Why a line passing through a point not belonging to an algebraic set can intersect finitely many of that algebraic set?

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Following is from Smith's Algebraic Geometry book:

If $V \subset \mathbb{A^n}$ is an algebraic set and $p \notin V$ is a point of $\mathbb{A^n}$, show that any line, $\ell$, through $p$ intersects $V$ in a finite number of points (if it intersects it at all).

If a line passes through the point $p \notin V$ and a point $q \in V$ then we have $$\dfrac{x^1_p}{x^1_q} = \dfrac{x^2_p}{x^2_q} = \dots = \dfrac{x^n_p}{x^n_q}.$$ How does one concludes that there are finitely many solutions for this if $x_q$'s are zeros of a set of polynomials?

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The intersection of the line and $V$ is a closed subset of both in the Zariski topology. The closed sets of a line are finite collections of points, the empty set, and the whole line, and we can safely exclude the last two possibilities.

Also, I don't understand your system of equalities. Are you assuming that the line passes through the origin?