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Perpendicular lines

  • Thread starter bacon
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  • #1
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Determine a constant real number k such that the lines AB and CD are perpendicular.
A(1,2), B(4,0), C(k,2), D(1,-3). (answer given is k=-3/2)

If two lines are perpendicular the product of their slopes is -1.

The slope of AB is [tex]\frac{0-2}{4-1}[/tex] = [tex]\frac{-2}{3}[/tex]

The slope of CD is [tex]\frac{-3-2}{1-k}[/tex] = [tex]\frac{-5}{1-k}[/tex]
I set the product of these slopes equal to -1 and solve for k.

[tex]\frac{-2}{3}[/tex] x [tex]\frac{-5}{1-k}[/tex]=-1
10=3(k-1)
10=3k-3
13=3k
13/3=k
This is not the answer given and I am not seeing my error. Any help would be appreciated.
Latex question. The = sign and x symbol don't line up well with the rest of the equations in the first half of my post. How can I correct that?
 

Answers and Replies

  • #2
1,750
1
Your k value is correct ...

[tex]\frac{0-2}{4-1}=\frac{-2}{3}[/tex]

Check my LaTeX ...
 
Last edited:
  • #3
67
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Thanks for the help.
 
  • #4
63
0
Verified using an alternate method (direction cosines and perpendicularity) to confirm that your answer is correct.
 
  • #5
You are 100% correct, reporting the text book error might be helpful for the rest using the same book...
 

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