A line is drawn through the point \((1,2)\) to meet the co-ordinate axes at \(\mathrm{P}\) and \(\mathrm{Q}\) such that it forms a \(\triangle O P Q\), where \(O\) is the origin. If the area of \(\triangle O P Q\) is least, then the slope of the line PQ is

The equation of line PQ passing through \((1,2)\) is \(y-2=\mathrm{m}(x-1)\)

|(\mathrm{A}(\triangle \mathrm{OPQ})=\frac{1}{2} \times \mathrm{OP} \times \mathrm{OQ} \)
\( =\frac{1}{2}\left(1-\frac{2}{\mathrm{~m}}\right)(2-\mathrm{m}) \)
\( =\frac{1}{2}\left(4-\mathrm{m}-\frac{4}{\mathrm{~m}}\right) \)
\( \mathrm{A} =2-\frac{\mathrm{m}}{2}-\frac{2}{\mathrm{~m}} \)
\( \therefore \quad \frac{\mathrm{dA}}{\mathrm{dm}} =-\frac{1}{2}+\frac{2}{\mathrm{~m}^2} \)
\( \mathrm{Now} \frac{\mathrm{dA}}{\mathrm{dm}}=0 \)
\( \Rightarrow- \frac{1}{2}+\frac{2}{\mathrm{~m}^2}=0 \)
\( \Rightarrow \mathrm{m}^2 =4 \)
\( \Rightarrow \mathrm{m} = \pm 2 \)
\( \frac{\mathrm{d}^2 \mathrm{~A}}{\mathrm{dm}^2} =-\frac{4}{\mathrm{~m}^3} \)
\( \mathrm{At} \mathrm{m} =2 \)
\( \frac{\mathrm{d}^2 \mathrm{~A}}{\mathrm{dm}^2} < 0\)
\(
\begin{aligned}
\text { At } \mathrm{m} & =-2, \\
\frac{\mathrm{d}^2 \mathrm{~A}}{\mathrm{dm}^2} & >0
\end{aligned}
\)
\(\therefore \quad\) Area of \(\triangle \mathrm{OPQ}\) will be least at \(\mathrm{m}=-2\) \(\Rightarrow\) Slope of the line \(P Q=-2\)