Do Reflections About Lines in 2-D Preserve Angle Measure for Figures?












0














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.



enter image description here



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. enter image description here






geometry







share|cite|improve this question





























    0














    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.



    enter image description here



    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. enter image description here






    geometry







    share|cite|improve this question



























      0












      0








      0







      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.



      enter image description here



      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. enter image description here






      geometry







      share|cite|improve this question















      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.



      enter image description here



      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. enter image description here






      geometry




      geometry






      share|cite|improve this question















      share|cite|improve this question













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      edited Dec 28 '18 at 14:31







      W. G.

















      asked Dec 28 '18 at 14:26









      W. G.W. G.

      5641516




      5641516






















          1 Answer
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          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.






          share|cite|improve this answer





















          • Thank you! I will check my algebra.
            – W. G.
            Dec 28 '18 at 14:34










          • I already found an error. This answers my question.
            – W. G.
            Dec 28 '18 at 14:37












          • Lol love the cardboard explanation, because it’s true.
            – Randall
            Dec 28 '18 at 14:44











          Your Answer





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          1 Answer
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          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          3














          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.






          share|cite|improve this answer





















          • Thank you! I will check my algebra.
            – W. G.
            Dec 28 '18 at 14:34










          • I already found an error. This answers my question.
            – W. G.
            Dec 28 '18 at 14:37












          • Lol love the cardboard explanation, because it’s true.
            – Randall
            Dec 28 '18 at 14:44
















          3














          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.






          share|cite|improve this answer





















          • Thank you! I will check my algebra.
            – W. G.
            Dec 28 '18 at 14:34










          • I already found an error. This answers my question.
            – W. G.
            Dec 28 '18 at 14:37












          • Lol love the cardboard explanation, because it’s true.
            – Randall
            Dec 28 '18 at 14:44














          3












          3








          3






          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.






          share|cite|improve this answer












          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.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 28 '18 at 14:30









          Ethan BolkerEthan Bolker

          41.8k547110




          41.8k547110












          • Thank you! I will check my algebra.
            – W. G.
            Dec 28 '18 at 14:34










          • I already found an error. This answers my question.
            – W. G.
            Dec 28 '18 at 14:37












          • Lol love the cardboard explanation, because it’s true.
            – Randall
            Dec 28 '18 at 14:44


















          • Thank you! I will check my algebra.
            – W. G.
            Dec 28 '18 at 14:34










          • I already found an error. This answers my question.
            – W. G.
            Dec 28 '18 at 14:37












          • Lol love the cardboard explanation, because it’s true.
            – Randall
            Dec 28 '18 at 14:44
















          Thank you! I will check my algebra.
          – W. G.
          Dec 28 '18 at 14:34




          Thank you! I will check my algebra.
          – W. G.
          Dec 28 '18 at 14:34












          I already found an error. This answers my question.
          – W. G.
          Dec 28 '18 at 14:37






          I already found an error. This answers my question.
          – W. G.
          Dec 28 '18 at 14:37














          Lol love the cardboard explanation, because it’s true.
          – Randall
          Dec 28 '18 at 14:44




          Lol love the cardboard explanation, because it’s true.
          – Randall
          Dec 28 '18 at 14:44


















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