If $P_1,\ldots,P_5\in \mathbb{P}^1\times\mathbb{P}^1$ are five arbitrary points, how can I prove that there exists a curve $C\subseteq\mathbb{P}^1\times\mathbb{P}^1$ of bidegree $(1, 2)$ such that $P_1,\ldots,P_5\in C$?
First of all, I think that if we denote the coordinates of $\mathbb{P}^1\times\mathbb{P}^1$ as $\big( [x_0:x_1], [y_0: y_1]\big)$, then a curve $C\subseteq\mathbb{P}^1\times\mathbb{P}^1\subseteq \mathbb{P}^3$ has bidegree $(r, s)$ if $C$ is the projective zero set of a homogeneous polynomial $G=G(x_0, x_1, y_0, y_1)$ of degree $r+s$ such that $G$ is homogeneous of degree $r$ WRT $x_0, x_1$ and homogeneous of degree $s$ WRT to $y_0, y_1$ (lets say then that $G$ is bihomogeneous of bidegree $(r, s)$).
Now, I can't think of a feasible way of solving such problem. I was thinking about classical problems such as Five points determine a conic and There is a cubic passing through 9 points, but I'm unable to use these techniques for this problem. I also don't have much tools to tackle this (just basic algebraic geometry up to Bézout's theorem, not any fancy stuff about divisors, linear systems and that all).
Bonus: When can we say that $C$ is uniquely determined?
Regards