Suppose we have $n+1$ control points $P_0 =(a_0, b_0),$ $P_1=(a_1, b_1),$ ... $P_n=(a_n, b_n).$ We will represent the corresponding Bézier curve by points $(x_t,y_t)$ where $t$ runs from 0 to 1. In other words, for each $t$ between 0 and 1 we get a point $(x_t, y_t)$ and together these points form the curve. The formulae for $x_t$ and $y_t$ are \begin{eqnarray*}x_t $=$ \sum_{i=0}^n { n \choose i} (1-t)^{n-i} t^i a_i \\ y_t $=$ \sum_{i=0}^n { n \choose i} (1-t)^{n-i} t^i b_i.\end{eqnarray*} where $${ n \choose i} = \frac{n!}{ (n-i)! i!}$$ is the binomial coefficient.
Suppose we have $n+1$ control points $P_0 =(a_0, b_0),$ $P_1=(a_1, b_1),$ ... $P_n=(a_n, b_n).$ We will represent the corresponding Bézier curve by points $(x_t,y_t)$ where $t$ runs from 0 to 1. In other words, for each $t$ between 0 and 1 we get a point $(x_t, y_t)$ and together these points form the curve. The formulae for $x_t$ and $y_t$ are \begin{eqnarray*}x_t $=$ \sum_{i=0}^n { n \choose i} (1-t)^{n-i} t^i a_i \\ y_t $=$ \sum_{i=0}^n { n \choose i} (1-t)^{n-i} t^i b_i.\end{eqnarray*} where $${ n \choose i} = \frac{n!}{ (n-i)! i!}$$ is the binomial coefficient.