For $\lambda, \mu \in \mathbb{R},(x-2 y-1)+\lambda(3 x+2 y-11)=0$ and $(3 x+4 y-11)+\mu(-x+2 y-3)=0$ represent two families of lines. If the equation of the line common to both the families is $a x+b y-5=0$, then $2 a+b=$

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Aicharya · Accepted Answer

$(x-2 y-1)+\lambda(3 x+2 y-11)=0$
Point of interscetion of family of lines is $(3,1)$. $(3 x+4 y-11)+\mu(-x+2 y-3)=0$
Point of intersection is $(1,2)$. Line common to 2 families is line passing throug $(3,1),(1,2)$. $\begin{aligned} & \frac{y-1}{x-3}=\frac{(2-1)}{(1-3)} \Rightarrow-2(y-1)=x-3 \\ & x+2 y-5=0 \Rightarrow a=1, b-2 \Rightarrow 2 a+b=4 \end{aligned}$

For $\lambda, \mu \in \mathbb{R},(x-2 y-1)+\lambda(3 x+2 y-11)=0$ and $(3 x+4 y-11)+\mu(-x+2 y-3)=0$…

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