5.3 Given $a b c \neq 0$, and $\frac{a+b}{c}=\frac{b+c}{a}=\frac{c+a}{b}=p$. Then the line $y=p x+p$ must pass through
(A) the first and second quadrants.
(B) the second and third quadrants.
(C) the third and fourth quadrants.
(D) the first and fourth quadrants.
(China Junior High School Mathematics League, 1998)
［Solution］From the given conditions, we have:
$$
\left\{\begin{array}{l}
a+b=p c, \\
b+c=p a, \\
a+c=p b .
\end{array}\right.
$$

Adding the above three equations, we get $2(a+b+c)=p(a+b+c)$.
$\therefore \quad p=2$ or $a+b+c=0$.
When $p=2$, $y=2 x+2$. Thus, the line passes through the first, second, and third quadrants.
When $a+b+c=0$, let's assume $a+b=-c$, then
$$
p=\frac{a+b}{c}=-1(c \neq 0) .
$$
$\therefore y=-x-1$, thus the line passes through the second, third, and fourth quadrants.
In summary, the line must pass through the second and third quadrants.
Therefore, the answer is $(B)$.