Advanced Mathematical Concepts: Precalculus with Applications, Student Edition

1st Edition

ISBN: 9780078682278

Author: McGraw-Hill, Berchie Holliday

Publisher: Glencoe/McGraw-Hill

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Question

Chapter 5.7, Problem 45E

To determine

To Find: The standard form of the equation.

Expert Solution & Answer

Answer to Problem 45E

5x+2y=−22

Explanation of Solution

Given:The graph is perpendicular to the graph of 2x+5y=7 and passes through the point (−6,4)

Given: 2x+5y=7 and point (−6,4) .

Write the equation in slope intercept form.

y=25x+75slope=25

Now

We know that perpendicular lines have slopes that are negative reciprocal to each other.

Thus

Slope of the perpendicular line is

m1m2=−125m2=−1m2=−52

Thus

Equation of the line

y−y1=m(x−x1)y−4=−52(x−(−6)y−4=−52(x+6)solving2y−8=−5x−305x+2y=−22

Hence, the standard form of the equation is 5x+2y=−22 .

Chapter 5.7, Problem 44E

Chapter 5.7, Problem 46E

Chapter 5 Solutions

Advanced Mathematical Concepts: Precalculus with Applications, Student Edition

Ch. 5.1 - Prob. 1CFU Ch. 5.1 - Prob. 2CFU Ch. 5.1 - Prob. 3CFU Ch. 5.1 - Prob. 4CFU Ch. 5.1 - Prob. 5CFU Ch. 5.1 - Prob. 6CFU Ch. 5.1 - Prob. 7CFU Ch. 5.1 - Prob. 8CFU Ch. 5.1 - Prob. 9CFU Ch. 5.1 - Prob. 10CFU

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