Question

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`
A. `a+b-c gt0`
B. `a-b+clt0`
C. `a-b+c gt 0`
D. `a+b-c lt 0`

Answer 1

Correct Answer - A
The lines `ax+by+c=0` and `bx+ay+c=0` intersect at `(-(c)/(a+b),(c)/(a+b))` . It is given that the distance between (1,1) and the point of intersection of the given lines is less than `2sqrt(2)`.
`therefore sqrt((1+(c)/(a+b))^(2)+(1+(c)/(a+b))^(2)) lt 2sqrt(2)`
`implies |(a+b+c)/(a+b)|sqrt(2) lt 2sqrt(2)`
`implies (a+b+c)/(a+b) lt 2" "[because a+b+c gt 0 and a+bgt 0]`
`implies a+b+c lt 2(a+b) implies a+b-c gt 0`