Straight Line Question 120
Question: Let $ \alpha $ be the distance between the lines $ -x+y=2 $ and $ x-y=2 $ , and $ \beta $ be the distance between the lines $ 4x-3y=5 $ and $ 6y-8x=1 $ , then [J & K 2005]
Options:
A) $ 20\sqrt{2}\beta =11\alpha $
B) $ 20\sqrt{2}\alpha =11\beta $
C) $ 11\sqrt{2}\beta =20\alpha $
D)None of these
Correct Answer: A Distance between lines $ -x+y=2 $ and $ x-y=2 $ is $ \alpha =\frac{2}{\sqrt{2}}+\frac{2}{\sqrt{2}}=2\sqrt{2} $ Distance between lines $ 4x-3y=5 $ and $ 8x-6y=-1 $ is, $ \beta =\frac{1}{10}+\frac{5}{5}=\frac{11}{10} $ Therefore $ \frac{\alpha }{\beta }=\frac{2\sqrt{2}}{11/10}\Rightarrow 20\sqrt{2}\beta =11\alpha $ .
Show Answer
Answer:
Solution:
BETA
