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Q.
If the x-intercept of some line $L$ is double as that of the line, $3x+4y = 12$ and the $y$-intercept of $L$, is half as that of the same line, then the slope of $L$ is:
Given line $3x + 4y= 12$ can be rewritten as
$\frac{3x}{12}+\frac{4y}{12}=1 \Rightarrow \frac{x}{4}+\frac{y}{3}=1$
$\Rightarrow $ x-intercept=4 and y-intercept= 3 Let the required line be
$L : \frac{x}{a}+\frac{y}{b}=1$ where
$a = x-$intercept and $b=y-$ intercept
According to the question
$a = 4 \times 2 = 8$ and $b = 3/2$
$\therefore $ Required line is $\frac{x}{8}+\frac{2y}{3}=1$
$\Rightarrow 3x+ 16y=24$
$\Rightarrow y=\frac{-3}{16}x+\frac{24}{16}$
Hence, required slope $=\frac{-3}{16}.$