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

Let the required line meet the $$x$$-axis at $$P(a,0)$$ and the $$y$$-axis at $$Q(0,b)$$, where $$a \gt 0,\, b \gt 0$$.
The equation of the line in intercept form is

$$\frac{x}{a}+\frac{y}{b}=1 \qquad -(1)$$

Because the line passes through the given point $$(1,2)$$, substitute $$x = 1,\; y = 2$$ in $$-(1)$$:

$$\frac{1}{a}+\frac{2}{b}=1 \qquad -(2)$$

The area of the triangle $$OPQ$$ formed with the coordinate axes is

$$A=\frac{1}{2}\,|a|\,|b|=\frac{1}{2}\,ab \qquad -(3)$$

We have to minimise $$A$$ subject to the condition $$-(2)$$.

Solve $$-(2)$$ for $$a$$:

$$\frac{1}{a}=1-\frac{2}{b}=\frac{b-2}{b}\;\; \Longrightarrow\;\; a=\frac{b}{\,b-2\,}, \quad b \gt 2$$

Substitute this in $$-(3)$$ to express $$A$$ only in terms of $$b$$:

$$A(b)=\frac{1}{2}\,\frac{b}{\,b-2\,}\,b=\frac{1}{2}\,\frac{b^{2}}{\,b-2\,} \qquad -(4)$$

To find the minimum, differentiate:

$$\frac{dA}{db}=\frac{1}{2}\,\frac{(2b)(b-2)-b^{2}}{(b-2)^{2}}=\frac{1}{2}\,\frac{2b(b-2)-b^{2}}{(b-2)^{2}}=\frac{1}{2}\,\frac{b(b-4)}{(b-2)^{2}}$$

Set $$\dfrac{dA}{db}=0$$:

$$b(b-4)=0 \;\;\Longrightarrow\;\; b=0 \text{ or } b=4$$

The admissible range is $$b \gt 2$$, hence $$b=4$$.

With $$b=4$$, equation $$a=\dfrac{b}{b-2}$$ gives

$$a=\frac{4}{4-2}=2$$

Slope of the line $$PQ$$:

$$m=\frac{0-b}{a-0}=-\frac{b}{a}=-\frac{4}{2}=-2$$

Thus, the line for which the triangle $$OPQ$$ has the least area has slope $$-2$$.

Option C which is: $$-2$$