The algebraic sum of distances of the line `ax + by + 2 = 0` from `(1,2), (2,1) and (3,5)` is zero and the lines `bx - ay + 4 = 0` and `3x + 4y + 5=0` cut the coordinate axes at concyclic points. Then (a) `a+b=-2/7` (b) area of triangle formed by the line `ax+by+2=0` with coordinate axes is `14/5` (c) line `ax+by+3=0` always passes through the point `(-1,1)` (d) max `{a,b}=5/7`

Options

A. `a +b =- (2)/(7)`
B. area of the triangle formed by the line `ax +by +2 = 0` with coordinate axes is `(14)/(5)`
C. line `ax +by +3 = 0` always passes through the point `(-1,1)`
D. max `{a,b} = (5)/(7)`

Answer

C

Explanation

Line `ax +by +2 = 0` always passes through the point `(2,(8)/(3))`
`:. 6a +8b +6 = 0`
or `3a +4b +3 = 0`
Now `bx - ay +4 = 0` and `3x +4y +5 = 0` meet axis in concyclic points.
So, `m_(1)m_(2) =1`
`((b)/(a)).(-(3)/(4)) =1`
`rArr 4a +3b = 0` (ii)
Solving (i) and (ii), we get `a = 9//7, b =- 12//7`
`rArr` Line `ax +by +3 = 0` always passes through the point `(-1,1)`.