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Mathematics
A line passes through the point of intersection of the lines 100x+50y-1=0 and 75x+25y+ 3=0 and makes equal intercepts on the axes. Its equation is
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Q. A line passes through the point of intersection of the lines $ 100x+50y-1=0 $ and $ 75x+25y+ $ $ 3=0 $ and makes equal intercepts on the axes. Its equation is
KEAM
KEAM 2008
Straight Lines
A
$ 25x+25y-1=0 $
B
$ 5x-5y+3=0 $
C
$ 25x+25y-4=0 $
D
$ 25x-25y+6=0 $
E
$ 5x-5y+7=0 $
Solution:
Equation of intersection of lines is
$ (100x+50y-1)+\lambda (75x+25y+3)=0 $
$ \Rightarrow $ $ (100+75\lambda )x+(50+25\lambda )y=1-3\lambda $ ?..(i)
$ \Rightarrow $ $ \frac{x}{\frac{1-3\lambda }{100+75\lambda }}+\frac{y}{\frac{1-3\lambda }{50+25\lambda }}=1 $
According to the given condition
$ \frac{1-3\lambda }{100+75\lambda }=\frac{1-3\lambda }{50+25\lambda } $
$ \Rightarrow $ $ 50=-50\lambda \Rightarrow \lambda =-1 $
$ \therefore $ From Eq. (i), $ 25x+25y-4=0 $