Find the distance between the lines 3x + 2y + | vedclass

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(B) The given lines are $L_1: 3x + 2y + 7 = 0$ and $L_2: 6x + 4y + 3 = 0$. First,rewrite $L_2$ in the form $ax + by + c_2 = 0$ by dividing by $2$: $3x

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$\frac{11}{2\sqrt{13}}$

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(B) The given lines are $L_1: 3x + 2y + 7 = 0$ and $L_2: 6x + 4y + 3 = 0$. First,rewrite $L_2$ in the form $ax + by + c_2 = 0$ by dividing by $2$: $3x + 2y + 1.5 = 0$. The distance $d$ between two parallel lines $ax + by + c_1 = 0$ and $ax + by + c_2 = 0$ is given by $d = \frac{|c_1 - c_2|}{\sqrt{a^2 + b^2}}$. Here,$a = 3$,$b = 2$,$c_1 = 7$,and $c_2 = 1.5$. $d = \frac{|7 - 1.5|}{\sqrt{3^2 + 2^2}} = \frac{5.5}{\sqrt{9 + 4}} = \frac{11/2}{\sqrt{13}} = \frac{11}{2\sqrt{13}}$.

Find the distance between the lines $3x + 2y + 7 = 0$ and $6x + 4y + 3 = 0$.

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