Find the equation of the straight line passing through the point (3, –4) and making X and Y- intercepts which are in the ratio 2:3.

The equation of the st.line whose intercepts on co-ordinate axes as  a, b is $\frac{\mathrm{x}}{\mathrm{a}}+\frac{\mathrm{y}}{\mathrm{b}}=1 ...\left(1\right)$
equation(1) making x-intercept = a
and y-intercept = b
Given a : b = 2 : 3
$$\begin{gathered} \Rightarrow \frac{\mathrm{a}}{\mathrm{b}}=\frac{2}{3}\Rightarrow 3\mathrm{a}=2\mathrm{b}\Rightarrow \mathrm{a}=\frac{2\mathrm{b}}{3} \\ \left(1\right)\Rightarrow \frac{\mathrm{x}}{\frac{2\mathrm{b}}{3}}+\frac{\mathrm{y}}{\mathrm{b}}=1\Rightarrow \frac{3\mathrm{x}}{2\mathrm{b}}+\frac{\mathrm{y}}{\mathrm{b}}=1 ...\left(2\right) \end{gathered}$$
Equation (2) passing through (3, – 4) then 
$\frac{3\left(3\right)}{2\mathrm{b}}-\frac{4}{\mathrm{b}}=1\Rightarrow \frac{9}{2\mathrm{b}}-\frac{4}{\mathrm{b}}=1$
$$\begin{gathered} \Rightarrow 9-8=2\mathrm{b} \\ 1=2\mathrm{b} \\ \Rightarrow \mathrm{b}=\frac{1}{2} \\ \Rightarrow \mathrm{a}=\frac{1}{3} \end{gathered}$$
$\therefore \text{ equation of st.line is }3\mathrm{x}+2\mathrm{y}=1$