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 co-ordinate axes at concyclic points then

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area of the triangle formed by the line ax + by + 2 = 0 with coordinate axes is

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line ax + by + 3 = 0 always passes through the point (-1,1)

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max {a, b} =

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Solution

The correct option is C line ax + by + 3 = 0 always passes through the point (-1,1)
Line always passes through the point $(2, \frac{8}{3})$ hence $6 a + 8 b + 6 = 0 \Rightarrow 3 a + 4 b + 3 = 0 b x - a y + 4 = 0$ and $3 x + 4 y + 5 = 0$ are concyclic.
So, $m_{1} m_{2} = 1$
$\frac{b}{a} . - \frac{3}{4} = 1 \Rightarrow 4 a + 3 b = 0$
Solving $a = \frac{9}{7}, b = - \frac{12}{7}$

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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 co-ordinate axes at concyclic points then

The correct option is C line ax + by + 3 = 0 always passes through the point (-1,1)Line always passes through the point (2,83) hence 6a+8b+6=0⇒3a+4b+3=0bx–ay+4=0 and 3x+4y+5=0 are concyclic. So, m1m2=1 ba.−34=1⇒4a+3b=0 Solving a=97,b=−127