Straight Line Question 263

Question: For $ a>b>c>0, $ the distance between (1, 1) and the point of intersection of the lines $ ax+by+c=0 $ and $ bx+ay+c=0 $ is less than $ 2\sqrt{2}. $ Then

Options:

A) $ a+b-c>0 $

B) $ a-b+c<0 $

C) $ a-b+c>0 $

D) $ a+b-c<0 $

Show Answer

Answer:

Correct Answer: A

Solution:

  • [a] The intersection point of two lines is $ ( \frac{-c}{a+b},\frac{-c}{a+b} ) $ Distance between (1, 1) and $ ( \frac{-c}{a+b},\frac{-c}{a+b} )<2\sqrt{2} $

$ \Rightarrow 2{{( 1+\frac{c}{a+b} )}^{2}}<8 $

$ \Rightarrow 1+\frac{c}{a+b}<2\Rightarrow a+b-c>0 $



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