A straight line forms a triangle of area 24 sq. units with the co-ordinate axes in first quadrant. Find the equation of the straight line if it passes through (3, 4).

Let the equation of the line in
$\text{ intercept form is }\frac{\mathrm{x}}{\mathrm{a}}+\frac{\mathrm{y}}{\mathrm{b}}=1$   ...(1)
The area of the triangle with the coordinate axes $\frac{1}{2}\left|\mathrm{ab}\right|=24\Rightarrow \mathrm{ab}=48\Rightarrow \mathrm{b}=\frac{48}{\mathrm{a}}(\because \mathrm{a}>0,\mathrm{b}>0)$
$\text{ sub. ' }\mathrm{b}\text{ ' in eq'n (1) }\frac{\mathrm{x}}{\mathrm{a}}+\frac{\mathrm{y}}{{\displaystyle \frac{48}{\mathrm{a}}}}=1$
$\frac{\mathrm{x}}{\mathrm{a}}+\frac{\mathrm{ay}}{48}=1 ...\left(2\right)$
Eq’n (2) passes through (3, 4)
$\frac{3}{\mathrm{a}}+\frac{4\mathrm{a}}{48}=1\Rightarrow \frac{3}{\mathrm{a}}+\frac{\mathrm{a}}{12}=1\Rightarrow 36+{\mathrm{a}}^2=12\mathrm{a}$
$$\begin{gathered} \Rightarrow {\mathrm{a}}^2-12\mathrm{a}+36=0 \\ \Rightarrow (\mathrm{a}-6{)}^2=0 \\ \Rightarrow \mathrm{a}=6 \end{gathered}$$
$\therefore \text{ the eq'n of st.line }\frac{\mathrm{x}}{6}+\frac{\mathrm{y}}{8}=1\Rightarrow 4\mathrm{x}+3\mathrm{y}=24$