Finding the line containing a point at infinity and a point of the form $(x:y:1)$
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How do you find the line containing $(0:0:1)$ and $(1:2:0)$ that does not contain $(1:1:1)?$
I tried to use projective transformation, but I ended up getting the line $ell(x,y,z)=z$ which I know is wrong since it does not contain $(0:0:1)$.
Please help!
algebraic-geometry
asked Nov 26 at 4:49
Mashed Potato
866
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up vote
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down vote
favorite
How do you find the line containing $(0:0:1)$ and $(1:2:0)$ that does not contain $(1:1:1)?$
I tried to use projective transformation, but I ended up getting the line $ell(x,y,z)=z$ which I know is wrong since it does not contain $(0:0:1)$.
Please help!
algebraic-geometry
asked Nov 26 at 4:49
Mashed Potato
866
3
Hint: every line in $mathbb{P}^2$ is given by an equation $aX + bY + cZ = 0$. Plug in the points to determine $a,b,c$.
– André 3000
Nov 26 at 4:51
1
Thank you so much!
– Mashed Potato
Nov 26 at 4:53
Feel free to write an answer to your question below when you solve it.
– André 3000
Nov 26 at 4:54
A line in the projective plane is uniquely determined by two points. If that third point happens to be on it as well, then you’re out of luck: there aren’t any other choices for the line.
– amd
Nov 26 at 6:13
As for how to find the line, take the cross product of the two points.
– amd
Nov 26 at 6:15
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
How do you find the line containing $(0:0:1)$ and $(1:2:0)$ that does not contain $(1:1:1)?$
I tried to use projective transformation, but I ended up getting the line $ell(x,y,z)=z$ which I know is wrong since it does not contain $(0:0:1)$.
Please help!
algebraic-geometry
asked Nov 26 at 4:49
Mashed Potato
866
How do you find the line containing $(0:0:1)$ and $(1:2:0)$ that does not contain $(1:1:1)?$
I tried to use projective transformation, but I ended up getting the line $ell(x,y,z)=z$ which I know is wrong since it does not contain $(0:0:1)$.
Please help!
algebraic-geometry
algebraic-geometry
asked Nov 26 at 4:49
Mashed Potato
866
asked Nov 26 at 4:49
Mashed Potato
866
asked Nov 26 at 4:49
Mashed Potato
866
asked Nov 26 at 4:49
Mashed Potato
866
asked Nov 26 at 4:49
Mashed Potato
866
866
3
Hint: every line in $mathbb{P}^2$ is given by an equation $aX + bY + cZ = 0$. Plug in the points to determine $a,b,c$.
– André 3000
Nov 26 at 4:51
1
Thank you so much!
– Mashed Potato
Nov 26 at 4:53
Feel free to write an answer to your question below when you solve it.
– André 3000
Nov 26 at 4:54
A line in the projective plane is uniquely determined by two points. If that third point happens to be on it as well, then you’re out of luck: there aren’t any other choices for the line.
– amd
Nov 26 at 6:13
As for how to find the line, take the cross product of the two points.
– amd
Nov 26 at 6:15
add a comment |
3
Hint: every line in $mathbb{P}^2$ is given by an equation $aX + bY + cZ = 0$. Plug in the points to determine $a,b,c$.
– André 3000
Nov 26 at 4:51
1
Thank you so much!
– Mashed Potato
Nov 26 at 4:53
Feel free to write an answer to your question below when you solve it.
– André 3000
Nov 26 at 4:54
A line in the projective plane is uniquely determined by two points. If that third point happens to be on it as well, then you’re out of luck: there aren’t any other choices for the line.
– amd
Nov 26 at 6:13
As for how to find the line, take the cross product of the two points.
– amd
Nov 26 at 6:15
3
3
Hint: every line in $mathbb{P}^2$ is given by an equation $aX + bY + cZ = 0$. Plug in the points to determine $a,b,c$.
– André 3000
Nov 26 at 4:51
Hint: every line in $mathbb{P}^2$ is given by an equation $aX + bY + cZ = 0$. Plug in the points to determine $a,b,c$.
– André 3000
Nov 26 at 4:51
1
1
Thank you so much!
– Mashed Potato
Nov 26 at 4:53
Thank you so much!
– Mashed Potato
Nov 26 at 4:53
Feel free to write an answer to your question below when you solve it.
– André 3000
Nov 26 at 4:54
Feel free to write an answer to your question below when you solve it.
– André 3000
Nov 26 at 4:54
A line in the projective plane is uniquely determined by two points. If that third point happens to be on it as well, then you’re out of luck: there aren’t any other choices for the line.
– amd
Nov 26 at 6:13
A line in the projective plane is uniquely determined by two points. If that third point happens to be on it as well, then you’re out of luck: there aren’t any other choices for the line.
– amd
Nov 26 at 6:13
As for how to find the line, take the cross product of the two points.
– amd
Nov 26 at 6:15
As for how to find the line, take the cross product of the two points.
– amd
Nov 26 at 6:15
add a comment |
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3
Hint: every line in $mathbb{P}^2$ is given by an equation $aX + bY + cZ = 0$. Plug in the points to determine $a,b,c$.
– André 3000
Nov 26 at 4:51
1
Thank you so much!
– Mashed Potato
Nov 26 at 4:53
Feel free to write an answer to your question below when you solve it.
– André 3000
Nov 26 at 4:54
A line in the projective plane is uniquely determined by two points. If that third point happens to be on it as well, then you’re out of luck: there aren’t any other choices for the line.
– amd
Nov 26 at 6:13
As for how to find the line, take the cross product of the two points.
– amd
Nov 26 at 6:15