# Math Help - Isometry Preserves Straight Lines Proof?

1. ## Isometry Preserves Straight Lines Proof?

I'm asked to prove as a corollary to: the image of the line segment PQ under F is a line segment between F(P) and F(Q), that an isometry preserves straight lines.

Here is the proof I've written down:
Let F be an isometry.
Let F(P) be denoted by P'.
Let P,Q be arbitrary points on a line L. Let X be a point on the line segment PQ.

Since F preserves distances we know that: d(P,X)=d(P',X') and d(Q,X)=d(Q',X').
We have d(P,Q)=d(P,X)+d(X,Q).
By assumption of F preserving distance we also have d(P',Q')=d(P',X')+d(X',Q').

Thus the image of segment PQ is contained in its image, segment P'Q'; but since P,Q are arbitrary, the segment PQ may be considered indefinitely long. Thus the line L is mapped to a straight line L', since the arbitrarily large segments composing that line are straight.

Is it correct? Am I writing too much? Is there a better way? Am I missing a second part? ie. to prove to sets equal you must show each belongs to the other.

2. Originally Posted by Ultros88
I'm asked to prove as a corollary to: the image of the line segment PQ under F is a line segment between F(P) and F(Q), that an isometry preserves straight lines. Is it correct? Am I writing too much? Is there a better way?
It is difficult to give a definite answer to your questions because definitions and theorem orders differ. But generally you work is correct. You have shown that a line segment is mapped into an line segment. You may want to expand to include the entire line.

Recall that if three points, P Q & X, are collinear then P-X-Q, P-Q-X, or X-P-Q.
You have done the first, the segment. The others follow from what you have done.