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

The correct option is C $-2$

Let $m$ be the slope of the line $PQ$, then the equation of $PQ$ is $y-2=m(x-1)$

Now, $PQ$ meets the $X-$axis at $P(1-\frac{2}{m},0)$ and $y-$axis at $Q(0,2-m)$

$OP=1-\frac{2}{m}$ and $OQ=2-m$

Also, Area of $△OPQ=\frac{1}{2}\times OP\times OQ$

$=\frac{1}{2}\left|(1-\frac{2}{m})(2-m)\right|$

$=\frac{1}{2}|2-m-\frac{4}{m}+2|$

$=\frac{1}{2}|4-(m+\frac{4}{m})|$

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

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

Now, $f^{\prime }\left(m\right)=0$

$m=\pm 2$

$f\left(2\right)=0$

$f(-2)=8$

Since, the area cannot be zero, hence the required value of $m$ is $-2$