Locus problem for vertex of equilateral triangle











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Question



Given an equilateral triangle $PQR$ where $P(1,3)$ is a fixed point and $Q$ is a moving point on the line $x=3.$ Find the locus of $R.$



My attempt




  1. Take $Q$ as $(3,p)$ and $R(h, k).$ Then substitute to the slope of $PQ$ side and $PR$ side to be equal, taking $tan angle Q$ and $tanangle R.$


  2. Similarly for $angle P$ to equate all 3 angles. Got the equation.


  3. Also use the distance method and then finally get the required locus.



My question



Is there any other method which is easier?






algebraic-geometry triangle self-learning coordinate-systems locus







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    Please don't just delete your previous unanswered question, because someone (e.g. @peterwhy) might be working on it.
    – peterwhy
    Sep 30 at 23:18












  • Oh sorry for that.
    – jayant98
    Oct 1 at 9:15















up vote
1
down vote

favorite
2












Question



Given an equilateral triangle $PQR$ where $P(1,3)$ is a fixed point and $Q$ is a moving point on the line $x=3.$ Find the locus of $R.$



My attempt




  1. Take $Q$ as $(3,p)$ and $R(h, k).$ Then substitute to the slope of $PQ$ side and $PR$ side to be equal, taking $tan angle Q$ and $tanangle R.$


  2. Similarly for $angle P$ to equate all 3 angles. Got the equation.


  3. Also use the distance method and then finally get the required locus.



My question



Is there any other method which is easier?






algebraic-geometry triangle self-learning coordinate-systems locus







share|cite|improve this question




















  • 1




    Please don't just delete your previous unanswered question, because someone (e.g. @peterwhy) might be working on it.
    – peterwhy
    Sep 30 at 23:18












  • Oh sorry for that.
    – jayant98
    Oct 1 at 9:15













up vote
1
down vote

favorite
2









up vote
1
down vote

favorite
2






2





Question



Given an equilateral triangle $PQR$ where $P(1,3)$ is a fixed point and $Q$ is a moving point on the line $x=3.$ Find the locus of $R.$



My attempt




  1. Take $Q$ as $(3,p)$ and $R(h, k).$ Then substitute to the slope of $PQ$ side and $PR$ side to be equal, taking $tan angle Q$ and $tanangle R.$


  2. Similarly for $angle P$ to equate all 3 angles. Got the equation.


  3. Also use the distance method and then finally get the required locus.



My question



Is there any other method which is easier?






algebraic-geometry triangle self-learning coordinate-systems locus







share|cite|improve this question















Question



Given an equilateral triangle $PQR$ where $P(1,3)$ is a fixed point and $Q$ is a moving point on the line $x=3.$ Find the locus of $R.$



My attempt




  1. Take $Q$ as $(3,p)$ and $R(h, k).$ Then substitute to the slope of $PQ$ side and $PR$ side to be equal, taking $tan angle Q$ and $tanangle R.$


  2. Similarly for $angle P$ to equate all 3 angles. Got the equation.


  3. Also use the distance method and then finally get the required locus.



My question



Is there any other method which is easier?






algebraic-geometry triangle self-learning coordinate-systems locus




algebraic-geometry triangle self-learning coordinate-systems locus






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 5 at 12:38









user376343

2,7682822




2,7682822










asked Sep 30 at 22:46









jayant98

39515




39515








  • 1




    Please don't just delete your previous unanswered question, because someone (e.g. @peterwhy) might be working on it.
    – peterwhy
    Sep 30 at 23:18












  • Oh sorry for that.
    – jayant98
    Oct 1 at 9:15














  • 1




    Please don't just delete your previous unanswered question, because someone (e.g. @peterwhy) might be working on it.
    – peterwhy
    Sep 30 at 23:18












  • Oh sorry for that.
    – jayant98
    Oct 1 at 9:15








1




1




Please don't just delete your previous unanswered question, because someone (e.g. @peterwhy) might be working on it.
– peterwhy
Sep 30 at 23:18






Please don't just delete your previous unanswered question, because someone (e.g. @peterwhy) might be working on it.
– peterwhy
Sep 30 at 23:18














Oh sorry for that.
– jayant98
Oct 1 at 9:15




Oh sorry for that.
– jayant98
Oct 1 at 9:15










2 Answers
2






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up vote
1
down vote













I am lazy, so I will pretend $P$ is the origin using shifted coordinates $(x', y')$, and $Q$ is on the line $x' = 2$:



$$pmatrix{x'\y'} = pmatrix{x\y}-pmatrix{1\ 3}$$



Let the coordinates of $Q$ be $(x' = 2,y' = p')$. There are two possible vertices for an equilateral triangle, obtained by rotating $Q$ by $pmfrac {2pi}6$ about the origin $P$.



