1. Given the lines $l_{1}: a x+2 y+6=0$, $l_{2}: x+(a-1) y+a^{2}-1=0$.
If $l_{1} \perp l_{2}$, then $a=$ $\qquad$
\begin{array}{l}\text {-1. } \frac{2}{3} . \\ l_{1} \perp l_{2} \Leftrightarrow a \times 1+2(a-1)=0 \\ \Leftrightarrow a=\frac{2}{3} .\end{array}