Explanation Given info. Points are ( 1 , 2 ) and ( 5 , 4 ) Concept used: Product of slope of line perpendicular to each other is − 1 that means m 1 m 2 = − 1 . The midpoint of the segment joining ( a , b ) and ( c , d ) is ( a + c 2 , b + d 2 ) Calculation: Mid-point = ( 5 + 1 2 , 2 + 4 2 ) = ( 6 2 , 6 2 ) = ( 3 , 3 ) Slope of two points can be calculated as y 2 − y 1 x 2 − x 1 = 4 − 2 5 − 1

3rd Edition

ISBN: 9781464125263

Author: Jon Rogawski, Colin Adams

Publisher: W.H. Freeman & Co

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Question

Chapter 1.2, Problem 21E

To determine

The equation of the perpendicular bisector of the segment joining (1,2) and (5,4).

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Chapter 1.2, Problem 20E

Chapter 1.2, Problem 22E

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Chapter 1 Solutions

Calculus - Standalone book

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

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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Prob. 10E Ch. 1.2 - Prob. 11E Ch. 1.2 - Prob. 12E Ch. 1.2 - Prob. 13E Ch. 1.2 - Prob. 14E Ch. 1.2 - Prob. 15E Ch. 1.2 - Prob. 16E Ch. 1.2 - Prob. 17E Ch. 1.2 - Prob. 18E Ch. 1.2 - Prob. 19E Ch. 1.2 - Prob. 20E Ch. 1.2 - Prob. 21E Ch. 1.2 - Prob. 22E Ch. 1.2 - Prob. 23E Ch. 1.2 - Prob. 24E Ch. 1.2 - Prob. 25E Ch. 1.2 - Prob. 26E Ch. 1.2 - Prob. 27E Ch. 1.2 - Prob. 28E Ch. 1.2 - Prob. 29E Ch. 1.2 - Prob. 30E Ch. 1.2 - Prob. 31E Ch. 1.2 - Prob. 32E Ch. 1.2 - Prob. 33E Ch. 1.2 - Prob. 34E Ch. 1.2 - Prob. 35E Ch. 1.2 - Prob. 36E Ch. 1.2 - Prob. 37E Ch. 1.2 - Prob. 38E Ch. 1.2 - Prob. 39E Ch. 1.2 - Prob. 40E Ch. 1.2 - Prob. 41E Ch. 1.2 - Prob. 42E Ch. 1.2 - Prob. 43E Ch. 1.2 - Prob. 44E Ch. 1.2 - Prob. 45E Ch. 1.2 - Prob. 46E Ch. 1.2 - Prob. 47E Ch. 1.2 - Prob. 48E Ch. 1.2 - Prob. 49E Ch. 1.2 - Prob. 50E Ch. 1.2 - Prob. 51E Ch. 1.2 - Prob. 52E Ch. 1.2 - Prob. 53E Ch. 1.2 - Prob. 54E Ch. 1.2 - Prob. 55E Ch. 1.2 - Prob. 56E Ch. 1.2 - Prob. 57E Ch. 1.2 - Prob. 58E Ch. 1.2 - Prob. 59E Ch. 1.2 - Prob. 60E Ch. 1.2 - Prob. 61E Ch. 1.3 - Prob. 1PQ Ch. 1.3 - Prob. 2PQ Ch. 1.3 - Prob. 3PQ Ch. 1.3 - Prob. 4PQ Ch. 1.3 - Prob. 5PQ Ch. 1.3 - Prob. 1E Ch. 1.3 - Prob. 2E Ch. 1.3 - Prob. 3E Ch. 1.3 - Prob. 4E Ch. 1.3 - Prob. 5E Ch. 1.3 - Prob. 6E Ch. 1.3 - Prob. 7E Ch. 1.3 - Prob. 8E Ch. 1.3 - Prob. 9E Ch. 1.3 - Prob. 10E Ch. 1.3 - Prob. 11E Ch. 1.3 - Prob. 12E Ch. 1.3 - Prob. 13E Ch. 1.3 - Prob. 14E Ch. 1.3 - Prob. 15E Ch. 1.3 - Prob. 16E Ch. 1.3 - Prob. 17E Ch. 1.3 - Prob. 18E Ch. 1.3 - Prob. 19E Ch. 1.3 - Prob. 20E Ch. 1.3 - Prob. 21E Ch. 1.3 - Prob. 22E Ch. 1.3 - Prob. 23E Ch. 1.3 - Prob. 24E Ch. 1.3 - Prob. 25E Ch. 1.3 - Prob. 26E Ch. 1.3 - Prob. 27E Ch. 1.3 - Prob. 28E Ch. 1.3 - Prob. 29E Ch. 1.3 - Prob. 30E Ch. 1.3 - Prob. 31E