2. Given real numbers $a, b, c$ satisfy $a+b+c \neq 0$, and $\frac{a}{b+c}=\frac{b}{c+a}=\frac{c}{a+b}=k$. Then the line $y=k x$ -3 must pass through ( ).
(A) the first, second, and third quadrants
(B) the first, second, and fourth quadrants
(C) the second, third, and fourth quadrants
(D) the first, third, and fourth quadrants
2.D.

By the property of ratios, we have
$$
k=\frac{a+b+c}{b+c+c+a+a+b}=\frac{1}{2}.
$$

Thus, the line is $y=\frac{1}{2} x-3$.
Therefore, the line passes through the first, third, and fourth quadrants.