EP ALGEBRA 1-ETEXT ACCESS
EP ALGEBRA 1-ETEXT ACCESS
11th Edition
ISBN: 9780133216455
Author: Charles
Publisher: SAVVAS L
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Chapter 5.6, Problem 34PPE
To determine

To find:the value of k for which two graphs are parallel and also find the value of k for which two graphs perpendicular.

Expert Solution & Answer
Check Mark

Answer to Problem 34PPE

The value of k when these two graphs are parallel is k=32

The value of k when these two graphs are perpendicular is k=24

Explanation of Solution

Given info:

The two graphs straight line equation as follows:

  12y=3x+86y=kx5

Consider the slope of 12y=3x+8 equation is m1

Consider the slope of 6y=kx5 equation is m2

Formula used:

Show the slope- intercepts equation or straight line equation as follows:

  y=mx+b

Here, x,y are the coordinates in the graph, m is the slope of the straight line, and b is the constant (number)

If two straight line equations have the same slope valuethat is m1=m2 , then it is said to be a parallel line.

If two straight lines don’t have the same slope value and the slope value is a negative sign of the reciprocal of another equation or the product of two slopes is 1 i.e. m1×m2=1 , then it is said to be a perpendicular line.

Calculation:

Slope value is taken as coefficient of x

Find the slope m1 for the equation 12y=3x+8 .

  12y=3x+8y=312x+812y=mx+bm1=312

Find the slope m2 for the equation 6y=kx5 .

  6y=kx5y=k6x5y=mx+bm2=k6

Conditions for the two straight line equations are parallel to each other, m1=m2

Find the k value when two straight lines are parallel.

  m1=m2312=k6k=312×6k=32

Conditions for the two straight line equations are perpendicular to each other, m1×m2=1

Find the k value when two straight lines are perpendicular.

  m1×m2=1312×k6=13k12×6=1k=12×63k=24

Hence, the value of k when these two graphs are parallel is k=32

Hence, the value of k when these two graphs are perpendicular is k=24

Conclusion:

Thus, the value of k when these two graphs are parallel is k=32

Thus, the value of k when these two graphs are perpendicular is k=24

Chapter 5 Solutions

EP ALGEBRA 1-ETEXT ACCESS

Ch. 5.1 - Prob. 7LCCh. 5.1 - Prob. 8PPECh. 5.1 - Prob. 9PPECh. 5.1 - Prob. 10PPECh. 5.1 - Prob. 11PPECh. 5.1 - Prob. 12PPECh. 5.1 - Prob. 13PPECh. 5.1 - Prob. 14PPECh. 5.1 - Prob. 15PPECh. 5.1 - Prob. 16PPECh. 5.1 - Prob. 17PPECh. 5.1 - Prob. 18PPECh. 5.1 - Prob. 19PPECh. 5.1 - Prob. 20PPECh. 5.1 - Prob. 21PPECh. 5.1 - Prob. 22PPECh. 5.1 - Prob. 23PPECh. 5.1 - Prob. 24PPECh. 5.1 - Prob. 25PPECh. 5.1 - Prob. 26PPECh. 5.1 - Prob. 27PPECh. 5.1 - Prob. 28PPECh. 5.1 - Prob. 29PPECh. 5.1 - Prob. 30PPECh. 5.1 - Prob. 31PPECh. 5.1 - Prob. 32PPECh. 5.1 - Prob. 33PPECh. 5.1 - Prob. 34PPECh. 5.1 - Prob. 35PPECh. 5.1 - Prob. 36PPECh. 5.1 - Prob. 37PPECh. 5.1 - Prob. 38PPECh. 5.1 - Prob. 39PPECh. 5.1 - Prob. 40PPECh. 5.1 - Prob. 41PPECh. 5.1 - Prob. 42PPECh. 5.1 - Prob. 43PPECh. 5.1 - Prob. 44PPECh. 5.1 - Prob. 45PPECh. 5.1 - Prob. 46PPECh. 5.1 - Prob. 47PPECh. 5.1 - Prob. 48PPECh. 5.1 - Prob. 49PPECh. 5.1 - Prob. 50PPECh. 5.1 - Prob. 51PPECh. 5.1 - Prob. 52PPECh. 5.1 - Prob. 53PPECh. 5.1 - Prob. 54PPECh. 5.1 - 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