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Suppose I have an point $(3, 2)$ which I've seen has homogeneous coordinates $(3,2,1)$ and $k(3,2,1)$ where $k \neq 0$, and $(3,2,1)$ and $(3k,2k,k)$ both represent the same Cartesian point $(3, 2)$.

My question is that could I write $(3,2)$ in homogeneous coordinates as $(3,2,\frac{1}{2})$, $(3,2,\frac{1}{3})$, $\dots$ or equivalently $(6,4,1)$, $(9,6,1)$, $\dots$?

If so, are the homogeneous points $(6,4,1)$, $(9,6,1)$, $\dots$ the same as homogeneous points $(3,2,1)$, $(6,4,2)$, $(9,6,3)$, $\dots$, and hence $(6,4,2)$, $(9,6,3)$, $\dots$ are homogeneous coordinates of the Cartesian point $(3,2)$?

My last question is I need to see the sequence of points how $(3,2,1)$ approach to $(3,2,0)$?

Michael Albanese
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S. M.
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  • A better notation is that Cartesian coordinate point $,(x,y),$ has homogeneous projective representation $,[x:y:1],$ by defintion. If you have $,[x:y:z],$ then this is the point with coordinates $,(x/z,y/z).$ – Somos Jul 10 '21 at 13:08
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    "My last question is I need to see the sequence of points how (3,2,1) approach to (3,2,0)?" Can you explain this a bit more? I don't understand the question. – Michael Albanese Jul 10 '21 at 13:13
  • For instance, a point in Cartesian (1, 2) becomes (1, 2, 1) in Homogeneous. If a point, (1, 2), moves toward infinity, it becomes (∞,∞) in Cartesian coordinates. And it becomes (1, 2, 0) in Homogeneous coordinates, because of (1/0, 2/0) ≈ (∞,∞). Notice that we can express the point at infinity without using "∞".. – S. M. Jul 10 '21 at 13:24
  • I want to know the sequence of points (1,2,1) to (1,2,0) .how z coordinate become 1 to 0. – S. M. Jul 10 '21 at 13:27
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    I think the last paragraph of my now edited answer is what you're looking for. – Michael Albanese Jul 10 '21 at 13:28

1 Answers1

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I will use square brackets to denote homogeneous coordinates (this is standard notation), so $[a, b, c] = [ka, kb, kc]$ for all $k \neq 0$.

As you said, given a point $(a, b)$ in the plane, it has homogeneous coordinates $[a, b, 1]$, and a point with homogeneous coordinates $[a, b, 1]$ corresponds to the point $(a, b)$ in the Cartesian plane. Given a point with homogeneous coordinates $[a, b, c]$, then there are two cases to consider:

  1. $c \neq 0$, in which case $[a, b, c] = [\frac{a}{c}, \frac{b}{c}, 1]$ corresponds to the point $(\frac{a}{c}, \frac{b}{c})$, and
  2. $c = 0$, in which case $[a, b, 0]$ is a point on the line at infinity - in particular, it is not a point in the Cartesian plane.

For example, the point $(3, 2)$ has homogeneous coordinates $[3, 2, 1]$, but this is also equal to $[6, 4, 2]$, $[9, 6, 3]$, etc. On the other hand, $[3, 2, \frac{1}{2}]$ is equal to $[6, 4, 1]$ and hence corresponds to the point $(6, 4)$, not $(3, 2)$.

Finally, note that the line through the origin and a point $(a, b) \neq (0, 0)$ can be parameterised by $(at, bt)$ for $t \in \mathbb{R}$, which is given by $[at, bt, 1]$ in homogeneous coordinates. For $t \neq 0$, we have $[at, bt, 1] = [a, b, \frac{1}{t}]$. As $\lim\limits_{t\to\pm\infty}\frac{1}{t} = 0$, we see that the line through the origin and the point $(a, b)$ meets the line at infinity at the point $[a, b, 0]$.

Michael Albanese
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