Ch. 1.3 - Prob. 32E Ch. 1.3 - Prob. 33E Ch. 1.3 - Prob. 34E Ch. 1.3 - Prob. 35E Ch. 1.3 - Prob. 36E Ch. 1.3 - Prob. 37E Ch. 1.3 - Prob. 38E Ch. 1.3 - Prob. 39E Ch. 1.3 - Prob. 40E Ch. 1.3 - Prob. 41E Ch. 1.3 - Prob. 42E Ch. 1.3 - Prob. 43E Ch. 1.3 - Prob. 44E Ch. 1.3 - 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Prob. 43E Ch. 1.4 - Prob. 44E Ch. 1.4 - Prob. 45E Ch. 1.4 - Prob. 46E Ch. 1.4 - Prob. 47E Ch. 1.4 - Prob. 48E Ch. 1.4 - Prob. 49E Ch. 1.4 - Prob. 50E Ch. 1.4 - Prob. 51E Ch. 1.4 - Prob. 52E Ch. 1.4 - Prob. 53E Ch. 1.4 - Prob. 54E Ch. 1.4 - Prob. 55E Ch. 1.4 - Prob. 56E Ch. 1.4 - Prob. 57E Ch. 1.4 - Prob. 58E Ch. 1.4 - Prob. 59E Ch. 1.4 - Prob. 60E Ch. 1.4 - Prob. 61E Ch. 1.4 - Prob. 62E Ch. 1.4 - Prob. 63E Ch. 1.4 - Prob. 64E Ch. 1.4 - Prob. 65E Ch. 1.4 - Prob. 66E Ch. 1.5 - Prob. 1PQ Ch. 1.5 - Prob. 2PQ Ch. 1.5 - Prob. 3PQ Ch. 1.5 - Prob. 4PQ Ch. 1.5 - Prob. 1E Ch. 1.5 - Prob. 2E Ch. 1.5 - Prob. 3E Ch. 1.5 - Prob. 4E Ch. 1.5 - Prob. 5E Ch. 1.5 - Prob. 6E Ch. 1.5 - Prob. 7E Ch. 1.5 - Prob. 8E Ch. 1.5 - Prob. 9E Ch. 1.5 - Prob. 10E Ch. 1.5 - Prob. 11E Ch. 1.5 - Prob. 12E Ch. 1.5 - Prob. 13E Ch. 1.5 - Prob. 14E Ch. 1.5 - Prob. 15E Ch. 1.5 - Prob. 16E Ch. 1.5 - Prob. 17E Ch. 1.5 - Prob. 18E Ch. 1.5 - Prob. 19E Ch. 1.5 - Prob. 20E Ch. 1.5 - Prob. 21E Ch. 1.5 - Prob. 22E Ch. 1.5 - Prob. 23E Ch. 1.5 - Prob. 24E Ch. 1 - Prob. 1CRE Ch. 1 - Prob. 2CRE Ch. 1 - Prob. 3CRE Ch. 1 - Prob. 4CRE Ch. 1 - Prob. 5CRE Ch. 1 - Prob. 6CRE Ch. 1 - Prob. 7CRE Ch. 1 - Prob. 8CRE Ch. 1 - Prob. 9CRE Ch. 1 - Prob. 10CRE Ch. 1 - Prob. 11CRE Ch. 1 - Prob. 12CRE Ch. 1 - Prob. 13CRE Ch. 1 - Prob. 14CRE Ch. 1 - Prob. 15CRE Ch. 1 - Prob. 16CRE Ch. 1 - Prob. 17CRE Ch. 1 - Prob. 18CRE Ch. 1 - Prob. 19CRE Ch. 1 - Prob. 20CRE Ch. 1 - Prob. 21CRE Ch. 1 - Prob. 22CRE Ch. 1 - Prob. 23CRE Ch. 1 - Prob. 24CRE Ch. 1 - Prob. 25CRE Ch. 1 - Prob. 26CRE Ch. 1 - Prob. 27CRE Ch. 1 - Prob. 28CRE Ch. 1 - Prob. 29CRE Ch. 1 - Prob. 30CRE Ch. 1 - Prob. 31CRE Ch. 1 - Prob. 32CRE Ch. 1 - Prob. 33CRE Ch. 1 - Prob. 34CRE Ch. 1 - Prob. 35CRE Ch. 1 - Prob. 36CRE Ch. 1 - Prob. 37CRE Ch. 1 - Prob. 38CRE Ch. 1 - Prob. 39CRE Ch. 1 - Prob. 40CRE Ch. 1 - Prob. 41CRE Ch. 1 - Prob. 42CRE Ch. 1 - Prob. 43CRE Ch. 1 - Prob. 44CRE Ch. 1 - Prob. 45CRE Ch. 1 - Prob. 46CRE Ch. 1 - Prob. 47CRE Ch. 1 - Prob. 48CRE Ch. 1 - Prob. 49CRE Ch. 1 - Prob. 50CRE

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The equation of the perpendicular bisector of the segment joining ( 1 , 2 ) and ( 5 , 4 ) .

Solution Summary: The author explains that the equation of the perpendicular bisector is 2x+y-9=0.

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The equation of the perpendicular bisector of the segment joining (1,2) and (5,4).

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Chapter 1 Solutions