The minimum area of the triangle formed by the variable line \(3 \cos \theta \cdot x+4 \sin \theta \cdot y=12\) and the coordinate axes is

Given equation of line is
\(\Rightarrow \quad \frac{x \cdot 3 \cos \theta+4 \sin \theta y=12}{(4 / \cos \theta)}+\frac{y}{(3 / \sin \theta)}=1...(i)\)
It intereset the coordinate axes at \(A\left(\frac{4}{\cos \theta}, 0\right)\) and
\(B\left(0, \frac{3}{\sin \theta}\right) \text {. }\)



\(\therefore\) Area of \(\triangle O A B\),
\(\begin{aligned}
\Delta &=\frac{1}{2} \times \frac{4}{\cos \theta} \times \frac{3}{\sin \theta} \\
&=\frac{12}{\sin 2 \theta}...(ii)
\end{aligned}\)
Now, for area to be minimum, \(\sin 2 \theta\) should be maximum i.e.,
\(\sin 2 \theta=1 \quad(\because|\sin 2 \theta| \leq 1)\)
\(\therefore\) Minimum area
\(\Delta_{\min }=\frac{12}{1}=12\)