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Prove that there is exactly one plane through a line and a point not on the line

Let L be a line and P a point not on L. Prove that there is exactly one plane through P containing every point of L.


Proof. Existence follows from a previous exercise (Section 13.8, Exercise #12) by letting Q,R be any two distinct points on L, then there is a plane M containing P,Q,R and every point on the line L through Q and R.
Now, to prove there is only one such plane we apply Theorem 13.10 to the points P,Q,R to conclude the plane that contains P,Q,R is unique. But, if we choose any other two points on L, the plane M must still contain Q and R (since it contains every point on L); hence, it is the same plane M. Thus, there is exactly one plane M containing P and L. \qquad \blacksquare

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