Let $$\left( {h,\,k} \right)$$  be a fixed point where $$h > 0,\,k > 0.$$    A straight line passing through this point cuts the positive direction of the coordinate axes at the points $$P$$ and $$Q.$$ Then the minimum area of the $$\Delta OPQ.\,O$$   being the origin, is :

Solution :
Let the equation of any line passing through $$A\left( {h,\,k} \right){\text{ be }}y - k = m\left( {x - h} \right).$$

Let this line cut the $$x$$-axis and $$y$$-axis at $$P$$ and $$Q.$$
Then $$P \equiv \left( {h - \frac{k}{m},\,0} \right){\text{ and }}Q \equiv \left( {0,\,k - mh} \right).$$
Let $$S$$ be the area of $$\Delta OPQ,$$   then
$$\eqalign{ & S = \frac{1}{2}OP \times OQ \cr & \Rightarrow S = \frac{1}{2}\left( {h - \frac{k}{m}} \right)\left( {k - mh} \right) \cr & \Rightarrow S = \frac{1}{2}\frac{{\left( {mh - k} \right)\left( {k - mh} \right)}}{m} \cr & \Rightarrow 2mS = hkm - {k^2} - {h^2}{m^2} + khm \cr & \Rightarrow {h^2}{m^2} - 2\left( {hk - S} \right)m + {k^2} = 0 \cr} $$
Since, $$m$$ is real,
$$\therefore $$  its discriminant $$D \geqslant 0$$
$$\eqalign{ & \therefore \,4{\left( {hk - S} \right)^2} - 4{h^2}{k^2} \geqslant 0 \cr & \Rightarrow S - 2hk \geqslant 0 \cr & \Rightarrow S \geqslant 2hk \cr} $$
Hence, minimum value of $$S$$ is $$2hk$$  square units.