Quantitative Aptitude Ques 2350

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:

  • The graphs of $ (k-1),x+y-2=0 $ and $ (2-k),x-3y+1=0 $ are parallel. Two straight lines $ a _1x+b _1y+c _1=0 $ and $ a _2x+b _2y+c _2=0 $ are parallel, if $ \frac{a _1}{b _2}=\frac{b _1}{b _2}\ne \frac{c _1}{c _2}. $ It means that the given system of equations has no solution.

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

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

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

$ \Rightarrow $ $ -,2k=-1 $

$ \therefore $ $ k=\frac{1}{2} $

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