Equation of a line passing through $$\left( {{x_1},{y_1}} \right)$$ having
slope $$m$$ is given by $$y - {y_1} = m\left( {x - {x_1}} \right)$$
Since the line $$PQ$$ is passing through $$(1,2)$$ therefore its
equation is
$$\left( {y - 2} \right) = m\left( {x - 1} \right)$$
where $$m$$ is the slope of the line $$PQ$$.
Now, point $$P\left( {x,0} \right)$$ will also satisfy the equation of $$PQ$$
$$\therefore$$ $$y - 2 = m\left( {x - 1} \right)$$
$$ \Rightarrow 0 - 2 = m\left( {x - 1} \right)$$
$$ \Rightarrow - 2 = m\left( {x - 1} \right)$$
$$ \Rightarrow x - 1 = {{ - 2} \over m}$$
$$ \Rightarrow x = {{ - 2} \over m} + 1$$
Also, $$OP = \sqrt {\left( {x - 0} \right){}^2 + {{\left( {0 - 0} \right)}^2}} = x = {{ - 2} \over m} + 1$$
Similarly, point $$Q\left( {0,y} \right)$$ will satisfy equation of $$PQ$$
$$\therefore$$ $$y - 2 = m\left( {x - 1} \right)$$
$$ \Rightarrow y - 2 = m\left( { - 1} \right) \Rightarrow y = 2 - m$$b
and $$OQ = y = 2 - m$$
Area of $$\Delta POQ = {1 \over 2}\left( {OP} \right)\left( {OQ} \right) = {1 \over 2}\left( {1 - {2 \over m}} \right)\left( {2 - m} \right)$$
( As Area of $$\Delta = {1 \over 2} \times $$ base $$\,\, \times \,\,$$ height )
$$ = {1 \over 2}\left[ {2 - m - {4 \over m} + 2} \right] = {1 \over 2}\left[ {4 - \left( {m + {4 \over m}} \right)} \right]$$
$$ = 2 - {m \over 2} - {2 \over m}$$
Let Area $$ = f\left( m \right) = 2 - {m \over 2} - {2 \over m}$$
Now, $$f'\left( m \right) = {{ - 1} \over 2} + {2 \over {{m^2}}}$$
Put $$f'\left( m \right) = 0$$
$$ \Rightarrow {m^2} = 4 \Rightarrow m = \pm 2$$
Now, $$f''\left( m \right) = {{ - 4} \over {{m^3}}}$$
$${\left. {f''\left( m \right)} \right|_{m = 2}} = - {1 \over 2} < 0$$
$${\left. {f''\left( m \right)} \right|_{m = - 2}} = {1 \over 2} > 0$$
Area will be least at $$m=-2$$
Hence, slope of $$PQ$$ is $$-2.$$