For $i in {0, 1}$:



$$begin{align*}
R_i=pmatrix{x'=h'_i\y'=k'_i} &=
pmatrix{cosfrac {2pi}6 & -sinleft[(-1)^ifrac {2pi}6right]\
sinleft[(-1)^ifrac {2pi}6right] & cosfrac {2pi}6}
pmatrix{2\p'}\
&= pmatrix{frac12 & -(-1)^ifrac{sqrt3}2\
(-1)^ifrac{sqrt3}2 & frac12}
pmatrix{2\p'}\
&= pmatrix{1-(-1)^isqrt3frac{p'}2\ (-1)^isqrt3+frac{p'}2}
end{align*}$$



One way to eliminate the $p'$ is to note that



$$begin{align*}
frac{p'}2 &= k'_i-(-1)^isqrt3\
h'_i &= 1-(-1)^isqrt3frac{p'}2\
&= 1-(-1)^isqrt3left[k'_i-(-1)^isqrt3right]\
&= 4- (-1)^isqrt3 k'_i
end{align*}$$



i.e. the loci are $x'=4-(-1)^isqrt3 y'$.



The above $R_0, R_1$ will be in $x'y'$-coordinates, so translate them back to $xy$-coordinates.



$$begin{align*}
(x-1) &= 4-(-1)^isqrt3(y-3)\
(x-1) + (-1)^isqrt3(y-3) - 4 &= 0
end{align*}$$



Lastly, if you prefer having one equation representing two straight lines:



$$begin{align*}
left[(x-1) + sqrt3(y-3) - 4right]left[(x-1) - sqrt3(y-3) - 4right] &= 0\
(x-5)^2-3(y-3)^2 &= 0
end{align*}$$






share|cite|improve this answer






























    up vote
    1
    down vote













    Moving the origin so that $P = (0,0)$ instead of $P = (1,3)$ and taking as parameters (to eliminate) the side, say $s$, of the triangle and the angle $alpha$ determined by $R = (x, y)$, we have the equations
    $$begin{cases}(x-2)^2+(y-ssin(60^{circ}+alpha))^2=s^2\x=scos (alpha)spacespace y=ssin(alpha)end{cases}$$



    enter image description here



    from which
    $$(x-2)^2+left(frac{y-sqrt3x}{2}right)^2=x^2+y^2$$ or $$3x^2-3y^2-2sqrt3xy-16x+16=0$$ or
    $$(3x+sqrt3y-4)(x-sqrt3y-4)=0$$
    Then the locus is given by the two straight lines
    $$3x+sqrt3y-4=0\x-sqrt3y-4=0$$
    The task of returning to the original coordinate system is immediate.






    share|cite|improve this answer























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      2 Answers
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      active

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      2 Answers
      2






      active

      oldest

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      active

      oldest

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      active

      oldest

      votes








      up vote
      1
      down vote













      I am lazy, so I will pretend $P$ is the origin using shifted coordinates $(x', y')$, and $Q$ is on the line $x' = 2$:



      $$pmatrix{x'\y'} = pmatrix{x\y}-pmatrix{1\ 3}$$



      Let the coordinates of $Q$ be $(x' = 2,y' = p')$. There are two possible vertices for an equilateral triangle, obtained by rotating $Q$ by $pmfrac {2pi}6$ about the origin $P$.



      For $i in {0, 1}$:



      $$begin{align*}
      R_i=pmatrix{x'=h'_i\y'=k'_i} &=
      pmatrix{cosfrac {2pi}6 & -sinleft[(-1)^ifrac {2pi}6right]\
      sinleft[(-1)^ifrac {2pi}6right] & cosfrac {2pi}6}
      pmatrix{2\p'}\
      &= pmatrix{frac12 & -(-1)^ifrac{sqrt3}2\
      (-1)^ifrac{sqrt3}2 & frac12}
      pmatrix{2\p'}\
      &= pmatrix{1-(-1)^isqrt3frac{p'}2\ (-1)^isqrt3+frac{p'}2}
      end{align*}$$



      One way to eliminate the $p'$ is to note that



      $$begin{align*}
      frac{p'}2 &= k'_i-(-1)^isqrt3\
      h'_i &= 1-(-1)^isqrt3frac{p'}2\
      &= 1-(-1)^isqrt3left[k'_i-(-1)^isqrt3right]\
      &= 4- (-1)^isqrt3 k'_i
      end{align*}$$



      i.e. the loci are $x'=4-(-1)^isqrt3 y'$.



