## Isometry of line

Can someone verify that my proof is correct?

I have to prove that if $A,B$ are points of line $p$ and $A_1, B_1$ are points of line $p_1$ and if $AB=A_1B_1$, then there is isometry $I$ which transforms line $p$ into line $p_1$ and also is $I(A)=A_1, I(B)=B_1$.

If $AB=A_1B_1$ then exists $I(A) = A_1 \wedge I(B) = B_1$ from definition of isometry $I(A) = A_1 \wedge I(B) = B_1 \Rightarrow AB = A_1 B_1$.

Knowing that isometry preserves congruence, colinearity (hope grammar term is correct) and order of points then we can chose any point on line $p$ for example point $C$ and prove it that if $I(C) = C_1$ then $AC=A_1C_1$, so we can conclude that there is $I(p)=p_1$.