96-7e60d0fa73d8%2Ft0k2t3_processed.jpeg&w=3840&q=75)
Transcribed Image Text:## Finding the Distance Between Each Pair of Parallel Lines
### Given Equations:
1. \( y = 2x + 1 \)
2. \( y = 2x - 4 \)
### Parameters:
- Slope (\( m \)) = 2
- \( b = 1 \)
- Perpendicular slope (\( \perp m \)) = \(-\frac{1}{2}\)
- Equation of perpendicular line: \( y = -\frac{1}{2}x + 1 \)
### Graph:
A coordinate grid is provided but no specific points are plotted. The grid can be used for graphing the lines and finding their intersection.
### Instructions:
Set the perpendicular line equation equal to one of the given parallel line equations and solve the system:
1. Solve:
\[
\left(-\frac{1}{2}x + 1\right) = (2x - 4)
\]
2. Simplify:
\[
-\frac{1}{2}x + 1 = 2x - 4
\]
3. Isolate x:
\[
-\frac{1}{2}x + 1 + 1 = 2x - 4 + 1
\]
\[
-\frac{1}{2}x + 2 = 2x - 3
\]
4. Multiply throughout to clear fractions and solve for \( x \):
\[
-x + 4 = 4x - 4
\]
\[
8 = 6x
\]
\[
x = \frac{8}{6} = \frac{4}{3}
\]
5. Substitute \( x \) into one equation to find \( y \).
6. Calculate the perpendicular distance to determine the distance between the points.
### Conclusion:
This exercise demonstrates finding the distance between parallel lines using their equations and calculating the perpendicular distance between points of intersection. The method ensures understanding of algebraic manipulations and geometric relationships among parallel lines.
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