Exercise 1.3 in Harris' Algebraic Geometry I am a beginner in Algebraic Geometry and was recommended to read Harris' Book about the subject. I am, however, stuck on the first exercise:

Show that if $\Gamma$ consists of $d$ points and is not contained in a
line, then $\Gamma$ may be described as the zero locus of polynomials
of degree $d-1$ and less.

Upon further research, the only useful hint I have found is that I am supposed to use Lagrange Interpolation (Any set of $d$ points in a projective space is the zero locus of polynomials of degree $d-1$?). I have thus far been unable to apply it here. Could anyone please help me and explain it in an understandable manner?
Thank you!
 A: First consider the easier task of obtaining $\Gamma$ as the zero-locus of a family of polynomials of degree $d$ or less: If $q \not \in \Gamma$ and $p \in \Gamma$ there exists a hyperplane $H_p$ such that $p \in H_p$ and $q \not \in H_p$. Let $L_p$ be a homogenous polynomial of degree one that cuts out $H_p$. Choose such an $L_p$ for a every $p \in \Gamma$ and define a polynomial of degree $d$, $F_q = \Pi_{p \in \Gamma}L_p$. Then $\Gamma$ is the zero-locus of the family $(F_q)_{q \in \Bbb{P}^n - \Gamma}$.
Under the additional hypothesis that $\Gamma$ is not contained in a line the construction above can modified to yield a family of polynomials of degree $d-1$. Here follows a sketch of the steps:

*

*If $q \not \in \Gamma$ and $\Gamma$ is not contained in a line, show that there exists distinct points $r,s \in \Gamma$ such that $q$ is not contained in the line through $r$ and $s$.


*With $r, s, q$ as in 1, show that there is a hyperplane $H$ such that $r,s \in H$ and $q \not \in H$. This means there is a degree one polynomial $L$ such that $L(r)=L(s)=0$ and $L(q) \not = 0$.


*Use $L$ found in step 2 to construct a polynomial $\tilde{F}_q$ of degree $d-1$ such that $\tilde{F}_q(q) \not = 0$ and $\tilde{F}_q(p) = 0$ for all $p \in \Gamma$.
