Prove that if line $l$ is incident with plane $P$, then there is a point of $P$ that is not incident with $l$.

geometry

Here's my attempt at a proof.

Proof. Suppose line $l$ is incident with plane $P$. Then $P$ contains all points of $l$, that is at least two points. Any two points on $l$ are collinear. Since $P$ is a set of at least three distinct points that are not collinear, there exists a point $A$ in $P$ not incident with $l$, for if it was incident with $l$, it would be collinear with the points of $l$.

I feel like I really haven't proven it, but using the fact of collinearity maybe I have. Can someone validate this as correct or incorrect? Thank you!

Here are the axioms I can work with:

(1) A line is a set of points incident with at least two points.

(2) Two distinct points are incident with exactly one line.

(3) A plane is a set of at least three distinct points that are not collinear.

(4) Three distinct points that are not incident with the same line are incident with exactly one plane.

(5) If a plane is incident with two points of a line, itis incident with the whole line.

(6) If two planes are incident with one point, they are incident with a second point.

Best Answer

This is how I would write this proof:

Suppose line $l$ is incident with plane $P$. By axiom (3) there exist three points incident with $P$ which are not collinear. (At least) one of them is not incident with $l$, because otherwise they would be collinear.

Therefore in your proof this part

Then P contains all points of $l$, that is at least two points. Any two points on $l$ are collinear.

is unnecessary and this part

Since $P$ is a set of at least three distinct points that are not collinear, there exists a point $A$ in $P$ not incident with $l$, for if it was incident with $l$, it would be collinear with the points of $l$.

is in my opinion logically incorrect (you didn't actually prove the existence of this point)

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