      The above $R_0, R_1$ will be in $x'y'$-coordinates, so translate them back to $xy$-coordinates.



      $$begin{align*}
      (x-1) &= 4-(-1)^isqrt3(y-3)\
      (x-1) + (-1)^isqrt3(y-3) - 4 &= 0
      end{align*}$$



      Lastly, if you prefer having one equation representing two straight lines:



      $$begin{align*}
      left[(x-1) + sqrt3(y-3) - 4right]left[(x-1) - sqrt3(y-3) - 4right] &= 0\
      (x-5)^2-3(y-3)^2 &= 0
      end{align*}$$






      share|cite|improve this answer



























        up vote
        1
        down vote













        I am lazy, so I will pretend $P$ is the origin using shifted coordinates $(x', y')$, and $Q$ is on the line $x' = 2$:



        $$pmatrix{x'\y'} = pmatrix{x\y}-pmatrix{1\ 3}$$



        Let the coordinates of $Q$ be $(x' = 2,y' = p')$. There are two possible vertices for an equilateral triangle, obtained by rotating $Q$ by $pmfrac {2pi}6$ about the origin $P$.



        For $i in {0, 1}$:



        $$begin{align*}
        R_i=pmatrix{x'=h'_i\y'=k'_i} &=
        pmatrix{cosfrac {2pi}6 & -sinleft[(-1)^ifrac {2pi}6right]\
        sinleft[(-1)^ifrac {2pi}6right] & cosfrac {2pi}6}
        pmatrix{2\p'}\
        &= pmatrix{frac12 & -(-1)^ifrac{sqrt3}2\
        (-1)^ifrac{sqrt3}2 & frac12}
        pmatrix{2\p'}\
        &= pmatrix{1-(-1)^isqrt3frac{p'}2\ (-1)^isqrt3+frac{p'}2}
        end{align*}$$



        One way to eliminate the $p'$ is to note that



        $$begin{align*}
        frac{p'}2 &= k'_i-(-1)^isqrt3\
        h'_i &= 1-(-1)^isqrt3frac{p'}2\
        &= 1-(-1)^isqrt3left[k'_i-(-1)^isqrt3right]\
        &= 4- (-1)^isqrt3 k'_i
        end{align*}$$



        i.e. the loci are $x'=4-(-1)^isqrt3 y'$.



        The above $R_0, R_1$ will be in $x'y'$-coordinates, so translate them back to $xy$-coordinates.



        $$begin{align*}
        (x-1) &= 4-(-1)^isqrt3(y-3)\
        (x-1) + (-1)^isqrt3(y-3) - 4 &= 0
        end{align*}$$



        Lastly, if you prefer having one equation representing two straight lines:



        $$begin{align*}
        left[(x-1) + sqrt3(y-3) - 4right]left[(x-1) - sqrt3(y-3) - 4right] &= 0\
        (x-5)^2-3(y-3)^2 &= 0
        end{align*}$$






        share|cite|improve this answer

























          up vote
          1
          down vote










          up vote
          1
          down vote









          I am lazy, so I will pretend $P$ is the origin using shifted coordinates $(x', y')$, and $Q$ is on the line $x' = 2$:



          $$pmatrix{x'\y'} = pmatrix{x\y}-pmatrix{1\ 3}$$



          Let the coordinates of $Q$ be $(x' = 2,y' = p')$. There are two possible vertices for an equilateral triangle, obtained by rotating $Q$ by $pmfrac {2pi}6$ about the origin $P$.



