If the $x$-intercept of some line $L$ is double as that of the line, $3 x+4 y=12$ and the $y$-intercept of…
If the $x$-intercept of some line $L$ is double as that of the line, $3 x+4 y=12$ and the $y$-intercept of $L$ is half as that of the same line, then the slope of $L$ is :
$-3$
$-3 / 8$
$-3 / 2$
$-3 / 16$
Solution
Given line $3 x+4 y=12$ can be rewritten as
$
\begin{aligned}
& \frac{3 x}{12}+\frac{4 y}{12}=1 \Rightarrow \frac{x}{4}+\frac{y}{3}=1 \\
& \Rightarrow x \text {-intercept }=4 \text { and } y \text {-intercept }=3
\end{aligned}
$
Let the required line be
$\mathrm{L}: \frac{x}{a}+\frac{y}{b}=1$ 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 $\frac{x}{8}+\frac{2 y}{3}=1$
$\Rightarrow 3 x+16 y=24$
$\Rightarrow y=\frac{-3}{16} x+\frac{24}{16}$
Hence, required slope $=\frac{-3}{16}$