6. As shown in Figure 3, the graph of the function $y=m x-4 m$ intersects the $x$-axis and $y$-axis at points $N$ and $M$, respectively. Points $A$ and $B$ on $MN$ have projections on the $x$-axis at $A_{1}$ and $B_{1}$, respectively. If $O A_{1}+O B_{1}>4$, then the relationship between the area $S_{1}$ of $\triangle O A_{1} A$ and the area $S_{2}$ of $\triangle O B_{1} B$ is ( ).
(A) $S_{1}>S_{2}$
(B) $S_{1}=S_{2}$
(C) $S_{1}<S_{2}$
(D) Cannot be determined
6.A.

Let $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)$. Then
$$
\begin{array}{l}
y_{1}=m x_{1}-4 m, y_{2}=m x_{2}-4 m . \\
\text { Also, } S_{1}=\frac{1}{2} O A_{1} \cdot A A_{1}=\frac{1}{2} x_{1}\left(m x_{1}-4 m\right), \\
S_{2}=\frac{1}{2} O B_{1} \cdot B B_{1}=\frac{1}{2} x_{2}\left(m x_{2}-4 m\right),
\end{array}
$$

Then $S_{1}-S_{2}=\frac{1}{2} m\left(x_{1}^{2}-x_{2}^{2}\right)-2 m\left(x_{1}-x_{2}\right)$
$$
=\frac{1}{2} m\left(x_{1}-x_{2}\right)\left(x_{1}+x_{2}-4\right) \text {. }
$$

By the problem, $m>4$. Hence $S_{1}-S_{2}>0$.