10th Edition

ISBN: 9780134781945

Author: Demana

Publisher: PEARSON

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Expert Solution & Answer

Chapter P.4, Problem 44E

Solution

(a)

To Find:

Find an equation for the line passing through the point and parallel to the given line. Support your work graphically.

y=35x−135

Given information:

Point (6,1)

Line 3x−5y=15

Concept used:

We can write the equation of a line parallel to a given line if we know a point on the line and an equation of the given line. y=mx+b. Parallel lines have the same slope. The slope of the line with equation y=mx+b is b.

Calculation:

The given Point (6,1)

Line 3x−5y=15

3x−5y=155y=3x−15y=3x5−155y=3x5−3point (6,1)y=mx+b1=35(6)+b1=185+bb=1−185b=5−185b=−135so y=35x−135

So the equation for the line passing through the point (6,1) and parallel to the given line is y=35x−135

By graphically the graph of 3x−5y=15 and y=35x−135

(b)

To Find:

Find an equation for the line passing through the point and perpendicular to the given line. Support your work graphically.

y=−5x3+11

Given information:

Point (6,1)

Line 3x−5y=15

Concept used:

We can write the equation of a line perpendicular to a given line if we know a point on the line and an equation of the given line. y−y1=m(x−x1) .

Calculation:

The given Point (6,1)

Line 3x−5y=15

3x−5y=155y=3x−15y=3x5−155y=3x5−3

Here slope m-

m=−53m=−53

y=3x5−3point (6,1)y−y1=m(x−x1)y−1=−53(x−6)y−1=−5x3−53(−6)y−1=−5x3+10y=−5x3+10+1y=−5x3+11

So the equation for the line passing through the point and perpendicular to the given line is y=−5x3+11

By graphically the graph of 3x−5y=15 and y=−5x3+11

Chapter P.4, Problem 43E

Chapter P.4, Problem 45E

Chapter P Solutions

PRECALCULUS:GRAPHICAL,...-W/ACCESS

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

Ch. P.1 - Prob. 1E Ch. P.1 - Prob. 2E Ch. P.1 - Prob. 3E Ch. P.1 - Prob. 4E Ch. P.1 - Prob. 5E Ch. P.1 - Prob. 6E Ch. P.1 - Prob. 7E Ch. P.1 - Prob. 8E Ch. P.1 - Prob. 9E Ch. P.1 - Prob. 10E Ch. P.1 - Prob. 11E Ch. P.1 - Prob. 12E Ch. P.1 - Prob. 13E Ch. P.1 - Prob. 14E Ch. P.1 - Prob. 15E Ch. P.1 - Prob. 16E Ch. P.1 - Prob. 17E Ch. P.1 - Prob. 18E Ch. P.1 - Prob. 19E Ch. P.1 - Prob. 20E Ch. P.1 - Prob. 21E Ch. P.1 - Prob. 22E Ch. P.1 - Prob. 23E Ch. P.1 - Prob. 24E Ch. P.1 - Prob. 25E Ch. P.1 - Prob. 26E Ch. P.1 - Prob. 27E Ch. P.1 - Prob. 28E Ch. P.1 - Prob. 29E Ch. P.1 - Prob. 30E Ch. P.1 - Prob. 31E Ch. P.1 - Prob. 32E Ch. P.1 - Prob. 33E Ch. P.1 - Prob. 34E Ch. P.1 - Prob. 35E Ch. P.1 - Prob. 36E Ch. P.1 - Prob. 37E Ch. P.1 - Prob. 38E Ch. P.1 - Prob. 39E Ch. P.1 - Prob. 40E Ch. P.1 - Prob. 41E Ch. P.1 - Prob. 42E Ch. P.1 - Prob. 43E Ch. P.1 - Prob. 44E Ch. P.1 - Prob. 45E Ch. P.1 - Prob. 46E Ch. P.1 - Prob. 47E Ch. P.1 - In Exercises 4752, simplify the expression. Assume...Ch. 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Points, Lines, Planes, Segments, & Rays - Collinear vs Coplanar Points - Geometry; Author: The Organic Chemistry Tutor;https://www.youtube.com/watch?v=dDWjhRfBsKM;License: Standard YouTube License, CC-BY

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Solution Summary: The author explains how to find an equation for the line passing through the point and perpendicular to the given line.

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Chapter P Solutions