A line is drawn through the point $(1,2)$ to meet the coordinate axes at $P$ and $Q$ such that it forms a…

A line is drawn through the point $(1,2)$ to meet the coordinate axes at $P$ and $Q$ such that it forms a triangle $OPQ$, where $O$ is the origin. If the area of the triangle $OPQ$ is least, then the slope of the line $PQ$ is
  1. $-\frac{1}{4}$
  2. $-4$
  3. $-2$
  4. $-\frac{1}{2}$

Solution

Equation of line passing through $(1,2)$ with slope $m$ is $y-2=m(x-1)$ Area of $\Delta \mathrm{OPQ}=\frac{(m-2)^2}{2|m|}$ $\Delta=\frac{m^2+4-4 m}{2 m} \quad \Delta=\frac{m}{2}+\frac{2}{m}-2$ $\Delta$ is least if $\frac{m}{2}=\frac{2}{m} \quad \Rightarrow m^2=4 \quad \Rightarrow m=\pm 2 \quad \Rightarrow m=-2$

Asked in: JEE Main 2012 (Offline)

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