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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 22. then 

a
a+b−c>0
b
a−b+c<0
c
a−b+c>0
d
a+b−c<0

detailed solution

Correct option is A

Solving given lines for their point of intersection, we get the point of intersection as-ca+b,-ca+b, Its distance from ( 1 , 1) is1+ca+b2+1+ca+b2<22 or (a+b+c)2<4(a+b)2 or (a+b+c)2−(2a+2b)2<0 or (c−a−b)(c+3a+3b)<0 Since a>b>c>0,(c−a−b)<0 or a+b−c>0

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