Find the equation of straight lines which passes through the point (3,-2) and cuts off positive intercepts on the x-axis and y-axis, which are in the ratio 4:3.

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$$\begin{gathered} \mathrm{The} \mathrm{line} \mathrm{passing} \mathrm{through} \left(3,-2\right) \mathrm{will} \mathrm{be} \\ \mathrm{y}+2=\mathrm{m}\left(\mathrm{x}-3\right) \\ \mathrm{put} \mathrm{x}=0, \\ \mathrm{we} \mathrm{get}, \\ \mathrm{y}+2=-3\mathrm{m} \\ \mathrm{y}=-3\mathrm{m}-2 \\ \mathrm{put} \mathrm{y}=0, \mathrm{we} \mathrm{get} \\ 2=\mathrm{mx}-3\mathrm{m} \\ 2+3\mathrm{m}=\mathrm{mx} \\ \mathrm{x}=3+\frac{2}{\mathrm{m}} \\ \mathrm{now} \mathrm{given} \\ \frac{\mathrm{x}}{\mathrm{y}}=\frac{4}{3} \\ \frac{3+{\displaystyle \frac{2}{\mathrm{m}}}}{-3\mathrm{m}-2}=\frac{4}{3} \\ \frac{3\mathrm{m}+2}{-\mathrm{m}\left(3\mathrm{m}+2\right)}=\frac{4}{3} \\ \mathrm{m}=\frac{-3}{4} \\ \mathrm{hence} \mathrm{eqn} \mathrm{of} \mathrm{line} \mathrm{will} \mathrm{be}: \\ \mathrm{y}+2=\frac{-3}{4}\left(\mathrm{x}-3\right) \\ 4\mathrm{y}+8=-3\mathrm{x}+9 \\ 4\mathrm{y}+3\mathrm{x}=1 \end{gathered}$$

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