Let be a line and
a point not on
. Prove that there is exactly one plane through
containing every point of
.
Proof. Existence follows from a previous exercise (Section 13.8, Exercise #12) by letting be any two distinct points on
, then there is a plane
containing
and every point on the line
through
and
.
Now, to prove there is only one such plane we apply Theorem 13.10 to the points to conclude the plane that contains
is unique. But, if we choose any other two points on
, the plane
must still contain
and
(since it contains every point on
); hence, it is the same plane
. Thus, there is exactly one plane
containing
and