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Question 147073: For what value(s) of k the lines y=(k+1)x-3 and (2k-1)x+1 are parallel and are perpendicular?
Answer by ankor@dixie-net.com(22740)   (Show Source):
You can put this solution on YOUR website! For what value(s) of k the lines y=(k+1)x-3 and (2k-1)x+1 are parallel
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Recognizing this in the form y = mx + b
You are given 2 slopes: m1 = (k+x) and m2 = (2k-1)
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Parallel lines have equal slopes, therefore we can find k like this:
(2k-1) = (k+1)
2k - k = 1 + 1
k = 2
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Substitute 2 for k in both equation and you can see they have equal slopes
y = 3x - 3 and y = 3x + 1
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and are perpendicular?
The slope relationships of perpendicular lines are m1 * m2 = -1
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(k+1) * (2k-1) = -1
FOIL
2k^2 + k - 1 = -1
2k^2 + k - 1 + 1 = 0
2k^2 + k = 0
Factor out k
k(2k+1) = 0
Two solutions
k = 0
and
2k = -1
k = -.5
Find the two equations when k = 0
y = 1x - 3 and y = -1x + 1 are perpendicular
and
when k = -.5
The two equations: y = .5x - 3, and y = -2x + 1 also are perpendicular
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Did this make sense to you, any questions??
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