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,the terms are given by $a = p + q$,$b = p + 3q$,and $c = p + 6q$.The e

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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,the terms are given by $a = p + q$,$b = p + 3q$,and $c = p + 6q$.The equation of the line is $ax + by + c = 0$.Substituting the values of $a, b, c$:$(p + q)x + (p + 3q)y + (p + 6q) = 0$Rearranging the terms to group $p$ and $q$:$p(x + y + 1) + q(x + 3y + 6) = 0$For this line to pass through a fixed point for all $p$ and $q$,both expressions must be zero:$x + y + 1 = 0$  (Equation $1$)$x + 3y + 6 = 0$  (Equation $2$)Subtracting Equation $1$ from Equation $2$:$(x + 3y + 6) - (x + y + 1) = 0$$2y + 5 = 0 \Rightarrow y = -\frac{5}{2}$Substituting $y = -\frac{5}{2}$ into Equation $1$:$x - \frac{5}{2} + 1 = 0$ $\Rightarrow x - \frac{3}{2} = 0$ $\Rightarrow x = \frac{3}{2}$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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