Quantitative Aptitude Ques 571

Question: The value of k for which the graphs of $ (k-1)x+y-2=0 $ and $ (2-k),x-3y+1=0 $ are parallel, is

Options:

A) $ \frac{1}{2} $

B) $ -\frac{1}{2} $

C) 2

D) $ -,2 $

Show Answer

Answer:

Correct Answer: A

Solution:

  • $ (k-1)x+y-2=0 $ and $ (2-k),x-3y+1=0 $ $ m _1=1-k $ and $ m _2=\frac{(2-k)}{3} $
    Since, the two times are parallel.

$ \therefore $ $ m _1=m _2 $

$ \Rightarrow $ $ 1-k=\frac{2-k}{3} $

$ \Rightarrow $ $ 3-3k=2-k $

$ \Rightarrow $ $ -,2k=-1 $
$ \Rightarrow $ $ k=\frac{1}{2} $

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