Do Reflections About Lines in 2-D Preserve Angle Measure for Figures?
$begingroup$
I read that reflections preserve angle measure but not distance because they are "flips".
$underline{textbf{Question:}}$ So, is it true that a figure that is reflected about a line preserves angle measure?
I'm going to now explain why I do not think it is true.
I'm sure there is a more succinct way to write this, but $textbf{ANYTHING WRITTEN BELOW THIS IS NOT NEEDED TO ANSWER}$
$textbf{THIS QUESTION.}$ I'm just showing my work in case there is an error.
Explanation for Why Not True:
First, I'm going to come up with a valid equation that allows us to reflect a point $P:=(x_o, y_o)$ about a line $g(x):=mx+b$ where $mneq 0$ (see bullet for end result). First, let's consider the figure below.
Since lines $t$ and $g$ are $perp$, we know $t(x)=frac{-1}{m}x+c$ where $c$ is some constant. Since $t(x_o)=y_o$, we know $-frac{1}{m}x+c=y_oimplies c=y_o+frac{1}{m}$. Thus, $t(x)=frac{-1}{m}x+(y_o+frac{1}{m})$ substituting $c$ back into our original equation. Now, we know $t(x_1)=g(x_1)implies frac{-1}{m}x_1+(y_o+frac{1}{m})=mx_1+bimplies y_o+frac{1}{m}x_o-b=(m+frac{1}{m})x_1implies x_1=frac{y_o+frac{1}{m}x_o-b}{(m+frac{1}{m})}=frac{y_o+frac{1}{m}x_o-b}{frac{m^2+1}{m}}.$ Thus, $g(x_1)=m(frac{y_o+frac{1}{m}x_o-b}{frac{m^2+1}{m}})+b$. Now, we know looking at the midpoint that $(frac{x_o+x'}{2},frac{y_o+y'}{2})=(underbrace{frac{y_o+frac{1}{m}x_o-b}{frac{m^2+1}{m}}}_{=x_1},underbrace{frac{y_o+frac{1}{m}x_o-b}{frac{m^2+1}{m}}}_{=y_1})$.
This means $x'=2(frac{y_o+frac{1}{m}x_o-b}{frac{m^2+1}{m}})-x_o$ and $y'=2(frac{y_o+frac{1}{m}x_o-b}{frac{m^2+1}{m}})-y_o$.
$bullet$ Hence, we know $x'=2cdot (y_o+frac{1}{m}x_o-b)cdot (frac{m}{m^2+1})-x_o$ and $y'=2cdot (y_o+frac{1}{m}x_o-b)cdot (frac{m}{m^2+1})-y_o$ are our last two equations.
As I have now shown how to reflect a point across a given line, I should be able to see the same angles preserved below using the transformation above. However, I do not with this example which is clearly shown in red. 
geometry
asked Dec 28 '18 at 14:26
W. G.W. G.
6101716
$endgroup$
add a comment |
$begingroup$
I read that reflections preserve angle measure but not distance because they are "flips".
$underline{textbf{Question:}}$ So, is it true that a figure that is reflected about a line preserves angle measure?
I'm going to now explain why I do not think it is true.
I'm sure there is a more succinct way to write this, but $textbf{ANYTHING WRITTEN BELOW THIS IS NOT NEEDED TO ANSWER}$
$textbf{THIS QUESTION.}$ I'm just showing my work in case there is an error.
Explanation for Why Not True:
First, I'm going to come up with a valid equation that allows us to reflect a point $P:=(x_o, y_o)$ about a line $g(x):=mx+b$ where $mneq 0$ (see bullet for end result). First, let's consider the figure below.
Since lines $t$ and $g$ are $perp$, we know $t(x)=frac{-1}{m}x+c$ where $c$ is some constant. Since $t(x_o)=y_o$, we know $-frac{1}{m}x+c=y_oimplies c=y_o+frac{1}{m}$. Thus, $t(x)=frac{-1}{m}x+(y_o+frac{1}{m})$ substituting $c$ back into our original equation. Now, we know $t(x_1)=g(x_1)implies frac{-1}{m}x_1+(y_o+frac{1}{m})=mx_1+bimplies y_o+frac{1}{m}x_o-b=(m+frac{1}{m})x_1implies x_1=frac{y_o+frac{1}{m}x_o-b}{(m+frac{1}{m})}=frac{y_o+frac{1}{m}x_o-b}{frac{m^2+1}{m}}.$ Thus, $g(x_1)=m(frac{y_o+frac{1}{m}x_o-b}{frac{m^2+1}{m}})+b$. Now, we know looking at the midpoint that $(frac{x_o+x'}{2},frac{y_o+y'}{2})=(underbrace{frac{y_o+frac{1}{m}x_o-b}{frac{m^2+1}{m}}}_{=x_1},underbrace{frac{y_o+frac{1}{m}x_o-b}{frac{m^2+1}{m}}}_{=y_1})$.
