Straight Line Question 235
Question: Given a family of lines a $ a(2x+y+4)+b(x-2y-3)=0 $ the number of lines belonging to the family at a distance $ \sqrt{10} $ from $ P(2,-3) $ is
Options:
A) 0
B) 1
C) 2
D) 4
Correct Answer: B [b] The length of perpendicular from $ P(2,-3) $ on the given family of lines $ \frac{a(4-3+4)+b(2+6-3)}{\sqrt{{{(2a+b)}^{2}}+{{(a-2b)}^{2}}}}=\pm \sqrt{10} $ (Given)
$ \Rightarrow 5a+5b=\pm \sqrt{25a^{2}+25b^{2}} $ $ \Rightarrow 25{{(a+b)}^{2}}=25(a^{2}+2ab+b^{2}) $ $ \Rightarrow 25{{(a-b)}^{2}}=0\Rightarrow a=b $ For which we get only line $ 3x-y+1=0 $
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Answer:
Solution:
BETA
