Mathematics - Straight Lines Question with Solution | TestHub

Given line $3x+4y=12$ can be rewritten as $$\begin{array}{rl}& \frac{3x}{12}+\frac{4y}{12}=1\Rightarrow \frac{x}{4}+\frac{y}{3}=1\\ & \Rightarrow x\text{-intercept }=4\text{ and }y\text{-intercept }=3\end{array}$$ 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 $\frac{x}{8}+\frac{2y}{3}=1$ $\Rightarrow 3x+16y=24$ $\Rightarrow y=\frac{-3}{16}x+\frac{24}{16}$ Hence, required slope $=\frac{-3}{16}$.