This means $x'=2(frac{y_o+frac{1}{m}x_o-b}{frac{m^2+1}{m}})-x_o$ and $y'=2(frac{y_o+frac{1}{m}x_o-b}{frac{m^2+1}{m}})-y_o$.
$bullet$ Hence, we know $x'=2cdot (y_o+frac{1}{m}x_o-b)cdot (frac{m}{m^2+1})-x_o$ and $y'=2cdot (y_o+frac{1}{m}x_o-b)cdot (frac{m}{m^2+1})-y_o$ are our last two equations.
As I have now shown how to reflect a point across a given line, I should be able to see the same angles preserved below using the transformation above. However, I do not with this example which is clearly shown in red.
geometry
asked Dec 28 '18 at 14:26
W. G.W. G.
6101716
$endgroup$
add a comment |
$begingroup$
I read that reflections preserve angle measure but not distance because they are "flips".
$underline{textbf{Question:}}$ So, is it true that a figure that is reflected about a line preserves angle measure?
I'm going to now explain why I do not think it is true.
I'm sure there is a more succinct way to write this, but $textbf{ANYTHING WRITTEN BELOW THIS IS NOT NEEDED TO ANSWER}$
$textbf{THIS QUESTION.}$ I'm just showing my work in case there is an error.
Explanation for Why Not True:
First, I'm going to come up with a valid equation that allows us to reflect a point $P:=(x_o, y_o)$ about a line $g(x):=mx+b$ where $mneq 0$ (see bullet for end result). First, let's consider the figure below.
Since lines $t$ and $g$ are $perp$, we know $t(x)=frac{-1}{m}x+c$ where $c$ is some constant. Since $t(x_o)=y_o$, we know $-frac{1}{m}x+c=y_oimplies c=y_o+frac{1}{m}$. Thus, $t(x)=frac{-1}{m}x+(y_o+frac{1}{m})$ substituting $c$ back into our original equation. Now, we know $t(x_1)=g(x_1)implies frac{-1}{m}x_1+(y_o+frac{1}{m})=mx_1+bimplies y_o+frac{1}{m}x_o-b=(m+frac{1}{m})x_1implies x_1=frac{y_o+frac{1}{m}x_o-b}{(m+frac{1}{m})}=frac{y_o+frac{1}{m}x_o-b}{frac{m^2+1}{m}}.$ Thus, $g(x_1)=m(frac{y_o+frac{1}{m}x_o-b}{frac{m^2+1}{m}})+b$. Now, we know looking at the midpoint that $(frac{x_o+x'}{2},frac{y_o+y'}{2})=(underbrace{frac{y_o+frac{1}{m}x_o-b}{frac{m^2+1}{m}}}_{=x_1},underbrace{frac{y_o+frac{1}{m}x_o-b}{frac{m^2+1}{m}}}_{=y_1})$.
This means $x'=2(frac{y_o+frac{1}{m}x_o-b}{frac{m^2+1}{m}})-x_o$ and $y'=2(frac{y_o+frac{1}{m}x_o-b}{frac{m^2+1}{m}})-y_o$.
$bullet$ Hence, we know $x'=2cdot (y_o+frac{1}{m}x_o-b)cdot (frac{m}{m^2+1})-x_o$ and $y'=2cdot (y_o+frac{1}{m}x_o-b)cdot (frac{m}{m^2+1})-y_o$ are our last two equations.
As I have now shown how to reflect a point across a given line, I should be able to see the same angles preserved below using the transformation above. However, I do not with this example which is clearly shown in red.
geometry
asked Dec 28 '18 at 14:26
W. G.W. G.
6101716
$endgroup$
I read that reflections preserve angle measure but not distance because they are "flips".
