Proof:[br]Dual Axiom 1: This axiom can be proved from the other three axioms. Note, Axiom 2 tells us that any two distinct lines have exactly one point in common. Therefore, a point must lie on at least two lines. [br]Dual Axiom 2: A counterexample of this can be shown by the affine plane of order 2. In the affine plane of order two, a point lies on three lines, thus Axiom 1 is true. Any two distinct points share exactly one line, thus Axiom 3 is true. And, there is a set of four lines such that no three are concurrent, thus Axiom 4 is true. However, there are parallel lines in this system so Axiom 2 is not true. Thus, it is independent. [br]Dual Axiom 3: Consider the geometry in which there are 4 points and 4 lines in the shape of a square. Axiom 1 is true because a point lies on two lines. Axiom 2 is true because any two distinct lines have exactly one point in common. Axiom 4 is true because there are 4 lines, no three of which are concurrent. However, Axiom 3 is not true because not all distinct points share a line. Thus, it is independent. [br]Dual Axiom 4: Now, consider the geometry in which there are 3 points and 3 lines in the shape of a triangle. Dual Axiom 1 is true because any point lies on two lines. Dual Axiom 2 is true because any two distinct lines have exactly one point in common. Dual Axiom 3 is true because any two distinct points have exactly one line in common. However, there are only 3 lines in this geometry so Axiom 4 is untrue. Thus, it is independent.