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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√2. Then,

A

a+bc>0

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B

ab+c<0

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C

ab+c>0

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D

a+bc<0

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Solution

## The correct option is A a+b−c>0 As a>b>c>0 a−c>0 and b>0 ⇒a+b−c>0……(i) a−b>0 and c>0 a+c−b>0……(ii) From Eqs. (i) and (ii), option (b) and (d) are eliminated. Also, the point of intersection for ax+by+c=0 and bx+ay+c=0 i.e (−ca+b,−ca+b) The distance between (1, 1) and (−ca+b,−ca+b) i.e. less than 2√2 ⇒√(1+ca+b)2+(1+ca+b)2<2√2⇒(a+b+ca+b)√2<2√2⇒a+b+c<2a+2b⇒a+b−c>0 ∴ Option (a) is correct.

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