          For $i in {0, 1}$:



          $$begin{align*}
          R_i=pmatrix{x'=h'_i\y'=k'_i} &=
          pmatrix{cosfrac {2pi}6 & -sinleft[(-1)^ifrac {2pi}6right]\
          sinleft[(-1)^ifrac {2pi}6right] & cosfrac {2pi}6}
          pmatrix{2\p'}\
          &= pmatrix{frac12 & -(-1)^ifrac{sqrt3}2\
          (-1)^ifrac{sqrt3}2 & frac12}
          pmatrix{2\p'}\
          &= pmatrix{1-(-1)^isqrt3frac{p'}2\ (-1)^isqrt3+frac{p'}2}
          end{align*}$$



          One way to eliminate the $p'$ is to note that



          $$begin{align*}
          frac{p'}2 &= k'_i-(-1)^isqrt3\
          h'_i &= 1-(-1)^isqrt3frac{p'}2\
          &= 1-(-1)^isqrt3left[k'_i-(-1)^isqrt3right]\
          &= 4- (-1)^isqrt3 k'_i
          end{align*}$$



          i.e. the loci are $x'=4-(-1)^isqrt3 y'$.



          The above $R_0, R_1$ will be in $x'y'$-coordinates, so translate them back to $xy$-coordinates.



          $$begin{align*}
          (x-1) &= 4-(-1)^isqrt3(y-3)\
          (x-1) + (-1)^isqrt3(y-3) - 4 &= 0
          end{align*}$$



          Lastly, if you prefer having one equation representing two straight lines:



          $$begin{align*}
          left[(x-1) + sqrt3(y-3) - 4right]left[(x-1) - sqrt3(y-3) - 4right] &= 0\
          (x-5)^2-3(y-3)^2 &= 0
          end{align*}$$






          share|cite|improve this answer














          I am lazy, so I will pretend $P$ is the origin using shifted coordinates $(x', y')$, and $Q$ is on the line $x' = 2$:



          $$pmatrix{x'\y'} = pmatrix{x\y}-pmatrix{1\ 3}$$



          Let the coordinates of $Q$ be $(x' = 2,y' = p')$. There are two possible vertices for an equilateral triangle, obtained by rotating $Q$ by $pmfrac {2pi}6$ about the origin $P$.



          For $i in {0, 1}$:



          $$begin{align*}
          R_i=pmatrix{x'=h'_i\y'=k'_i} &=
          pmatrix{cosfrac {2pi}6 & -sinleft[(-1)^ifrac {2pi}6right]\
          sinleft[(-1)^ifrac {2pi}6right] & cosfrac {2pi}6}
          pmatrix{2\p'}\
          &= pmatrix{frac12 & -(-1)^ifrac{sqrt3}2\
          (-1)^ifrac{sqrt3}2 & frac12}
          pmatrix{2\p'}\
          &= pmatrix{1-(-1)^isqrt3frac{p'}2\ (-1)^isqrt3+frac{p'}2}
          end{align*}$$



          One way to eliminate the $p'$ is to note that



          $$begin{align*}
          frac{p'}2 &= k'_i-(-1)^isqrt3\
          h'_i &= 1-(-1)^isqrt3frac{p'}2\
          &= 1-(-1)^isqrt3left[k'_i-(-1)^isqrt3right]\
          &= 4- (-1)^isqrt3 k'_i
          end{align*}$$



          i.e. the loci are $x'=4-(-1)^isqrt3 y'$.



          The above $R_0, R_1$ will be in $x'y'$-coordinates, so translate them back to $xy$-coordinates.



          $$begin{align*}
          (x-1) &= 4-(-1)^isqrt3(y-3)\
          (x-1) + (-1)^isqrt3(y-3) - 4 &= 0
          end{align*}$$



          Lastly, if you prefer having one equation representing two straight lines:



          $$begin{align*}
          left[(x-1) + sqrt3(y-3) - 4right]left[(x-1) - sqrt3(y-3) - 4right] &= 0\
          (x-5)^2-3(y-3)^2 &= 0
          end{align*}$$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Oct 1 at 0:34

























          answered Sep 30 at 23:15









          peterwhy

          12k21228




          12k21228






















              up vote
              1
              down vote













              Moving the origin so that $P = (0,0)$ instead of $P = (1,3)$ and taking as parameters (to eliminate) the side, say $s$, of the triangle and the angle $alpha$ determined by $R = (x, y)$, we have the equations
              $$begin{cases}(x-2)^2+(y-ssin(60^{circ}+alpha))^2=s^2\x=scos (alpha)spacespace y=ssin(alpha)end{cases}$$



              enter image description here



              from which
              $$(x-2)^2+left(frac{y-sqrt3x}{2}right)^2=x^2+y^2$$ or $$3x^2-3y^2-2sqrt3xy-16x+16=0$$ or
              $$(3x+sqrt3y-4)(x-sqrt3y-4)=0$$
              Then the locus is given by the two straight lines
              $$3x+sqrt3y-4=0\x-sqrt3y-4=0$$
              The task of returning to the original coordinate system is immediate.






