12. The area of $\triangle A B C$ with vertices $A(4,8) 、 B(-1,4) 、 C(7,1)$ is


12. $23 \frac{1}{2}$ Hint: As shown in the figure, through the vertices of $\triangle A B C$, construct the rectangle $P Q C S$, $S_{\text {rectangle } P Q C S}=8 \times 7=56, S_{\triangle P A B}=\frac{1}{2} \times 5 \times 4=10, S_{\triangle B Q C}=\frac{1}{2} \times 8 \times 3=$
12, $S_{\triangle A S C}=\frac{1}{2} \times 3 \times 7=10 \frac{1}{2}, S_{\triangle A B C}=56-\left(10+12+10 \frac{1}{2}\right)=23 \frac{1}{2}$.