Let $C \subset \mathbb{P}^2_k$ an irreducible plane curve of degree $d >1$ 

 over algebraically closed field $k$. That is $C=V(f(x,y,z))$ where $f \in k[x,y,z]$ homogeneous of degree $d$. Let $\{p_1,...,p_n\}$ be the singular points of $C$ and let $m_i$ the multiplicity of $C$ at $p_i$.

We look at the linear system $\vert L \vert$ of all curves (in $\mathbb{P}^2$) of degree $d − 1$ that have
multiplicity $m_i − 1$ at every singular point $p_i \in C$. $\vert L \vert$ is not empty since e.g. the curve $V(\frac{\partial f}{\partial x})$ is contained in $\vert L \vert$. 

For any $L \in \vert L \vert$, the intersection $L \cap C$ consists of points $p_i$, each with multiplicity
$\ge m_i(m_i − 1)$ and a residual $R$. These build a linear system $\vert R \vert$ on $C$.

Two questions:

1. Why the multiplicity of $L \cap C$ in $p_i$ satisfies only the equality $\ge m_i(m_i − 1)$? Shouldn't it be strictly equal $m_i(m_i − 1)$?

2. Why following two equalities hold?

$$\dim \vert R \vert= \dim \vert L \vert = \binom{d+1}{2}-1 - \sum_i \binom{m_i}{2}$$

Recall that the dimension $\dim \vert D \vert$ of a linear system corresponding to a divisor $D$ is defined as dimension of projective variety $V_D= (\Gamma(X, \mathcal{L}(D))-\{0\}) / k^*$. 

About the second equation it is clear that the $\binom{d+1}{2}-1$ represents the linear system curves of degree $d − 1$ in $\mathbb{P}^2$. The question is why the additional condition to have
multiplicity $m_i − 1$ at every singular point $p_i \in C$ is encoded in $\sum_i \binom{m_i}{2}$?