Explanation Proof: The given figure is shown below in Figure 1, Figure (1) The points A ( 1 , b ) and O ( 0 , 0 ) are on the line L 1 . The slope of a line when two passing points are ( x 1 , y 1 ) and ( x 2 , y 2 ) is given by, m = y 2 − y 1 x 2 − x 1 (1) Substitute m 1 for m , 1 for x 1 , b for y 1 , 0 for x 2 , and 0 for y 2 in equation (1). The slope of the line L 1 is calculated as follows. m 1 = 0 − b 0 − 1 = − b − 1 = b Thus, the slope of line L 1 is b . The points B ( 1 , c ) and O ( 0 , 0 ) are on the line L 2 . Substitute m 2 for m , 1 for x 1 , c for y 1 , 0 for x 2 , and 0 for y 2 in equation (1). The slope of the line L 2 is calculated as follows. m 2 = 0 − c 0 − 1 = − c − 1 = c Thus, the slope of line L 2 is c . Use distance formula to find the distance between A ( 1 , b ) , O ( 0 , 0 ) and B ( 1 , c ) , O ( 0 , 0 ) . Distance between two points ( x 1 , y 1 ) and ( x 2 , y 2 ) is given by, d = ( y 2 − y 1 ) 2 + ( x 2 − x 1 ) 2 (2) Substitute 1 for x 1 , b for y 1 , 0 for x 2 , and 0 for y 2 in equation (2). The distance between A and O is calculated as follows. d = ( 0 − b ) 2 + ( 0 − 1 ) 2 = b 2 + 1 2 = 1 + b 2 The length of side AO is 1 + b 2

Applied Calculus for the Managerial, Life, and Social Sciences

10th Edition

ISBN: 9781305887831

Author: Tan

Publisher: CENGAGE CO

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Chapter 1.4, Problem 90E

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To show: If two lines namely L1 with slope m1 and L2 with slope m2 is perpendicular to each other then m1m2=−1 .

