Calculus 2012 Student Edition (by Finney/Demana/Waits/Kennedy)

4th Edition

ISBN: 9780133178579

Author: Ross L. Finney

Publisher: PEARSON

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Question

Chapter 2.4, Problem 9QR

To determine

To find:The equation of line that passes through (−1,3) andparallelto 2x+3y=5 .

Expert Solution & Answer

Answer to Problem 9QR

The equation of line passes through (−1,3) and parallel to 2x+3y=5 is y=−23x+73 .

Explanation of Solution

Given information:The given point is (−1,3) and the line parallel to it is 2x+3y=5 .

Formula used: The general equation of line passing through (x1,y1) and slope m is given by,

y−y1=m(x−x1)

Calculation:

Simplify the given equation of line.

2x+3y=53y=−2x+5y=−23x+53

Compare the given equation of line with general equation y=mx+c .

m=−23 and c=53

The given line is parallel to the required line, and then the slope of both the lines is equal.So, the slope of the line is −23 .

To find the required equation of line, substitute −1 for x1 , 3 for y1 and −23 for m .

y−y1=m(x−x1)y−3=−23(x−(−1))y−3=−23x−23

Further simplify.

y=−23x−23+3y=−23x+73

Therefore, the equation of line passes through (−1,3) and parallel to 2x+3y=5 is y=−23x+73 .

Chapter 2.4, Problem 8QR

Chapter 2.4, Problem 10QR

Chapter 2 Solutions

Calculus 2012 Student Edition (by Finney/Demana/Waits/Kennedy)

Ch. 2.1 - Prob. 1QR Ch. 2.1 - Prob. 2QR Ch. 2.1 - Prob. 3QR Ch. 2.1 - Prob. 4QR Ch. 2.1 - Prob. 5QR Ch. 2.1 - Prob. 6QR Ch. 2.1 - Prob. 7QR Ch. 2.1 - Prob. 8QR Ch. 2.1 - Prob. 9QR Ch. 2.1 - Prob. 10QR

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The equation of line that passes through ( − 1 , 3 ) andparallelto 2 x + 3 y = 5 .

Concept explainers

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To find:The equation of line that passes through (−1,3) andparallelto 2x+3y=5 .

Answer to Problem 9QR

Explanation of Solution

Chapter 2 Solutions

Additional Math Textbook Solutions