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

Answer 1

The correct option is D $-2$

Let ${}^{\prime }{m}^{\prime }$ be the slope of the line $PQ$, then the equation of $PQ$ is

$y-2=m(x-1)$

Now, $PQ$ meet $X-axis$ at $p\left(1-\frac{2}{m},0\right)$ and $Y-axis$ at $(0,2-m)$

$OP=1-\frac{2}{m}andOQ=2-m$

Also, area of $\Delta OPQ=\frac{1}{2}\left(OP\right)\left(OQ\right)$

Let $f\left(m\right)=4-\left(m+\frac{4}{m}\right)$

$f^1\left(m\right)=-1+\frac{4}{m^2}$

Now, $f^1\left(m\right)=0$

$m=\pm 2$

$f\left(2\right)=8$

$f(-2)=8$

Since, the area cannot be zero

Hence, the required value of $m$ is $-2$

So option $D$ is correct answer.