If L: ax + by + c = 0 is a variable straight | vedclass

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(B) Let the first term of the $AP$ be $p$ and the common difference be $q$.Then,$a = p + q$,$b = p + 3q$,and $c = p + 6q$.The equation of the line is

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$\left( \frac{3}{2}, -\frac{5}{2} \right)$

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(B) Let the first term of the $AP$ be $p$ and the common difference be $q$.Then,$a = p + q$,$b = p + 3q$,and $c = p + 6q$.The equation of the line is $(p + q)x + (p + 3q)y + (p + 6q) = 0$.Rearranging the terms in terms of $p$ and $q$:$p(x + y + 1) + q(x + 3y + 6) = 0$.For the line to pass through a fixed point regardless of $p$ and $q$,both expressions must be zero:$x + y + 1 = 0$ and $x + 3y + 6 = 0$.Subtracting the first equation from the second: $(x + 3y + 6) - (x + y + 1) = 0$ $\Rightarrow 2y + 5 = 0$ $\Rightarrow y = -\frac{5}{2}$.Substituting $y = -\frac{5}{2}$ into $x + y + 1 = 0$: $x - \frac{5}{2} + 1 = 0 \Rightarrow x = \frac{3}{2}$.Thus,the fixed point is $\left( \frac{3}{2}, -\frac{5}{2} \right)$.

If $L: ax + by + c = 0$ is a variable straight line,where $a, b$ and $c$ are the second,fourth and seventh terms of an $AP$ respectively,then $L$ passes through the fixed point -

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