# Isometry of line

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• Aug 13th 2006, 03:08 PM
OReilly
Isometry of line
Can someone verify that my proof is correct?

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

If \$\displaystyle AB=A_1B_1\$ then exists \$\displaystyle I(A) = A_1 \wedge I(B) = B_1 \$ from definition of isometry \$\displaystyle 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 \$\displaystyle p\$ for example point \$\displaystyle C\$ and prove it that if \$\displaystyle I(C) = C_1 \$ then \$\displaystyle AC=A_1C_1\$, so we can conclude that there is \$\displaystyle I(p)=p_1\$.