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} $
BETA
