PRECALCULUS:GRAPH...-NASTA ED.(FLORIDA)
PRECALCULUS:GRAPH...-NASTA ED.(FLORIDA)
10th Edition
ISBN: 9780136174318
Author: Demana
Publisher: PEARSON
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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

  PRECALCULUS:GRAPH...-NASTA ED.(FLORIDA), Chapter P.4, Problem 43E , 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=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

  PRECALCULUS:GRAPH...-NASTA ED.(FLORIDA), Chapter P.4, Problem 43E , additional homework tip 2

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Chapter P.4, Problem 42E
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Chapter P.4, Problem 44E

Chapter P Solutions

PRECALCULUS:GRAPH...-NASTA ED.(FLORIDA)

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