Proof that the images of parallel lines under an isometry are parallel

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Homework Statement

Let L, K be two parallel lines, and let F be an isometry. Prove that F(L) and F(K) are parallel.

Relevant Equations

F

The proof in the solutions was done much differently then mine (much simpler), so I would like feedback on whether my proof is valid or not.

Assume that F(L) and F(K) are not parallel, then they necessarily intersect. Let X' be the point of intersection. Then X' lies on both F(L) and F(K).

Let Q_L', P_L' and Q_K', P_K' be points such that the former pair lie on L' and the later on K' and such that X' lies on the segments Q_L'P_L' and Q_K'P_K'

There must be the points Q_L, P_L, and Q_K, P_K on L and K respectively such that their image under F is the corresponding points above.

By Seg1 X where F(X) = 'X must lie on both of the segments Q_LP_L and Q_KP_K and therefore on the unique line that passes through Q_Land P_L which is L, and the unique line which passes through Q_K and P_K which is K, thus X must lie on the point of intersection between L and K but this is impossible because L and K are parallel.

Seg 1: Let P,Q, M be points. We have d(P, M) = d(P,Q) + d(Q, M) if and only if Q lies on the segment between P and M.

 

[pasmith]

How do you define isometry? You speak of parallel lines, so you must be in an inner product space - say \mathbb{R}^n. Is that corrrect? Are you using the euclidean inner product?

If T is an isometry, then \begin{split}<br /> \|x\|^2 - 2\langle x,y \rangle + \|y\|^2 &amp;= \|x - y\|^2 \\ <br /> &amp;= \|T(x) - T(y)\|^2 \\<br /> &amp;= \|T(x)\|^2 - 2\langle T(x), T(y)\rangle + \|T(y)\|^2 \end{split} What can you say about \langle T(x), T(y)\rangle? What does that imply about the direction and separation of the images of two parallel lines?

 

[clone]

How do you define isometry? You speak of parallel lines, so you must be in an inner product space - say \mathbb{R}^n. Is that corrrect? Are you using the euclidean inner product?

If T is an isometry, then \begin{split}<br /> \|x\|^2 - 2\langle x,y \rangle + \|y\|^2 &amp;= \|x - y\|^2 \\<br /> &amp;= \|T(x) - T(y)\|^2 \\<br /> &amp;= \|T(x)\|^2 - 2\langle T(x), T(y)\rangle + \|T(y)\|^2 \end{split} What can you say about \langle T(x), T(y)\rangle? What does that imply about the direction and separation of the images of two parallel lines?

I don't really understand what any of that means to be honest. I guess I should have been more specific about the context. This is from Serge Lang's basic mathematics, and in the section titled "Intuitive Geometry". We are just doing 2d geometry, drawing diagrams on paper. The definition of isometry given is:

Let F be a mapping of the plane into itself. We say that F preserves distances, or is distance preserving, if and only if for every pair of points P,Q in the plane, the distance between P and Q is the same as the distance between F(P) and F(Q).

This is the proof given in the solutions for reference:

The lines F(L) and F(K) cannot have a point in common, otherwise this point would be of the form F(P) = F(Q) for some point P on L and Q on K. But the distance between P and Q is the same as the distance between F(P) and F(Q), and it would then follow that d(P, Q) = 0, so P = Q, which is impossible. Hence F(L) and F(K) have no point in common, and are therefore parallel.

Thanks for your help.

 

[Gavran]

The proof in the solutions was done much differently then mine (much simpler), so I would like feedback on whether my proof is valid or not.

Your proof is not valid because it is not complete. Based on your proof, I can not see that $X_1=X_2$ for $F(X_1)=F(X_2)=X’$. I can only see those $X_1$ lies on the segment $Q_LP_L$ and $X_2$ lies on the segment $Q_KP_K$ where $F(X_1)=F(X_2)=X’$. You have one additional statement $F(X_1)=F(X_2)\iff X_1=X_2$ which also should be proved.

The proof given in the solution (the post #3) is simple and understandable.

 

[clone]

Your proof is not valid because it is not complete. Based on your proof, I can not see that $X_1=X_2$ for $F(X_1)=F(X_2)=X’$. I can only see those $X_1$ lies on the segment $Q_LP_L$ and $X_2$ lies on the segment $Q_KP_K$ where $F(X_1)=F(X_2)=X’$. You have one additional statement $F(X_1)=F(X_2)\iff X_1=X_2$ which also should be proved.

The proof given in the solution (the post #3) is simple and understandable.

Thanks! I see now that I was looking at what actually needed to be proved from completely the wrong way