1. Given sets
$$
\begin{array}{l}
A=\left\{(x, y)\left\{\frac{y-3}{x}-2=a+1\right\},\right. \\
B=\left\{(x, y) \mid\left(a^{2}-1\right) x+(a-1) y=15\right\} .
\end{array}
$$

If $A \cap B=\varnothing$, then all possible values of $a$ are ( )
(A) $-1,1$
(B) $-1, \frac{1}{2}$
(C) $\pm 1, \frac{1}{2}$
(D) $\pm 1,-4, \frac{5}{2}$
- (D).
From the problem, when $a=1$, $B=\varnothing$, then we have $A \cap B=\varnothing$.
When $a=-1$, $a+1=-(a+1)=0$, and $A=$ $\{(x, 3) \mid x \neq 2, x \in \mathbf{R}\}, B=\left\{\left.\left(x,-\frac{15}{2}\right) \right\rvert\, x \in \mathbf{R}\right\}$, then we have $A \cap B=\varnothing$.

When $a, t \pm 1$, we get $A=C /\{(2,3)\}$, where $C=$ $\{(x, y):(y-3)=(a+1)(x-2) \mid, B=$ $\left.\{(x, y)\} y=-(a+1) x+\frac{15}{a-1}\right\}$, from the fact that the two lines are not parallel, we know $C \cap B$ is a single point, but $A \cap B=\varnothing \Rightarrow C \cap B=\{(2,3)\}$ $\Rightarrow 3=-(a+1) \times 2+\frac{15}{a-1} \Rightarrow a=-4$ or $\frac{5}{2}$.