Locus problem for vertex of equilateral triangle
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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
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 $tanangle R.$
Similarly for $angle P$ to equate all 3 angles. Got the equation.
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
edited Dec 5 at 12:38
user376343
2,7682822
asked Sep 30 at 22:46
jayant98
39515
add a comment |
up vote
1
down vote
favorite
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
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 $tanangle R.$
Similarly for $angle P$ to equate all 3 angles. Got the equation.
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
edited Dec 5 at 12:38
user376343
2,7682822
asked Sep 30 at 22:46
jayant98
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
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
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
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 $tanangle R.$
Similarly for $angle P$ to equate all 3 angles. Got the equation.
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
edited Dec 5 at 12:38
user376343
2,7682822
asked Sep 30 at 22:46
jayant98
39515
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
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 $tanangle R.$
Similarly for $angle P$ to equate all 3 angles. Got the equation.
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
edited Dec 5 at 12:38
user376343
2,7682822
asked Sep 30 at 22:46
jayant98
39515
edited Dec 5 at 12:38
user376343
2,7682822
asked Sep 30 at 22:46
jayant98
39515
edited Dec 5 at 12:38
user376343
2,7682822
edited Dec 5 at 12:38
user376343
2,7682822
edited Dec 5 at 12:38
user376343
2,7682822
2,7682822
asked Sep 30 at 22:46
jayant98
39515
asked Sep 30 at 22:46
jayant98
39515
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
add a comment |
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
add a comment |
2 Answers
2
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:
$$begin{align*}
left[(x-1) + sqrt3(y-3) - 4right]left[(x-1) - sqrt3(y-3) - 4right] &= 0\
(x-5)^2-3(y-3)^2 &= 0
end{align*}$$
answered Sep 30 at 23:15
peterwhy
12k21228
add a comment |
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
$$begin{cases}(x-2)^2+(y-ssin(60^{circ}+alpha))^2=s^2\x=scos (alpha)spacespace y=ssin(alpha)end{cases}$$
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.
answered Oct 2 at 16:41
Piquito
17.8k31436
add a comment |
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
votes
active
oldest
votes
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:
$$begin{align*}
left[(x-1) + sqrt3(y-3) - 4right]left[(x-1) - sqrt3(y-3) - 4right] &= 0\
(x-5)^2-3(y-3)^2 &= 0
end{align*}$$
answered Sep 30 at 23:15
peterwhy
12k21228
add a comment |
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:
$$begin{align*}
left[(x-1) + sqrt3(y-3) - 4right]left[(x-1) - sqrt3(y-3) - 4right] &= 0\
(x-5)^2-3(y-3)^2 &= 0
end{align*}$$
answered Sep 30 at 23:15
peterwhy
12k21228
add a comment |
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:
$$begin{align*}
left[(x-1) + sqrt3(y-3) - 4right]left[(x-1) - sqrt3(y-3) - 4right] &= 0\
(x-5)^2-3(y-3)^2 &= 0
end{align*}$$
answered Sep 30 at 23:15
peterwhy
12k21228
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:
$$begin{align*}
left[(x-1) + sqrt3(y-3) - 4right]left[(x-1) - sqrt3(y-3) - 4right] &= 0\
(x-5)^2-3(y-3)^2 &= 0
end{align*}$$
answered Sep 30 at 23:15
peterwhy
12k21228
edited Oct 1 at 0:34
answered Sep 30 at 23:15
peterwhy
12k21228
answered Sep 30 at 23:15
peterwhy
12k21228
answered Sep 30 at 23:15
peterwhy
12k21228
12k21228
add a comment |
add a comment |
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
$$begin{cases}(x-2)^2+(y-ssin(60^{circ}+alpha))^2=s^2\x=scos (alpha)spacespace y=ssin(alpha)end{cases}$$
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.
answered Oct 2 at 16:41
Piquito
17.8k31436
add a comment |
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
$$begin{cases}(x-2)^2+(y-ssin(60^{circ}+alpha))^2=s^2\x=scos (alpha)spacespace y=ssin(alpha)end{cases}$$
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.
answered Oct 2 at 16:41
Piquito
17.8k31436
add a comment |
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
$$begin{cases}(x-2)^2+(y-ssin(60^{circ}+alpha))^2=s^2\x=scos (alpha)spacespace y=ssin(alpha)end{cases}$$
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.
answered Oct 2 at 16:41
Piquito
17.8k31436
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
$$begin{cases}(x-2)^2+(y-ssin(60^{circ}+alpha))^2=s^2\x=scos (alpha)spacespace y=ssin(alpha)end{cases}$$
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.
answered Oct 2 at 16:41
Piquito
17.8k31436
edited Oct 2 at 18:36
answered Oct 2 at 16:41
Piquito
17.8k31436
answered Oct 2 at 16:41
Piquito
17.8k31436
answered Oct 2 at 16:41
Piquito
17.8k31436
17.8k31436
add a comment |
add a comment |
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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