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