Triangle $A B C$ has vertices $A(1,2), B(4,0)$ and $C(1,-4)$. The area of $\triangle A B C$ is
(A) 18
(D) 10
(B) 12
(E) 9
(C) 8

![](https://cdn.mathpix.com/cropped/2024_04_20_6ed09463f225f8ba1f07g-129.jpg?height=450&width=458&top_left_y=1317&top_left_x=1300)
Side $A C$ of $\triangle A B C$ is parallel to the $y$-axis and so is perpendicular to the $x$-axis, and crosses the $x$-axis at $(1,0)$. So we can think of $A C$ as the base of the triangle (length 6) and the part of the $x$-axis inside $\triangle A B C$ as its height (length 3). So the area of $\triangle A B C$ is $\frac{1}{2}(6)(3)=9$.

ANSWER: (E)