# How to prove that in PG(2,R), if a projectivity takes 3 points to themselves, then the projectivity takes all points to themselves?

If I have $$(P,Q,R,X,Y,Z,...)\barwedge(P,Q,R,X',Y',Z',...)$$ and all points lie on line $$l$$,

How do I go about proving $$(X,Y,Z,...)=(X',Y',Z',...)$$?

By a theorem, there exists a unique projectivity taking P,Q,R to itself.

If I define it by $$f$$, such that, $$f(a)=a$$,

Can I say, since there is no other projectivity apart from $$f$$ that takes P,Q,R to itself, in any tuple taking these points to themselves, the only possible projectivity is f. hence $$f(X,Y,Z) = X', Y', Z'$$

I can understand that it has to be true, but what is the correct way to prove this?

• PLease clarify: do you need help proving the theorem you quote, about the unique projectivity, or can you assume its truth and just need to apply it? Also, what's your definition of a projectivity? – MvG Oct 21 '18 at 16:02