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

The value of k such that the two lines 2 x + k y = 3 and x + y = 1 are parallel.

The value of k such that the two lines 2 x + k y = 3 and x + y = 1 are parallel.

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

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 1.1, Problem 41E

(a)

To determine

To find: The value of k such that the two lines 2x+ky=3 and x+y=1 are parallel.

(a)

Expert Solution

Answer to Problem 41E

The value of k is 2 .

Explanation of Solution

Given information: The equations are 2x+ky=3 and x+y=1 .

Calculation:

Write the equation 2x+ky=3 in slope intercept form.

2x+ky=3ky=−2x+3y=−2kx+3k

Write the equation x+y=1 in slope intercept form.

x+y=1y=−x+1

Compare the equations y=−2kx+3 , and y=−x+1 with y=mx+b to find the slope.

m1=−2km2=−1

It is given that the lines are parallel. So, m1=m2 .

−2k=−1k=2

Therefore, the value of k is 2 .

(b)

To determine

To find: The value of k such that the two lines 2x+ky=3 , and x+y=1 are perpendicular.

(b)

Expert Solution

Answer to Problem 41E

The value of k is −2 .

Explanation of Solution

Given information: The equations are 2x+ky=3 , and x+y=1 .

Calculation:

Write the equation 2x+ky=3 in slope intercept form.

2x+ky=3ky=−2x+3y=−2kx+3k

Write the equation x+y=1 in slope intercept form.

x+y=1y=−x+1

Compare the equations y=−2kx+3 , and y=−x+1 with y=mx+b to find the slope.

m1=−2km2=−1

It is given that the lines are perpendicular. So, m1⋅m2=−1 .

(−2k)(−1)=−1k=−2

Therefore, the value of k is −2 .

Chapter 1.1, Problem 40E

Chapter 1.1, Problem 42E

Chapter 1 Solutions

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

Ch. 1.1 - Prob. 1QR Ch. 1.1 - Prob. 2QR Ch. 1.1 - Prob. 3QR Ch. 1.1 - Prob. 4QR Ch. 1.1 - Prob. 5QR Ch. 1.1 - Prob. 6QR Ch. 1.1 - Prob. 7QR Ch. 1.1 - Prob. 8QR Ch. 1.1 - Prob. 9QR Ch. 1.1 - Prob. 10QR

Ch. 1.1 - Prob. 1E Ch. 1.1 - Prob. 2E Ch. 1.1 - Prob. 3E Ch. 1.1 - Prob. 4E Ch. 1.1 - Prob. 5E Ch. 1.1 - Prob. 6E Ch. 1.1 - Prob. 7E Ch. 1.1 - Prob. 8E Ch. 1.1 - Prob. 9E Ch. 1.1 - Prob. 10E Ch. 1.1 - Prob. 11E Ch. 1.1 - Prob. 12E Ch. 1.1 - Prob. 13E Ch. 1.1 - Prob. 14E Ch. 1.1 - Prob. 15E Ch. 1.1 - Prob. 16E Ch. 1.1 - Prob. 17E Ch. 1.1 - Prob. 18E Ch. 1.1 - Prob. 19E Ch. 1.1 - Prob. 20E Ch. 1.1 - Prob. 21E Ch. 1.1 - Prob. 22E Ch. 1.1 - Prob. 23E Ch. 1.1 - Prob. 24E Ch. 1.1 - Prob. 25E Ch. 1.1 - Prob. 26E Ch. 1.1 - Prob. 27E Ch. 1.1 - Prob. 28E Ch. 1.1 - Prob. 29E Ch. 1.1 - Prob. 30E Ch. 1.1 - Prob. 31E Ch. 1.1 - Prob. 32E Ch. 1.1 - Prob. 33E Ch. 1.1 - Prob. 34E Ch. 1.1 - Prob. 35E Ch. 1.1 - Prob. 36E Ch. 1.1 - Prob. 37E Ch. 1.1 - Prob. 38E Ch. 1.1 - Prob. 39E Ch. 1.1 - Prob. 40E Ch. 1.1 - Prob. 41E Ch. 1.1 - Prob. 42E Ch. 1.1 - Prob. 43E Ch. 1.1 - Prob. 44E Ch. 1.1 - Prob. 45E Ch. 1.1 - Prob. 46E Ch. 1.1 - Prob. 47E Ch. 1.1 - Prob. 48E Ch. 1.1 - Prob. 49E Ch. 1.1 - Prob. 50E Ch. 1.1 - 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