PRECALCULUS:GRAPHICAL,...-W/ACCESS
PRECALCULUS:GRAPHICAL,...-W/ACCESS
10th Edition
ISBN: 9780134781945
Author: Demana
Publisher: PEARSON
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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

  PRECALCULUS:GRAPHICAL,...-W/ACCESS, Chapter P.4, Problem 44E , additional homework tip 1

(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

  PRECALCULUS:GRAPHICAL,...-W/ACCESS, Chapter P.4, Problem 44E , additional homework tip 2

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Chapter P Solutions

PRECALCULUS:GRAPHICAL,...-W/ACCESS

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