First time, I found a line associated with antipodal points, detail:
Let $ABC$ be a triangle, $(C)$ is circumconic of $ABC$. $P$ and $P'$ are two antipodal points. Construct three lines through $P'$ and parallel to $PA$, $PB$, $PC$ meets $BC$, $CA$, $AB$ respectively at three collinear points, the new line through the center of circumconic.
Question: Is a line associated with antipodal points above known?
Update: But the fact, the result is generalization of the Simson line, I reformulate as follows:
Let $ABC$ be a triangle, $P$ be a point in the plane, let $(C)$ is the Nine point conic of $A$, $B$, $C$, $P$. Let $O$ be arbitrary point on $C$, $P'$ is the reflection of $P$ in $O$. Then three lines through $P'$ and parallel to $PA$, $PB$, $PC$ meet three lines $BC$, $CA$, $AB$ respectively at three collinear point.
When $P$ is the orthorcenter, the line is the Simson line of $P'$
Question again: The generalization of the Simson line above is known?
See also:
Best Answer
This is not an answer, but I'd like to point out that the concepts in question are projective, although they have a special Euclidean case.
Consider the diagram below. Start with a conic $\gamma$(green), a triangle $ABC$ inscribed in $\gamma$, and a line $\omega$ (black dot-dashed). Let $X$ be the polar of $\omega$ wrt the conic, and draw a line (dotted) through $X$ that meets $\omega$ at $P,P'$. Let the dashed lines through $P'$ meet the respective lines from $P$ to $A,B,C$ at $\omega$. Then the dashed lines meet the triangle sides at collinear points (red), and $X$ lies on this line.
The OP is the special case when $\omega$ is the projective line at infinity.
So, if the line in OP Question 1 is known, and anybody is trying to hunt it down, it may be in the projective geometry literature.