If a, b and c are in AP, then the straight li | vedclass

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(A) Given that $a, b$ and $c$ are in $AP$. Therefore, $2b = a + c$ or $c = 2b - a$. The equation of the straight line is $ax + 2by + c = 0$. Substitut

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$(1, -1)$

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(A) Given that $a, b$ and $c$ are in $AP$. Therefore, $2b = a + c$ or $c = 2b - a$. The equation of the straight line is $ax + 2by + c = 0$. Substituting the value of $c$, we get $ax + 2by + (2b - a) = 0$. Rearranging the terms, we have $a(x - 1) + 2b(y + 1) = 0$. For this line to pass through a fixed point for all values of $a$ and $b$, the coefficients must be zero independently. Thus, $x - 1 = 0 \Rightarrow x = 1$ and $y + 1 = 0 \Rightarrow y = -1$. Hence, the fixed point is $(1, -1)$.

If $a, b$ and $c$ are in $AP$, then the straight line $ax + 2by + c = 0$ will always pass through a fixed point whose coordinates are

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