I have a 2-Dimensional Closed Convex Compact Body $\mathbb{S}$(a set, for eg, a circular disc)). Assume, it has a non-zero intersection with the all-negative quadrant. Consider the following 2-D point $P_1$
\begin{align}
x_{min}= \min_{(x,y)\in \mathbb{S}}x \\
y'= \min_{(x_{min},y)\in \mathbb{S}}y\\
P_1=(x_{min},y')
\end{align}
and also the point $P_2$
\begin{align}
y_{min}= \min_{(x,y)\in \mathbb{S}}y \\
x'= \min_{(x,y_{min})\in \mathbb{S}}x\\
P_2=(x',y_{min})
\end{align}
Simply put, $P_1$ is the left-most vertical edge ( west-most point) and $P_2$ is the bottom-most horizontal edge (south-most point). (its ok if you assume both edges are kind-of sharp). The question is
*Given $P_1$ and $P_2$, and also using properties of $\mathbb{S}$ (convex,compact,closed), how can we tell, whether the given body $\mathbb{S}$ have an intersection with $x=y$ line in the negative quadrant. *
You can't. Let $A=(-2,-3)$, $B=(-1,-4)$, $C=(-1,0)$. Both the line segment $AB$ and the triangle $ABC$ have the same $P_1=A$ and $P_2=B$, but only the latter intersects the line $x=y$.