If the line 3x + 4y + lambda = 0 divides the | vedclass

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(B) The distance between two parallel lines $ax + by + c_1 = 0$ and $ax + by + c_2 = 0$ is given by $d = \frac{|c_1 - c_2|}{\sqrt{a^2 + b^2}}$.Let the

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$2$

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(B) The distance between two parallel lines $ax + by + c_1 = 0$ and $ax + by + c_2 = 0$ is given by $d = \frac{|c_1 - c_2|}{\sqrt{a^2 + b^2}}$.Let the lines be $L_1: 3x + 4y + 5 = 0$ and $L_2: 3x + 4y - 5 = 0$.The line $L: 3x + 4y + \lambda = 0$ divides the distance between $L_1$ and $L_2$ in the ratio $3:7$.The distance between $L$ and $L_1$ is $d_1 = \frac{|\lambda - 5|}{\sqrt{3^2 + 4^2}} = \frac{|\lambda - 5|}{5}$.The distance between $L$ and $L_2$ is $d_2 = \frac{|\lambda - (-5)|}{\sqrt{3^2 + 4^2}} = \frac{|\lambda + 5|}{5}$.Given $\frac{d_1}{d_2} = \frac{3}{7}$,we have $\frac{|\lambda - 5|}{|\lambda + 5|} = \frac{3}{7}$.Assuming $-5 $7(5 - \lambda) = 3(\lambda + 5) \Rightarrow 35 - 7\lambda = 3\lambda + 15$.$10\lambda = 20 \Rightarrow \lambda = 2$.

If the line $3x + 4y + \lambda = 0$ divides the distance between the lines $3x + 4y + 5 = 0$ and $3x + 4y - 5 = 0$ in the ratio of $3:7$,then a value of $\lambda$ is

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