For lambda, mu in R,(x-2y-1)+lambda(3x+2y-11) | vedclass

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(C) The first family of lines $(x-2y-1)+\lambda(3x+2y-11)=0$ passes through the intersection of $x-2y-1=0$ and $3x+2y-11=0$. Solving these,we get $4x=

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$4$

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(C) The first family of lines $(x-2y-1)+\lambda(3x+2y-11)=0$ passes through the intersection of $x-2y-1=0$ and $3x+2y-11=0$. Solving these,we get $4x=12 \Rightarrow x=3$ and $y=1$. So,the point is $(3,1)$.The second family of lines $(3x+4y-11)+\mu(-x+2y-3)=0$ passes through the intersection of $3x+4y-11=0$ and $-x+2y-3=0$. Solving these,we get $3x+4y-11=0$ and $-3x+6y-9=0$. Adding them,$10y=20 \Rightarrow y=2$ and $x=1$. So,the point is $(1,2)$.The line common to both families passes through $(3,1)$ and $(1,2)$.The equation of the line passing through $(3,1)$ and $(1,2)$ is given by $\frac{y-1}{x-3} = \frac{2-1}{1-3} = \frac{1}{-2}$.$-2(y-1) = x-3$ $\Rightarrow -2y+2 = x-3$ $\Rightarrow x+2y-5=0$.Comparing $x+2y-5=0$ with $ax+by-5=0$,we get $a=1$ and $b=2$.Therefore,$2a+b = 2(1)+2 = 4$.

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

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