Prove that there is exactly one perpendicular line

Consider an absolute plane (i.e. it satisfies all axioms except the parallel axiom).
Let $g$ be a straight line and $P$ a point not in $g$. Then there is a unique straight line going through $P$ which is perpendicular on $g$.

Existence: I try to construct the line perpendicular to $g$ that goes through $P$. For this let $A, B$ be two distinct points on $g$. Then $d(A,P)=r$ and $d(B,P)=r’.$
By a previous result, we can find a unique point $P’$ in the half-plane that does not contain $P$ such that $d(A,P’)=r$ and $d(B,P’)=r’$.

Now consider the line going through $P$ and $P’$. This will be the line that we’re looking for, but how do I prove that it is perpendicular to $g$ now?

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