Straight Line Question 47
Question: If the lines $ ax+y+1=0,x+by+1=0 $ and $ x+y+c=0 $ (a, b, c being distinct and different from 1) are concurrent, then $ \frac{1}{1-a}+\frac{1}{1-b}+\frac{1}{1-c}= $
Options:
A)0
B)1
C) $ \frac{1}{a+b+c} $
D)None of these
Correct Answer: B
Show Answer
Answer:
Solution:
Þ $ a(b-1)(c-1)-(b-1)(1-a)-(c-1)(1-a)=0 $
Þ $ \frac{a}{1-a}+\frac{1}{1-b}+\frac{1}{1-c}=0 $ {Divide by $ (1-a)(1-b)(1-c) $ }
Þ $ \frac{1}{1-a}+\frac{1}{1-b}+\frac{1}{1-c}=1 $ .
BETA
