The paper of Preparata and Hong says:

>  We let $l_A$ and $r_A$ be two points of A such that $x_2(l_A) = \min_i {x_2(a_i)}$ and $x_2(r_A) = \max_i{x_2(a_i)}$; similarly $l_B$ and $r_B$ are defined in B.

and:

> Without loss of generality, we shall also assume that $r_A = a_1$ and $r_B = b_1$.

So, in this notation, vertices in your two polygons will be numbered differently:

$$A=((1.2,1),(0.2,0.2),(0.3,1.3))$$
$$B=((3,2),(2,2),(2,3))$$

In simple words, the algorithm starts from rightmost vertices on both polygons and continues clockwise potentially until leftmost vertices - so, only lower sides of both polygons will be involved. Tangential vertices $(0.2,0.2)$ and $(3,2)$ will be found.

This paper has been published long time ago, and since then this algorithm has been clarified and simplified. The newest version of it starts from the **rightmost** vertex on the **left** polygon and the **leftmost** vertex on the **right** polygon, and continues clockwise for the left polygon and counterclockwise for the right polygon (to find *lower* tangential line). Also they usually check a vertex location relative to a line using the dot product instead of calculating and comparing slopes - because the tangential vertex has to have both its neighbors on the same side of the tangential line. For more information, pictures and code - please see [this monograph][1], Section 3.8.


  [1]: https://www.amazon.com/Computational-Geometry-Cambridge-Theoretical-Paperback/dp/0521649765