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

Answer 1

Let m be the slope of the line PQ, then the equation of PQ is

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

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

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

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

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

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

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.