$underline{textbf{Question:}}$ So, is it true that a figure that is reflected about a line preserves angle measure?
I'm going to now explain why I do not think it is true.
I'm sure there is a more succinct way to write this, but $textbf{ANYTHING WRITTEN BELOW THIS IS NOT NEEDED TO ANSWER}$
$textbf{THIS QUESTION.}$ I'm just showing my work in case there is an error.
Explanation for Why Not True:
First, I'm going to come up with a valid equation that allows us to reflect a point $P:=(x_o, y_o)$ about a line $g(x):=mx+b$ where $mneq 0$ (see bullet for end result). First, let's consider the figure below.
Since lines $t$ and $g$ are $perp$, we know $t(x)=frac{-1}{m}x+c$ where $c$ is some constant. Since $t(x_o)=y_o$, we know $-frac{1}{m}x+c=y_oimplies c=y_o+frac{1}{m}$. Thus, $t(x)=frac{-1}{m}x+(y_o+frac{1}{m})$ substituting $c$ back into our original equation. Now, we know $t(x_1)=g(x_1)implies frac{-1}{m}x_1+(y_o+frac{1}{m})=mx_1+bimplies y_o+frac{1}{m}x_o-b=(m+frac{1}{m})x_1implies x_1=frac{y_o+frac{1}{m}x_o-b}{(m+frac{1}{m})}=frac{y_o+frac{1}{m}x_o-b}{frac{m^2+1}{m}}.$ Thus, $g(x_1)=m(frac{y_o+frac{1}{m}x_o-b}{frac{m^2+1}{m}})+b$. Now, we know looking at the midpoint that $(frac{x_o+x'}{2},frac{y_o+y'}{2})=(underbrace{frac{y_o+frac{1}{m}x_o-b}{frac{m^2+1}{m}}}_{=x_1},underbrace{frac{y_o+frac{1}{m}x_o-b}{frac{m^2+1}{m}}}_{=y_1})$.
This means $x'=2(frac{y_o+frac{1}{m}x_o-b}{frac{m^2+1}{m}})-x_o$ and $y'=2(frac{y_o+frac{1}{m}x_o-b}{frac{m^2+1}{m}})-y_o$.
$bullet$ Hence, we know $x'=2cdot (y_o+frac{1}{m}x_o-b)cdot (frac{m}{m^2+1})-x_o$ and $y'=2cdot (y_o+frac{1}{m}x_o-b)cdot (frac{m}{m^2+1})-y_o$ are our last two equations.
As I have now shown how to reflect a point across a given line, I should be able to see the same angles preserved below using the transformation above. However, I do not with this example which is clearly shown in red.
geometry
geometry
asked Dec 28 '18 at 14:26
W. G.W. G.
6101716
asked Dec 28 '18 at 14:26
W. G.W. G.
6101716
edited Dec 28 '18 at 14:31
W. G.
asked Dec 28 '18 at 14:26
W. G.W. G.
6101716
asked Dec 28 '18 at 14:26
W. G.W. G.
6101716
asked Dec 28 '18 at 14:26
W. G.W. G.
6101716
6101716
add a comment |
add a comment |
1 Answer
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$begingroup$
Reflection across a line is an isometry - a rigid motion. It preserves all distances between points and (therefore) all angles.
You don't really need formulas to see this. Cut a triangle out of cardboard. Now turn it over. All lengths and angles are the same. The only difference is that "clockwise" becomes "counterclockwise".
There must be an error somewhere in your linear algebra - the obtuse triangle can't be a reflection of the acute one.
answered Dec 28 '18 at 14:30
Ethan BolkerEthan Bolker
43.5k551116
$endgroup$
$begingroup$
Thank you! I will check my algebra.
$endgroup$
– W. G.
Dec 28 '18 at 14:34
$begingroup$
I already found an error. This answers my question.
$endgroup$
– W. G.
Dec 28 '18 at 14:37
$begingroup$
Lol love the cardboard explanation, because it’s true.
$endgroup$
– Randall
Dec 28 '18 at 14:44
add a comment |
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1 Answer
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$begingroup$
Reflection across a line is an isometry - a rigid motion. It preserves all distances between points and (therefore) all angles.
You don't really need formulas to see this. Cut a triangle out of cardboard. Now turn it over. All lengths and angles are the same. The only difference is that "clockwise" becomes "counterclockwise".
