If the x-intercept of some line L is double t | vedclass

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(D) The given line is $3x + 4y = 12$. Dividing by $12$,we get $\frac{x}{4} + \frac{y}{3} = 1$. Thus,the $x-$intercept is $4$ and the $y-$intercept is

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$-3/16$

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(D) The given line is $3x + 4y = 12$. Dividing by $12$,we get $\frac{x}{4} + \frac{y}{3} = 1$. Thus,the $x-$intercept is $4$ and the $y-$intercept is $3$. Let the line $L$ have $x-$intercept $a$ and $y-$intercept $b$. According to the problem,$a = 2 	imes 4 = 8$ and $b = \frac{3}{2}$. The equation of line $L$ in intercept form is $\frac{x}{a} + \frac{y}{b} = 1$,which becomes $\frac{x}{8} + \frac{y}{3/2} = 1$. This simplifies to $\frac{x}{8} + \frac{2y}{3} = 1$. Multiplying by $24$,we get $3x + 16y = 24$. Rewriting in slope-intercept form $y = mx + c$,we have $16y = -3x + 24$,or $y = -\frac{3}{16}x + \frac{24}{16}$. Therefore,the slope of $L$ is $-\frac{3}{16}$.

If the $x-$intercept of some line $L$ is double that of the line $3x + 4y = 12$ and the $y-$intercept of $L$ is half that of the same line,then the slope of $L$ is

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