solution verification for finding ideal point(point at infinity) of a line

Question

I have the line l represented in the following way:

$$ l: \left\{ \begin{array}{c} x+y-2t=0 \\ 2x+z-3t-0 \end{array} \right. $$

My idea is the following:

Since it's an point at infinity then $t=0$. I choose $x=1 \implies y=-1 \implies z=-2$. Therefore a point at infinity should be the point $P(1,-1, -2, 0)$.

Is that correct?

Answer 1

You idea is correct.

The plane at infinity has equation $t=0$. Then you have the system of equations:

\begin{cases} x+y=0\\ 2x+z=0 \end{cases}

You can rewrite it as

\begin{cases} y=-x\\ z=-2x \end{cases}

So every solution has the form $(x,-x,-2x,0)$. Since the system of coordinates we are using is homogeneous, we can choose a non zero value for $x$. By putting $x=1$, for example, we obtain $ P(1,-1,-2,0). $