Let d be the distance between the parallel li | vedclass

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(A) The distance $d$ between parallel lines $Ax + By + C_1 = 0$ and $Ax + By + C_2 = 0$ is given by $d = \frac{|C_1 - C_2|}{\sqrt{A^2 + B^2}}$.For $3x

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$(3x - 2y)^2 + 24(3x - 2y) + 143 = 0$

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(A) The distance $d$ between parallel lines $Ax + By + C_1 = 0$ and $Ax + By + C_2 = 0$ is given by $d = \frac{|C_1 - C_2|}{\sqrt{A^2 + B^2}}$.For $3x - 2y + 5 = 0$ and $3x - 2y + 5 + 2\sqrt{13} = 0$,$d = \frac{|5 - (5 + 2\sqrt{13})|}{\sqrt{3^2 + (-2)^2}} = \frac{2\sqrt{13}}{\sqrt{13}} = 2$.Given $L_1: 3x - 2y + k_1 = 0$ is at distance $\frac{4d}{\sqrt{13}} = \frac{8}{\sqrt{13}}$ from $3x - 2y + 5 = 0$,we have $\frac{|k_1 - 5|}{\sqrt{13}} = \frac{8}{\sqrt{13}} \Rightarrow |k_1 - 5| = 8$. Since $k_1 > 0$,$k_1 - 5 = 8 \Rightarrow k_1 = 13$.Given $L_2: 3x - 2y + k_2 = 0$ is at distance $\frac{3d}{\sqrt{13}} = \frac{6}{\sqrt{13}}$ from $3x - 2y + 5 = 0$,we have $\frac{|k_2 - 5|}{\sqrt{13}} = \frac{6}{\sqrt{13}} \Rightarrow |k_2 - 5| = 6$. Since $k_2 > 0$,$k_2 - 5 = 6 \Rightarrow k_2 = 11$.The combined equation is $(3x - 2y + 13)(3x - 2y + 11) = 0$.Let $u = 3x - 2y$. Then $(u + 13)(u + 11) = u^2 + 24u + 143 = 0$.Substituting back,we get $(3x - 2y)^2 + 24(3x - 2y) + 143 = 0$.

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