Let $k$ be a field. Let $y_1,\cdots,y_n$ be homogeneous coordinates in $\mathbb{P}^{n-1}(k)$ and $x_1,\cdots,x_n$ be coordinates in $k^n$. Let $\Gamma$ be the variety defined by the $(y_1,\cdots,y_n)$-homogeneous polynomials $x_i y_j - x_j y_i$. Let $\pi : \Gamma \mapsto k^n$ be the projection map.
Let $L$ be a line parameterized by $tv$ for $v \in k^n - \{0\}$, and $w$ the corresponding point in $\mathbb{P}^{n-1}(k)$. Since we can pick a representative $t'v$ for $w$, it is easy to see that $(w,tv)$ belongs in $\Gamma$.
How can we see that such points actually describe a curve in $\Gamma$? How can we understand the assertion that "distinct lines through the origin in $k^n$ give distinct points in $\pi^{-1}(0)$"?
For reference, this is from Exercise 14 in Chapter 9, Section 7 of Cox-Little-O'Shea.
The variety $\Gamma$ sits in~$k^n \times \mathbb{P}^{n-1}$. Consider the projection $$ f \colon \Gamma \to \mathbb{P}^{n-1}; $$ this is a morphism of algebraic varieties. Your curves are the fibers of $f$, hence they are algebraic curves. And it also follows that they do not intersect.