Minimum area of a triangle Question: A line is drawn through the point (1, 2) to meet the coordinate axis at P And Q such that OPQ is a triangle with O as the origin. If area of triangle OPQ is least, then what is the slope of line PQ.
I have been given four options
A(-1/4), B(-4), C(-2), D (-1/2)
So I found all four equations of line with given slopes, found the length of intercepts of each and then found the area of each triangle (b = base, h = height) using
$$area(b, h) = \frac1{2}bh$$
My answer came out to be slope = -2
But I was wondering if this is least area out of all possible slopes  of line passing throught (1, 2) or just these four slopes, also is there any concept which we can use to directly obtain the slope of least area without checking each option.
Additional information: I have found slope 1.1 which gives area less than given by slope -2.

Like Gabor,
$$\dfrac{y-2}{x-1}=m\iff\dfrac x{(2-m)/m}+\dfrac y{2-m}=1$$
$$\implies A=\left|\dfrac{(2-m)^2}{-2m}\right|$$
Now if $m>0,$ $$\dfrac{(2-m)^2}m=\dfrac4m+m-4\ge2\sqrt{\dfrac4m\cdot m}-4\ge0$$ the equality occurs of $m=2$
if $m<0, n=-m$(say),
$$\dfrac{(2-m)^2}{-m}=\dfrac{(2+n)^2}n=\dfrac4n+n+4\ge2\sqrt{\dfrac4n\cdot n}+4\ge8$$ the equality occurs of $n=2$
