Let $\lambda$ denote the Lebesgue measure on $\mathbb R^2$, $\Delta\subseteq\mathbb R^2$ be the triangle spanned by $(0,0)$, $(1,0)$ and $(0,1)$ and $$\mathbb P_r(\Delta):=\left\{p:\Delta\to\mathbb R\mid p(x)=\sum_{|\alpha|\le r}\lambda_\alpha x^\alpha\text{ for all }x\in\Delta\text{ for some }(\lambda_\alpha)_{|\alpha|\le r}\subseteq\mathbb R\right\}$$ for some $r\in\mathbb N_0$. Using the approach $$\int_\Delta f\:{\rm d}\lambda\approx\frac12\sum_{i=1}^nw_if(x_i)\;\;\;\text{for all }f\in\mathcal L^1\left(\left.\lambda\right|_\Delta\right)\;,\tag1$$ with $n\in\mathbb N$, $w_1,\ldots,w_n\in\mathbb R$ and $x_1,\ldots,x_n\in\Delta$ to be determined, we're able to find a quadrature rule which is exact for all $f\in\mathbb P_r(\Delta)$.
If, for example, $r=2$, we're able to find a solution (i.e. a choice for $n$, $w_1,\ldots,w_n$ and $x_1,\ldots,x_n$) with $n=2$ (there are $6$ equations and $6$ unknowns). However, I've read that, in practice, a solution with $n=3$ is preferred (such a solution is not unique, since there are $6$ equations and $9$ unknowns).
In those readings, they say that the reason for that is that, with $n=3$, the obtained solutions are "symmetric".
About which kind of "symmetry" are they talking and why is it a beneficial feature?