Question

A line is drawn through the point $(1,2)$ to meet the coordinates axes at P and Q such that it forms a triangle OP Q, where O is the origin, if the area of the triangle OPQ is less, 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 $m$ be slope of line $PQ$ then equation of $PQ$ is

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

$PQ$ meet $x-$axis at $(1-\frac{2}{m},0)$

& $y-$axis at $(0,2-m)$

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

$ar△OPQ=\frac{1}{2}OP.OQ$

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

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

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

$f^{\prime }\left(m\right)=0\Rightarrow m=\pm 2$

$f\left(2\right)=0,f(-2)=8$

since area can not be zero this required value of $m16-2$