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Chapter 1.4, Problem 89E

Chapter 1, Problem 1CRQ

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Chapter 1 Solutions

Applied Calculus for the Managerial, Life, and Social Sciences

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

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A furniture...Ch. 1.3 - Prob. 38E Ch. 1.3 - MINIMIZING SHIPPING COSTS FOR A FLEET OF AUTOS...Ch. 1.3 - Prob. 40E Ch. 1.3 - Prob. 41E Ch. 1.3 - Prob. 42E Ch. 1.3 - Prob. 43E Ch. 1.3 - WATCHING A ROCKET LAUNCH At a distance of 4000 ft...Ch. 1.3 - Prob. 45E Ch. 1.3 - A SCAVENGER HUNT A tree is located 20 yd to the...Ch. 1.3 - Prob. 47E Ch. 1.3 - Prob. 48E Ch. 1.3 - Prob. 49E Ch. 1.3 - Prob. 50E Ch. 1.3 - Prob. 51E Ch. 1.3 - Prob. 52E Ch. 1.4 - Prob. 1CQ Ch. 1.4 - Prob. 2CQ Ch. 1.4 - Prob. 3CQ Ch. 1.4 - Prob. 4CQ Ch. 1.4 - Prob. 1E Ch. 1.4 - Prob. 2E Ch. 1.4 - Prob. 3E Ch. 1.4 - Prob. 4E Ch. 1.4 - Prob. 5E Ch. 1.4 - Prob. 6E Ch. 1.4 - Prob. 7E Ch. 1.4 - Prob. 8E Ch. 1.4 - Prob. 9E Ch. 1.4 - Prob. 10E Ch. 1.4 - In Exercises 11-16, find the slope of the line...Ch. 1.4 - Prob. 12E Ch. 1.4 - In Exercises 11-16, find the slope of the line...Ch. 1.4 - Prob. 14E Ch. 1.4 - Prob. 15E Ch. 1.4 - Prob. 16E Ch. 1.4 - Prob. 17E Ch. 1.4 - Prob. 18E Ch. 1.4 - Prob. 19E Ch. 1.4 - Prob. 20E Ch. 1.4 - Prob. 21E Ch. 1.4 - Prob. 22E Ch. 1.4 - Prob. 23E Ch. 1.4 - Prob. 24E Ch. 1.4 - Prob. 25E Ch. 1.4 - Prob. 26E Ch. 1.4 - Prob. 27E Ch. 1.4 - Prob. 28E Ch. 1.4 - Prob. 29E Ch. 1.4 - Prob. 30E Ch. 1.4 - Prob. 31E Ch. 1.4 - Prob. 32E Ch. 1.4 - Prob. 33E Ch. 1.4 - Prob. 34E Ch. 1.4 - Prob. 35E Ch. 1.4 - Prob. 36E Ch. 1.4 - Prob. 37E Ch. 1.4 - Prob. 38E Ch. 1.4 - Prob. 39E Ch. 1.4 - Prob. 40E Ch. 1.4 - Prob. 41E Ch. 1.4 - Prob. 42E Ch. 1.4 - Prob. 43E Ch. 1.4 - Prob. 44E Ch. 1.4 - Prob. 45E Ch. 1.4 - Find an equation of the line that passes through...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. 1.4 - Prob. 67E Ch. 1.4 - Prob. 68E Ch. 1.4 - Prob. 69E Ch. 1.4 - Prob. 70E Ch. 1.4 - Prob. 71E Ch. 1.4 - Prob. 72E Ch. 1.4 - Prob. 73E Ch. 1.4 - Prob. 74E Ch. 1.4 - Prob. 75E Ch. 1.4 - Prob. 76E Ch. 1.4 - Prob. 77E Ch. 1.4 - Prob. 78E Ch. 1.4 - SPENDING ON EQUIPMENT AND SOFTWARE As the United...Ch. 1.4 - SALES GROWTH Metro Department Stores annual sales...Ch. 1.4 - Prob. 81E Ch. 1.4 - Prob. 82E Ch. 1.4 - Prob. 83E Ch. 1.4 - Prob. 84E Ch. 1.4 - Prob. 85E Ch. 1.4 - Prob. 86E Ch. 1.4 - Prob. 87E Ch. 1.4 - Prob. 88E Ch. 1.4 - Prob. 89E Ch. 1.4 - Prob. 90E Ch. 1 - Prob. 1CRQ Ch. 1 - Prob. 2CRQ Ch. 1 - Prob. 3CRQ Ch. 1 - Prob. 4CRQ Ch. 1 - Prob. 5CRQ Ch. 1 - Prob. 6CRQ Ch. 1 - Prob. 7CRQ Ch. 1 - Prob. 8CRQ Ch. 1 - Prob. 1RE Ch. 1 - Prob. 2RE Ch. 1 - Prob. 3RE Ch. 1 - Prob. 4RE Ch. 1 - Prob. 5RE Ch. 1 - Prob. 6RE Ch. 1 - Prob. 7RE Ch. 1 - Prob. 8RE Ch. 1 - Prob. 9RE Ch. 1 - Prob. 10RE Ch. 1 - Prob. 11RE Ch. 1 - Prob. 12RE Ch. 1 - Prob. 13RE Ch. 1 - Prob. 14RE Ch. 1 - Prob. 15RE Ch. 1 - Prob. 16RE Ch. 1 - Prob. 17RE Ch. 1 - Prob. 18RE Ch. 1 - Prob. 19RE Ch. 1 - Prob. 20RE Ch. 1 - Prob. 21RE Ch. 1 - Prob. 22RE Ch. 1 - Prob. 23RE Ch. 1 - Prob. 24RE Ch. 1 - Prob. 25RE Ch. 1 - Prob. 26RE Ch. 1 - Prob. 27RE Ch. 1 - Prob. 28RE Ch. 1 - In Exercises 2932, find the value(s) of x that...Ch. 1 - Prob. 30RE Ch. 1 - Prob. 31RE Ch. 1 - Prob. 32RE Ch. 1 - Prob. 33RE Ch. 1 - Prob. 34RE Ch. 1 - Prob. 35RE Ch. 1 - Prob. 36RE Ch. 1 - Prob. 37RE Ch. 1 - Prob. 38RE Ch. 1 - Prob. 39RE Ch. 1 - Prob. 40RE Ch. 1 - Prob. 41RE Ch. 1 - Prob. 42RE Ch. 1 - Prob. 43RE Ch. 1 - Prob. 44RE Ch. 1 - Prob. 45RE Ch. 1 - Prob. 46RE Ch. 1 - Prob. 47RE Ch. 1 - Prob. 48RE Ch. 1 - Prob. 49RE Ch. 1 - Prob. 50RE Ch. 1 - Prob. 51RE Ch. 1 - Prob. 52RE Ch. 1 - Prob. 53RE Ch. 1 - Prob. 54RE Ch. 1 - Prob. 55RE Ch. 1 - Prob. 56RE Ch. 1 - Prob. 57RE Ch. 1 - Prob. 58RE Ch. 1 - Prob. 59RE Ch. 1 - Prob. 60RE Ch. 1 - Prob. 61RE Ch. 1 - Prob. 62RE Ch. 1 - Prob. 1BM Ch. 1 - Prob. 2BM Ch. 1 - Prob. 3BM Ch. 1 - Prob. 4BM Ch. 1 - Prob. 5BM Ch. 1 - Prob. 6BM Ch. 1 - Prob. 7BM Ch. 1 - Prob. 8BM Ch. 1 - Prob. 9BM Ch. 1 - Prob. 10BM

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If two lines namely L 1 with slope m 1 and L 2 with slope m 2 is perpendicular to each other then m 1 m 2 = − 1 .

Solution Summary: The author illustrates how the slope of a line when two passing points are (x_1,y

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To show: If two lines namely L1 with slope m1 and L2 with slope m2 is perpendicular to each other then m1m2=−1 .

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Chapter 1 Solutions