There must be an error somewhere in your linear algebra - the obtuse triangle can't be a reflection of the acute one.
answered Dec 28 '18 at 14:30
Ethan BolkerEthan Bolker
43.5k551116
$endgroup$
$begingroup$
Thank you! I will check my algebra.
$endgroup$
– W. G.
Dec 28 '18 at 14:34
$begingroup$
I already found an error. This answers my question.
$endgroup$
– W. G.
Dec 28 '18 at 14:37
$begingroup$
Lol love the cardboard explanation, because it’s true.
$endgroup$
– Randall
Dec 28 '18 at 14:44
add a comment |
$begingroup$
Reflection across a line is an isometry - a rigid motion. It preserves all distances between points and (therefore) all angles.
You don't really need formulas to see this. Cut a triangle out of cardboard. Now turn it over. All lengths and angles are the same. The only difference is that "clockwise" becomes "counterclockwise".
There must be an error somewhere in your linear algebra - the obtuse triangle can't be a reflection of the acute one.
answered Dec 28 '18 at 14:30
Ethan BolkerEthan Bolker
43.5k551116
$endgroup$
$begingroup$
Thank you! I will check my algebra.
$endgroup$
– W. G.
Dec 28 '18 at 14:34
$begingroup$
I already found an error. This answers my question.
$endgroup$
– W. G.
Dec 28 '18 at 14:37
$begingroup$
Lol love the cardboard explanation, because it’s true.
$endgroup$
– Randall
Dec 28 '18 at 14:44
add a comment |
$begingroup$
Reflection across a line is an isometry - a rigid motion. It preserves all distances between points and (therefore) all angles.
You don't really need formulas to see this. Cut a triangle out of cardboard. Now turn it over. All lengths and angles are the same. The only difference is that "clockwise" becomes "counterclockwise".
There must be an error somewhere in your linear algebra - the obtuse triangle can't be a reflection of the acute one.
answered Dec 28 '18 at 14:30
Ethan BolkerEthan Bolker
43.5k551116
$endgroup$
Reflection across a line is an isometry - a rigid motion. It preserves all distances between points and (therefore) all angles.
You don't really need formulas to see this. Cut a triangle out of cardboard. Now turn it over. All lengths and angles are the same. The only difference is that "clockwise" becomes "counterclockwise".
There must be an error somewhere in your linear algebra - the obtuse triangle can't be a reflection of the acute one.
answered Dec 28 '18 at 14:30
Ethan BolkerEthan Bolker
43.5k551116
answered Dec 28 '18 at 14:30
Ethan BolkerEthan Bolker
43.5k551116
answered Dec 28 '18 at 14:30
Ethan BolkerEthan Bolker
43.5k551116
answered Dec 28 '18 at 14:30
Ethan BolkerEthan Bolker
43.5k551116
43.5k551116
$begingroup$
Thank you! I will check my algebra.
$endgroup$
– W. G.
Dec 28 '18 at 14:34
$begingroup$
I already found an error. This answers my question.
$endgroup$
– W. G.
Dec 28 '18 at 14:37
$begingroup$
Lol love the cardboard explanation, because it’s true.
$endgroup$
– Randall
Dec 28 '18 at 14:44
add a comment |
$begingroup$
Thank you! I will check my algebra.
$endgroup$
– W. G.
Dec 28 '18 at 14:34
$begingroup$
I already found an error. This answers my question.
$endgroup$
– W. G.
Dec 28 '18 at 14:37
$begingroup$
Lol love the cardboard explanation, because it’s true.
$endgroup$
– Randall
Dec 28 '18 at 14:44
$begingroup$
Thank you! I will check my algebra.
$endgroup$
– W. G.
Dec 28 '18 at 14:34
$begingroup$
Thank you! I will check my algebra.
$endgroup$
– W. G.
Dec 28 '18 at 14:34
$begingroup$
I already found an error. This answers my question.
$endgroup$
– W. G.
Dec 28 '18 at 14:37
$begingroup$
I already found an error. This answers my question.
$endgroup$
– W. G.
Dec 28 '18 at 14:37
$begingroup$
Lol love the cardboard explanation, because it’s true.
$endgroup$
– Randall
Dec 28 '18 at 14:44
$begingroup$
Lol love the cardboard explanation, because it’s true.
$endgroup$
– Randall
Dec 28 '18 at 14:44
add a comment |
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