The algebraic sum of distances of the line $a x + b y + 2 = 0$ from $(1, 2), (2, 1)$ and $(3, 5)$ is zero and the lines $b x - a y + 4 = 0$ and $3 x + 4 y + 5 = 0$ cut the co-ordinate axes at concyclic points. Then

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max ${a, b} = \frac{5}{7}$

line $a x + b y + 3 = 0$ always passes through the point $(- 1, 1)$ .

$a + b = - \frac{2}{7}$

area of the triangle formed by the line $a x + b y + 2 = 0$ with coordinate axes is $\frac{14}{5}$

answer is C.

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Detailed Solution

Line $a x + b y + 2 = 0$ always passes through the point $(2, \frac{8}{3})$
$∴ 6 a + 8 b + 6 = 0 o r 3 a + 4 b + 3 = 0____(i)$
Now $b x - a y + 4 = 0$ and $3 x + 4 y + 5 = 0$ meet axis in concyclic points.
So, $m_{1} m_{2} = 1$
$(\frac{b}{a}) . (\frac{- 3}{4}) = 1$
$\Rightarrow 4 a + 3 b = 0_____(i i)$
Solving (i) and (ii) we get $a = 9 / 7, b = - 12 / 7$
$\Rightarrow$ Line $a x + b y + 3 = 0$ always passes through the point $(- 1, 1)$ .