36 Given the lines $l_{1}: A x+B y+C_{1}=0, l_{2}: A x+B y+C_{2}=0\left(C_{1} \neq C_{2}\right)$, let $A-$ $B+C_{1}+C_{2}=0$, line $l$ passes through the point $H(-1,1)$ and the midpoint $M$ of the segment intercepted by the parallel lines $l_{1}, l_{2}$ lies on the line $x-y-1=0$, then the equation of line $l$ is $\qquad$ .
$36 x+y=0$. Let the line $l^{\prime}$ passing through point $M$ and parallel to $l_{1}$ be $A x+B y+$ $\frac{C_{1}+C_{2}}{2}=0$. The line system passing through the intersection of $l^{\prime}$ and $x-y-1=0$ is $A x+B y+\frac{C_{1}+C_{2}}{2}+$ $\lambda(x-y-1)=0$. Suppose this line passes through point $H(-1,1)$, we get $\lambda=\frac{B-A}{2}$. Therefore, the equation of $l$ is $x+y=0$.