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...
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![**Problem Statement:**
Find the value of \( c \) so that the lines \( 5x + cy = 4 \) and \( x - 3y = 9 \) are perpendicular.
**Options:**
- (A) \( c = -3 \)
- (B) \( c = 15 \)
- (C) \( c = -15 \)
- (D) \( c = -\frac{5}{3} \)
- (E) \( c = \frac{5}{3} \)
**Explanation:**
To determine the value of \( c \) that makes the lines perpendicular, we need to find the slopes of the two lines and then use the property that states two lines are perpendicular if the product of their slopes is -1.
1. The equation of the first line is: \( 5x + cy = 4 \).
To find its slope, we convert it to slope-intercept form \( y = mx + b \):
\[ cy = -5x + 4 \]
\[ y = -\frac{5}{c}x + \frac{4}{c} \]
Hence, the slope \( m_1 \) of the first line is \( -\frac{5}{c} \).
2. The equation of the second line is: \( x - 3y = 9 \).
Again, convert it to slope-intercept form:
\[ -3y = -x + 9 \]
\[ y = \frac{1}{3}x - 3 \]
Hence, the slope \( m_2 \) of the second line is \( \frac{1}{3} \).
3. For the lines to be perpendicular, the product of their slopes \( m_1 \) and \( m_2 \) must be -1:
\[ m_1 \cdot m_2 = -1 \]
\[ -\frac{5}{c} \cdot \frac{1}{3} = -1 \]
\[ -\frac{5}{3c} = -1 \]
\[ \frac{5}{3c} = 1 \]
\[ 5 = 3c \]
\[ c = \frac{5}{3} \]
Hence, the correct option is **](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fedc716be-e066-4234-bd1a-1a86837575c3%2F1cd4e954-29a9-4871-8270-c7959c4d861e%2Fy388qc_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
Find the value of \( c \) so that the lines \( 5x + cy = 4 \) and \( x - 3y = 9 \) are perpendicular.
**Options:**
- (A) \( c = -3 \)
- (B) \( c = 15 \)
- (C) \( c = -15 \)
- (D) \( c = -\frac{5}{3} \)
- (E) \( c = \frac{5}{3} \)
**Explanation:**
To determine the value of \( c \) that makes the lines perpendicular, we need to find the slopes of the two lines and then use the property that states two lines are perpendicular if the product of their slopes is -1.
1. The equation of the first line is: \( 5x + cy = 4 \).
To find its slope, we convert it to slope-intercept form \( y = mx + b \):
\[ cy = -5x + 4 \]
\[ y = -\frac{5}{c}x + \frac{4}{c} \]
Hence, the slope \( m_1 \) of the first line is \( -\frac{5}{c} \).
2. The equation of the second line is: \( x - 3y = 9 \).
Again, convert it to slope-intercept form:
\[ -3y = -x + 9 \]
\[ y = \frac{1}{3}x - 3 \]
Hence, the slope \( m_2 \) of the second line is \( \frac{1}{3} \).
3. For the lines to be perpendicular, the product of their slopes \( m_1 \) and \( m_2 \) must be -1:
\[ m_1 \cdot m_2 = -1 \]
\[ -\frac{5}{c} \cdot \frac{1}{3} = -1 \]
\[ -\frac{5}{3c} = -1 \]
\[ \frac{5}{3c} = 1 \]
\[ 5 = 3c \]
\[ c = \frac{5}{3} \]
Hence, the correct option is **
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