QJEEM12Mac9a4: JEE Main 2012 Mathematics - Class 12 JEEM12Mac9a4 Solution: A line is drawn through the point to meet the coordinate axes…

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 $-2.$