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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![### Writing the Equation of a Parallel Line
**Problem:**
Write an equation for a line parallel to \( y = -2x + 3 \) and passing through the point \( (4, -6) \).
**Solution:**
1. **Identify the slope of the given line:**
The equation of the given line is \( y = -2x + 3 \).
The slope-intercept form of a line is \( y = mx + b \) where \( m \) is the slope and \( b \) is the y-intercept.
From the given equation, the slope \( m \) is \(-2\).
2. **Use the slope for the parallel line:**
Parallel lines have the same slope. Therefore, the slope of our new line will also be \(-2\).
3. **Form the equation of the line:**
The point-slope form of a line is \( y - y_1 = m(x - x_1) \), where \( (x_1, y_1) \) is a point on the line and \( m \) is the slope.
4. **Substitute the given point and slope into the point-slope form:**
Given point is \( (4, -6) \). So, \( x_1 = 4 \) and \( y_1 = -6 \).
The slope \( m \) is \(-2\).
Substituting these values into the point-slope form:
\[
y - (-6) = -2(x - 4)
\]
Simplify the equation:
\[
y + 6 = -2(x - 4)
\]
\[
y + 6 = -2x + 8
\]
\[
y = -2x + 8 - 6
\]
\[
y = -2x + 2
\]
Therefore, the equation for the line parallel to \( y = -2x + 3 \) and passing through the point \( (4, -6) \) is:
\[
y = -2x + 2
\]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb98bdb5b-53bd-47c2-bd74-197e9ff57506%2F8f9d7667-fe30-4845-b02d-cab09755d85d%2Ftgwpn6p_processed.png&w=3840&q=75)
Transcribed Image Text:### Writing the Equation of a Parallel Line
**Problem:**
Write an equation for a line parallel to \( y = -2x + 3 \) and passing through the point \( (4, -6) \).
**Solution:**
1. **Identify the slope of the given line:**
The equation of the given line is \( y = -2x + 3 \).
The slope-intercept form of a line is \( y = mx + b \) where \( m \) is the slope and \( b \) is the y-intercept.
From the given equation, the slope \( m \) is \(-2\).
2. **Use the slope for the parallel line:**
Parallel lines have the same slope. Therefore, the slope of our new line will also be \(-2\).
3. **Form the equation of the line:**
The point-slope form of a line is \( y - y_1 = m(x - x_1) \), where \( (x_1, y_1) \) is a point on the line and \( m \) is the slope.
4. **Substitute the given point and slope into the point-slope form:**
Given point is \( (4, -6) \). So, \( x_1 = 4 \) and \( y_1 = -6 \).
The slope \( m \) is \(-2\).
Substituting these values into the point-slope form:
\[
y - (-6) = -2(x - 4)
\]
Simplify the equation:
\[
y + 6 = -2(x - 4)
\]
\[
y + 6 = -2x + 8
\]
\[
y = -2x + 8 - 6
\]
\[
y = -2x + 2
\]
Therefore, the equation for the line parallel to \( y = -2x + 3 \) and passing through the point \( (4, -6) \) is:
\[
y = -2x + 2
\]
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