If a, b, c are in Arithmetic Progression (AP) | vedclass

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(A) Given that $a, b, c$ are in Arithmetic Progression $(AP)$,we have $2b = a + c$,which implies $a - 2b + c = 0$. Comparing this with the equation of

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$(1, -2)$

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(A) Given that $a, b, c$ are in Arithmetic Progression $(AP)$,we have $2b = a + c$,which implies $a - 2b + c = 0$. Comparing this with the equation of the line $ax + by + c = 0$,we can rewrite the equation as $a(1) + b(x) + c = 0$ is not correct,let us substitute $c = 2b - a$ into the line equation: $ax + by + (2b - a) = 0$ $a(x - 1) + b(y + 2) = 0$ For this equation to hold for all $a$ and $b$,the coefficients must be zero: $x - 1 = 0 \implies x = 1$ $y + 2 = 0 \implies y = -2$ Thus,the fixed point is $(1, -2)$.

If $a, b, c$ are in Arithmetic Progression $(AP)$,then the line $ax + by + c = 0$ always passes through a fixed point. The coordinates of this point are:

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