A line is drawn through the point $(1,2)$ to meet the co-ordinate axes at $\mathrm{P}$ and $\mathrm{Q}$ such…

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
  1. -2
  2. 2
  3. $\frac{-1}{2}$
  4. $\frac{1}{2}$

Solution

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

Asked in: MHT CET 2023 (13 May Shift 1)

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