How to show that there are exactly two motions mapping $Pto P'$ and $Qto Q'$ given that $d(P,Q)=d(P',Q')$
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Given points $P,Q,P',Q'inmathbb{E}^2$ with $d(P,Q)=d(P',Q')$. Show there exists exactly two motions (isometries) $T_{1},T_{2}$ that maps $Pto P'$ and $Qto Q'$
How exactly would I attempt proving this? I know that I can choose an appropriate coordinate system so that $P=(0,0)$ and $Q=(1,0)$ but how would I show that there are exactly two? Would it be that you split the two mappings where one is a direct mapping and the other is indirect? Or could you just construct the two mappings by composition (you take $Pto P'$ first and then take $Qto Q'$ and then the other way around - or would those mappings not be unique?)
Thanks!
geometry euclidean-geometry isometry
asked Nov 18 at 10:38
BigWig
1019
add a comment |
up vote
1
down vote
favorite
Given points $P,Q,P',Q'inmathbb{E}^2$ with $d(P,Q)=d(P',Q')$. Show there exists exactly two motions (isometries) $T_{1},T_{2}$ that maps $Pto P'$ and $Qto Q'$
How exactly would I attempt proving this? I know that I can choose an appropriate coordinate system so that $P=(0,0)$ and $Q=(1,0)$ but how would I show that there are exactly two? Would it be that you split the two mappings where one is a direct mapping and the other is indirect? Or could you just construct the two mappings by composition (you take $Pto P'$ first and then take $Qto Q'$ and then the other way around - or would those mappings not be unique?)
Thanks!
geometry euclidean-geometry isometry
asked Nov 18 at 10:38
BigWig
1019
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Given points $P,Q,P',Q'inmathbb{E}^2$ with $d(P,Q)=d(P',Q')$. Show there exists exactly two motions (isometries) $T_{1},T_{2}$ that maps $Pto P'$ and $Qto Q'$
How exactly would I attempt proving this? I know that I can choose an appropriate coordinate system so that $P=(0,0)$ and $Q=(1,0)$ but how would I show that there are exactly two? Would it be that you split the two mappings where one is a direct mapping and the other is indirect? Or could you just construct the two mappings by composition (you take $Pto P'$ first and then take $Qto Q'$ and then the other way around - or would those mappings not be unique?)
Thanks!
geometry euclidean-geometry isometry
asked Nov 18 at 10:38
BigWig
1019
Given points $P,Q,P',Q'inmathbb{E}^2$ with $d(P,Q)=d(P',Q')$. Show there exists exactly two motions (isometries) $T_{1},T_{2}$ that maps $Pto P'$ and $Qto Q'$
How exactly would I attempt proving this? I know that I can choose an appropriate coordinate system so that $P=(0,0)$ and $Q=(1,0)$ but how would I show that there are exactly two? Would it be that you split the two mappings where one is a direct mapping and the other is indirect? Or could you just construct the two mappings by composition (you take $Pto P'$ first and then take $Qto Q'$ and then the other way around - or would those mappings not be unique?)
Thanks!
geometry euclidean-geometry isometry
geometry euclidean-geometry isometry
asked Nov 18 at 10:38
BigWig
1019
asked Nov 18 at 10:38
BigWig
1019
edited Nov 18 at 11:16
asked Nov 18 at 10:38
BigWig
1019
asked Nov 18 at 10:38
BigWig
1019
asked Nov 18 at 10:38
BigWig
1019
1019
add a comment |
add a comment |
1 Answer
1
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0
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Observe that any isometry $T$ sending the pair $(P,Q)$ to $(P',Q')$ also maps the line between $l(P,Q)$ in the line $l(P',Q')$ ($l(A,B)$ being the unique line between the two points $A$ and $B$). Then, if $Xnotin l(P,Q)$, there is only another point $X'$ such that $d(P,X)=d(P,X')$ and $d(Q,X)=d(P,X')$ (namely the reflection with respect to the line). This means that every point has at most two possible images. Once such an image is fixed for one point not on $l(P,Q)$, as you were saying you can assume to have a complete frame of reference. Hence there are at most two isometries.
answered Nov 18 at 10:49
Leo163
1,540512
so would it be that a mapping takes $l(P,Q)to l(P',Q')$ and $X$ can be mapped to two possible $X'$ and its just a matter of which one to choose?
– BigWig
Nov 18 at 11:00
Exactly: a point $X$ not on $l(P,Q)$ can be mapped to exactly two points $X_1,X_2$ such that $d(P,X)=d(P',X_1)=d(P',X_2)$ and $d(Q,X)=d(Q,X_1)=d(Q,X_2)$.
– Leo163
Nov 18 at 11:05
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
Observe that any isometry $T$ sending the pair $(P,Q)$ to $(P',Q')$ also maps the line between $l(P,Q)$ in the line $l(P',Q')$ ($l(A,B)$ being the unique line between the two points $A$ and $B$). Then, if $Xnotin l(P,Q)$, there is only another point $X'$ such that $d(P,X)=d(P,X')$ and $d(Q,X)=d(P,X')$ (namely the reflection with respect to the line). This means that every point has at most two possible images. Once such an image is fixed for one point not on $l(P,Q)$, as you were saying you can assume to have a complete frame of reference. Hence there are at most two isometries.
answered Nov 18 at 10:49
Leo163
1,540512
so would it be that a mapping takes $l(P,Q)to l(P',Q')$ and $X$ can be mapped to two possible $X'$ and its just a matter of which one to choose?
