For `a gt b gt c gt 0`, if the distance between `(1,1)` and the point of intersection of the line `ax+by-c=0` and `bx+ay+c=0` is less than `2sqrt2` then, (A) `a+b-cgt0` (B) `a-b+clt0` (C) `a-b+cgt0` (D) `a+b-clt0`

Options

A. `a+b-c gt 0`
B. `a-b+c lt 0`
C. `a-b+c gt0`
D. `a+b-c lt 0`

Answer

A

Explanation

Solving given lines for their point of intersection, we get the point of intersection as (-c/(a+b), -c/(a+b)).
Its distance from (1, 1) is
`sqrt((1+(c)/(a+b))^(2) + (1+(c)/(a+b))^(2)) lt 2sqrt(2) " " ("given")`
`"or " (a+b+c)^(2) lt 4(a+b)^(2) " or " (a+b+c)^(2)-(2a+2b)^(2) lt 0`
`"or " (c-a-b)(c+3a+3b) lt 0`
`"Since "a gt b gt c gt 0, (c-a-b) lt 0 " or "a+b-c gt 0`