$2 \cdot 117$ If $\frac{x^{2}-b x}{a x-c}=\frac{m-1}{m+1}$ has roots that are equal in magnitude but opposite in sign, then the value of $m$ should be
(A) $\frac{a-b}{a+b}$.
(B) $\frac{a+b}{a-b}$.
(C) $c$.
(D) $\frac{1}{c}$.
(E) 1 .
(3rd American High School Mathematics Examination, 1952)
[Solution] By the given, when $a x-c \neq 0$, we have
or $\quad(m+1) x^{2}-[a(m-1)+b(m+1)] x+c(m-1)=0$.
For the original equation to have equal but opposite roots, it should satisfy:
$m+1 \neq 0$, and $a(m-1)+b(m+1)=0$, which gives
$$
m=\frac{a-b}{a+b} \quad(a+b \neq 0) .
$$

Therefore, the answer is $(A)$.