Explanation Result used: (1) The point-slope form of the equation of a straight line is y − y 1 = m ( x − x 1 ) , where ( x 1 , y 1 ) is the point on a line. (2) “The relation between slopes of perpendicular lines m 1 = − 1 m 2 ”. Calculation: The required equation is perpendicular to 6.0 x − 2.4 y − 3.9 = 0 . Therefore, the slope of the required line is m = − ( 1 2.5 ) . That is, m = − 0.4 . In order to find the equation of the line, substitute ( x 1 , y 1 ) = ( 7.5 , − 4.7 ) and m = − 0

Basic Technical Mathematics with Calculus (11th Edition)

11th Edition

ISBN: 9780134437736

Author: Allyn J. Washington, Richard Evans

Publisher: PEARSON

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Question

Chapter 21.2, Problem 18E

To determine

The equation of the line perpendicular to the line with slope 6.0x−2.4y−3.9=0 and it is passes through (7.5,−4.7).

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Chapter 21.2, Problem 17E

Chapter 21.2, Problem 19E

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Chapter 21 Solutions

Basic Technical Mathematics with Calculus (11th Edition)

Ch. 21.1 - Find the distance between (−2, 6) and (5, −3). Ch. 21.1 - Prob. 2PE Ch. 21.1 - Prob. 1E Ch. 21.1 - Prob. 2E Ch. 21.1 - Prob. 5E Ch. 21.1 - In Exercises 5–14, find the distance between the...Ch. 21.1 - Prob. 7E Ch. 21.1 - Prob. 8E Ch. 21.1 - Prob. 9E Ch. 21.1 - Prob. 10E

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The equation of the line perpendicular to the line with slope 6.0 x − 2.4 y − 3.9 = 0 and it is passes through ( 7.5 , − 4.7 ) .

The equation of the line perpendicular to the line with slope 6.0x−2.4y−3.9=0 and it is passes through (7.5,−4.7).

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Chapter 21 Solutions