L_1 equiv ax-3y+5=0 and L_2 equiv 4x-6y+8=0 a | vedclass

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(B) Since $L_1$ and $L_2$ are parallel,their slopes are equal.For $L_2 \equiv 4x-6y+8=0$,the slope is $m_2 = -\frac{4}{-6} = \frac{2}{3}$.Since $L_1$

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$2x+3y=0$

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(B) Since $L_1$ and $L_2$ are parallel,their slopes are equal.For $L_2 \equiv 4x-6y+8=0$,the slope is $m_2 = -\frac{4}{-6} = \frac{2}{3}$.Since $L_1$ is parallel to $L_2$,the slope of $L_1 \equiv ax-3y+5=0$ is $\frac{a}{3} = \frac{2}{3}$,which gives $a=2$.For $L_1: 2x-3y+5=0$,the $X$-intercept $p$ is found by setting $y=0$,so $2p+5=0 \implies p = -\frac{5}{2}$.The $Y$-intercept $q$ is found by setting $x=0$,so $-3q+5=0 \implies q = \frac{5}{3}$. Point $(p, q) = (-\frac{5}{2}, \frac{5}{3})$.For $L_2: 4x-6y+8=0$,the $X$-intercept $m$ is found by setting $y=0$,so $4m+8=0 \implies m = -2$.The $Y$-intercept $n$ is found by setting $x=0$,so $-6n+8=0 \implies n = \frac{4}{3}$. Point $(m, n) = (-2, \frac{4}{3})$.The slope of the line passing through $(p, q)$ and $(m, n)$ is $M = \frac{\frac{4}{3} - \frac{5}{3}}{-2 - (-\frac{5}{2})} = \frac{-\frac{1}{3}}{\frac{1}{2}} = -\frac{2}{3}$.The equation of the line is $y - \frac{4}{3} = -\frac{2}{3}(x + 2) \implies 3y - 4 = -2x - 4 \implies 2x + 3y = 0$.

$L_1 \equiv ax-3y+5=0$ and $L_2 \equiv 4x-6y+8=0$ are two parallel lines. If $p, q$ are the intercepts made by $L_1=0$ and $m, n$ are the intercepts made by $L_2=0$ on the $X$ and $Y$ coordinate axes respectively,then the equation of the line passing through the points $(p, q)$ and $(m, n)$ is

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