11. Given the line $a x+b y+c=0$ where $a, b, c$ are 3 different elements taken from the set $\{-3,-2,-1,0,1,2,3\}$, and the inclination angle of the line is acute. Then, the number of such lines is $\qquad$ .
Assume $a>0, b<0$, then the number of ways to choose $(a, b, c)$ is $3 \times 3 \times 5=45$. Note that $(1,-1,0),(2,-2,0),(3,-3,0)$ represent the same line, so the total number of different lines is $45-3+1=43$.