Locus problem for vertex of equilateral triangle
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
2,7012822
asked Sep 30 at 22:46
jayant98
35414
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
2,7012822
asked Sep 30 at 22:46
jayant98
35414
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
2,7012822
asked Sep 30 at 22:46
jayant98
35414
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
2,7012822
asked Sep 30 at 22:46
jayant98
35414
edited Dec 5 at 12:38
user376343
2,7012822
asked Sep 30 at 22:46
jayant98
35414
edited Dec 5 at 12:38
user376343
2,7012822
edited Dec 5 at 12:38
user376343
2,7012822
edited Dec 5 at 12:38
user376343
2,7012822
2,7012822
asked Sep 30 at 22:46
jayant98
35414
asked Sep 30 at 22:46
jayant98
35414
asked Sep 30 at 22:46
jayant98
35414
35414
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 |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2937291%2flocus-problem-for-vertex-of-equilateral-triangle%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
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 |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2937291%2flocus-problem-for-vertex-of-equilateral-triangle%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
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