Precalculus: Graphical, Numerical, Algebraic Common Core 10th Edition

10th Edition

ISBN: 9780134673301

Author: Waits, Foley, Kennedy, Bock Demana

Publisher: PEARSON

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Expert Solution & Answer

Chapter P.4, Problem 43E

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=−23x+3

Given information:

Point (3,1)

Line 2x+3y=12

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 (3,1)

Line 2x+3y=12

2x+3y=123y=12−2xy=12−2x3y=4−2x3y=−2x3+4point (3,1)y=mx+b1=−23(3)+b1=−2+bb=1+2b=3so y=−23x+3

So the equation for the line passing through the point (3,1) and parallel to the given line is y=−2x−1

By graphically the graph of 2x+3y=12 and y=−23x+3

(b)

To Find:

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

y=3x2−72

Given information:

Point (3,1)

Line 2x+3y=12

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 (3,1)

Line 2x+3y=12

2x+3y=123y=12−2xy=12−2x3y=4−2x3y=−2x3+4

Here slope m-

m=−3−2m=32

y=−2x3+4point (3,1)y−y1=m(x−x1)y−1=32(x−3)y−1=3x2−92y=3x2−92+1y=3x2−9+22y=3x2−9+22y=3x2−72

So the equation for the line passing through the point and perpendicular to the given line is y=3x2−72

By graphically the graph of 2x+3y=12 and y=3x2−72

Chapter P.4, Problem 42E

Chapter P.4, Problem 44E

Chapter P Solutions

Precalculus: Graphical, Numerical, Algebraic Common Core 10th Edition

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Chapter P Solutions