$\begin {array} {1 1} (A)\;(1,-2) & \quad (B)\;(1,2) \\ (C)\;(-1, 2) & \quad (D)\;(-1, -2) \end {array}$

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- If a,b and c are in A.P then $ \large\frac{a+c}{2}$$=b \quad \Rightarrow a-2b=-c \quad $ (i.e) $2b-a=c$

Step 1 :

It is given that a, b and c are in A.P

$ \therefore c = 2b-a$

Substituting for c in the equation.

$ax+by+c=0$, we get,

$ax+by+2b-a=0$

$ \Rightarrow a(x-1)+b(y+2)=0$

Let $ \lambda = \large\frac{b}{a}$

$ \therefore (x-1)+ \lambda (y+2)=0$

This equation is of the form $ L_1+\lambda L_2=0$ which represents a straight line through the intersection of the $L_1=0$ and $L_2=0$.

(i.e) $x-1=0 \quad x=1$

$y+2=0 \quad y=-2$

Hence the straight line always passes through (1, -2).

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