Question

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 :

• A. $-\frac{1}{4}$
• B. -4
• C. -2
• D. $-\frac{1}{2}$

Answer 1

Answer: (C)

Equation of a line passing through $(x_1,y_1)$ having slope m is given by $y-y_1=m(x-x_1)$
Since the line PQ is passing through (1,2) therefore its equation is $(y-2)=m(x-1)$
where m is the slope of the line PQ.
Now, point P (x,0) will also satisfy the equation of PQ
$\therefore y-2=m(x-1)\Rightarrow 0-2=m(x-1)$
$\Rightarrow -2=m(x-1)\Rightarrow x-1=\frac{-2}{m}$
$\Rightarrow x=\frac{-2}{m}+1$
Also , $OP=\sqrt{(x-0{)}^2+(0-0{)}^2}=x$
$=\frac{-2}{m}+1$
Similarly, point Q (0,y) will satisfy equation of PQ
$\therefore y-2=m\left(x-1\right)$
$\Rightarrow y-2=m\left(-1\right)$
$\Rightarrow y=2-m$ and $OQ=y=2-m$
Area of $\Delta POQ=\frac{1}{2}\left(OP\right)\left(OQ\right)$
$=\frac{1}{2}\left(1-\frac{2}{m}\right)\left(2-m\right)$
$\left(\because \text{Area}\text{of}\Delta =\frac{1}{2}\times \text{base}\times \text{height}\right)$
$=\frac{1}{2}\left[2-m-\frac{4}{m}+2\right]=\frac{1}{2}\left[4-\left(m+\frac{4}{m}\right)\right]$
$=2-\frac{m}{2}-\frac{2}{m}$

Let Area $=f\left(m\right)=2-\frac{m}{2}-\frac{2}{m}$
Now, $f^{\prime }\left(m\right)=\frac{-1}{2}+\frac{2}{m^2}$
Put $f^{\prime }\left(m\right)=0$
$\Rightarrow m^2=4\Rightarrow m=\pm 2$
Now, $f^{\prime \prime }\left(m\right)=\frac{-4}{m^3}$
$f^{\prime \prime }\left(m\right){|}_{m=2}=-\frac{1}{2}<0$
$f^{\prime \prime }\left(m\right){|}_{m=-2}=\frac{1}{2}<0$
Area will be least at m = -2
Hence, slope of PQ is -2.