Multiple methods would work

The equation of that line cuts the x-asis at $(a,0)$ and the y-axis at $(0,b)$ . Hence, the area is $S=\dfrac{ab}{2}$ .

Also, since $(\alpha,\beta)$ belongs to that line, $\dfrac{\alpha}{a}+\dfrac{\beta}{b}=1$ .

By the AM-GM inequality we know that:

$\dfrac{\alpha}{a}+\dfrac{\beta}{b} \geq 2\sqrt{\dfrac{\alpha}{a}\cdot\dfrac{\beta}{b}}$

$1\geq 2\sqrt{\dfrac{\alpha}{a}\cdot\dfrac{\beta}{b}}$

$\dfrac{1}{4} \geq \dfrac{\alpha \beta}{ab}$

$ab \geq 4 \alpha \beta$

$\dfrac{ab}{2} \geq 2 \alpha \beta$

$S \geq \boxed{2 \alpha \beta}$

By drawing the graph of this line $\dfrac{x}{a} +\dfrac{y}{b} = 1$

we can see that the area of the triangle formed is $\dfrac{1}{2} ab$

Hence, $S = \dfrac{1}{2} ab$

Expressing $b$ in terms of $S$ and $a$ we get:

$b = \dfrac{2S}{a}$

Substituting the point $(\alpha,\beta)$ in the line equation and that $b = \dfrac{2S}{a}$ , we get:

$\dfrac{\alpha}{a} + \dfrac{a\beta}{2S} = 1$

Take L.C.M and conver it into a quadratic in $a$ , which is

$a^{2}\beta - 2Sa + 2S\alpha = 0$

Since $a$ is real,

Discriminant is $\geq 0$ ,

$2^2 S^2 - 4\times\beta\times 2S\alpha \geq 0$

$4S^2 \geq 8\times S \alpha \times \beta$

$\Rightarrow S_{\text{min}} = \boxed{2\alpha\beta}$