– BigWig
Nov 18 at 11:00
Exactly: a point $X$ not on $l(P,Q)$ can be mapped to exactly two points $X_1,X_2$ such that $d(P,X)=d(P',X_1)=d(P',X_2)$ and $d(Q,X)=d(Q,X_1)=d(Q,X_2)$.
– Leo163
Nov 18 at 11:05
add a comment |
up vote
0
down vote
accepted
Observe that any isometry $T$ sending the pair $(P,Q)$ to $(P',Q')$ also maps the line between $l(P,Q)$ in the line $l(P',Q')$ ($l(A,B)$ being the unique line between the two points $A$ and $B$). Then, if $Xnotin l(P,Q)$, there is only another point $X'$ such that $d(P,X)=d(P,X')$ and $d(Q,X)=d(P,X')$ (namely the reflection with respect to the line). This means that every point has at most two possible images. Once such an image is fixed for one point not on $l(P,Q)$, as you were saying you can assume to have a complete frame of reference. Hence there are at most two isometries.
answered Nov 18 at 10:49
Leo163
1,540512
so would it be that a mapping takes $l(P,Q)to l(P',Q')$ and $X$ can be mapped to two possible $X'$ and its just a matter of which one to choose?
– BigWig
Nov 18 at 11:00
Exactly: a point $X$ not on $l(P,Q)$ can be mapped to exactly two points $X_1,X_2$ such that $d(P,X)=d(P',X_1)=d(P',X_2)$ and $d(Q,X)=d(Q,X_1)=d(Q,X_2)$.
– Leo163
Nov 18 at 11:05
add a comment |
up vote
0
down vote
accepted
up vote
0
down vote
accepted
Observe that any isometry $T$ sending the pair $(P,Q)$ to $(P',Q')$ also maps the line between $l(P,Q)$ in the line $l(P',Q')$ ($l(A,B)$ being the unique line between the two points $A$ and $B$). Then, if $Xnotin l(P,Q)$, there is only another point $X'$ such that $d(P,X)=d(P,X')$ and $d(Q,X)=d(P,X')$ (namely the reflection with respect to the line). This means that every point has at most two possible images. Once such an image is fixed for one point not on $l(P,Q)$, as you were saying you can assume to have a complete frame of reference. Hence there are at most two isometries.
answered Nov 18 at 10:49
Leo163
1,540512
Observe that any isometry $T$ sending the pair $(P,Q)$ to $(P',Q')$ also maps the line between $l(P,Q)$ in the line $l(P',Q')$ ($l(A,B)$ being the unique line between the two points $A$ and $B$). Then, if $Xnotin l(P,Q)$, there is only another point $X'$ such that $d(P,X)=d(P,X')$ and $d(Q,X)=d(P,X')$ (namely the reflection with respect to the line). This means that every point has at most two possible images. Once such an image is fixed for one point not on $l(P,Q)$, as you were saying you can assume to have a complete frame of reference. Hence there are at most two isometries.
answered Nov 18 at 10:49
Leo163
1,540512
answered Nov 18 at 10:49
Leo163
1,540512
answered Nov 18 at 10:49
Leo163
1,540512
answered Nov 18 at 10:49
Leo163
1,540512
1,540512
so would it be that a mapping takes $l(P,Q)to l(P',Q')$ and $X$ can be mapped to two possible $X'$ and its just a matter of which one to choose?
– BigWig
Nov 18 at 11:00
Exactly: a point $X$ not on $l(P,Q)$ can be mapped to exactly two points $X_1,X_2$ such that $d(P,X)=d(P',X_1)=d(P',X_2)$ and $d(Q,X)=d(Q,X_1)=d(Q,X_2)$.
– Leo163
Nov 18 at 11:05
add a comment |
so would it be that a mapping takes $l(P,Q)to l(P',Q')$ and $X$ can be mapped to two possible $X'$ and its just a matter of which one to choose?
– BigWig
Nov 18 at 11:00
Exactly: a point $X$ not on $l(P,Q)$ can be mapped to exactly two points $X_1,X_2$ such that $d(P,X)=d(P',X_1)=d(P',X_2)$ and $d(Q,X)=d(Q,X_1)=d(Q,X_2)$.
– Leo163
Nov 18 at 11:05
so would it be that a mapping takes $l(P,Q)to l(P',Q')$ and $X$ can be mapped to two possible $X'$ and its just a matter of which one to choose?
– BigWig
Nov 18 at 11:00
so would it be that a mapping takes $l(P,Q)to l(P',Q')$ and $X$ can be mapped to two possible $X'$ and its just a matter of which one to choose?
– BigWig
Nov 18 at 11:00
Exactly: a point $X$ not on $l(P,Q)$ can be mapped to exactly two points $X_1,X_2$ such that $d(P,X)=d(P',X_1)=d(P',X_2)$ and $d(Q,X)=d(Q,X_1)=d(Q,X_2)$.
– Leo163
Nov 18 at 11:05
Exactly: a point $X$ not on $l(P,Q)$ can be mapped to exactly two points $X_1,X_2$ such that $d(P,X)=d(P',X_1)=d(P',X_2)$ and $d(Q,X)=d(Q,X_1)=d(Q,X_2)$.
– Leo163
Nov 18 at 11:05
add a comment |
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