              share|cite|improve this answer



























                up vote
                1
                down vote













                Moving the origin so that $P = (0,0)$ instead of $P = (1,3)$ and taking as parameters (to eliminate) the side, say $s$, of the triangle and the angle $alpha$ determined by $R = (x, y)$, we have the equations
                $$begin{cases}(x-2)^2+(y-ssin(60^{circ}+alpha))^2=s^2\x=scos (alpha)spacespace y=ssin(alpha)end{cases}$$



                enter image description here



                from which
                $$(x-2)^2+left(frac{y-sqrt3x}{2}right)^2=x^2+y^2$$ or $$3x^2-3y^2-2sqrt3xy-16x+16=0$$ or
                $$(3x+sqrt3y-4)(x-sqrt3y-4)=0$$
                Then the locus is given by the two straight lines
                $$3x+sqrt3y-4=0\x-sqrt3y-4=0$$
                The task of returning to the original coordinate system is immediate.






                share|cite|improve this answer

























                  up vote
                  1
                  down vote










                  up vote
                  1
                  down vote









                  Moving the origin so that $P = (0,0)$ instead of $P = (1,3)$ and taking as parameters (to eliminate) the side, say $s$, of the triangle and the angle $alpha$ determined by $R = (x, y)$, we have the equations
                  $$begin{cases}(x-2)^2+(y-ssin(60^{circ}+alpha))^2=s^2\x=scos (alpha)spacespace y=ssin(alpha)end{cases}$$



                  enter image description here



                  from which
                  $$(x-2)^2+left(frac{y-sqrt3x}{2}right)^2=x^2+y^2$$ or $$3x^2-3y^2-2sqrt3xy-16x+16=0$$ or
                  $$(3x+sqrt3y-4)(x-sqrt3y-4)=0$$
                  Then the locus is given by the two straight lines
                  $$3x+sqrt3y-4=0\x-sqrt3y-4=0$$
                  The task of returning to the original coordinate system is immediate.






                  share|cite|improve this answer














                  Moving the origin so that $P = (0,0)$ instead of $P = (1,3)$ and taking as parameters (to eliminate) the side, say $s$, of the triangle and the angle $alpha$ determined by $R = (x, y)$, we have the equations
                  $$begin{cases}(x-2)^2+(y-ssin(60^{circ}+alpha))^2=s^2\x=scos (alpha)spacespace y=ssin(alpha)end{cases}$$



                  enter image description here



                  from which
                  $$(x-2)^2+left(frac{y-sqrt3x}{2}right)^2=x^2+y^2$$ or $$3x^2-3y^2-2sqrt3xy-16x+16=0$$ or
                  $$(3x+sqrt3y-4)(x-sqrt3y-4)=0$$
                  Then the locus is given by the two straight lines
                  $$3x+sqrt3y-4=0\x-sqrt3y-4=0$$
                  The task of returning to the original coordinate system is immediate.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Oct 2 at 18:36

























                  answered Oct 2 at 16:41









                  Piquito

                  17.8k31436




                  17.8k31436






























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                      Cabo Verde

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                        Nota: Cabo-verdiano redireciona para este artigo. Para a língua falada em Cabo Verde, veja crioulo cabo-verdiano. Para outros significados, veja Cabo Verde (desambiguação). Coordenadas: 14° 55' N 23° 31' O Cabo Verde República de Cabo Verde Bandeira Brasão de Armas Lema: Unidade, Trabalho, Progresso Hino nacional: Cântico da Liberdade Gentílico: cabo-verdiano [ 1 ] Capital Praia 14° 55' N 23° 31' O Cidade mais populosa Praia Língua oficial Português Governo República unitária semipresidencialista [ 2 ]  - Presidente Jorge Carlos Fonseca  - Primeiro-ministro Ulisses Correia e Silva História    - Descoberta 1460   - Independência 00 • de Portugal 5 de Julho de 1975   - Multipartidarismo 13 de Janeiro de 1990  Área    - Total 4 033 km² (146.º)  - Água (%) 0,08 População    - Estimativa para 2017 560 899 [ 3 ] hab. (173.º) ...
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                      Gyllenstierna

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