The lines $y=-2 x+8$ and $y=\frac{1}{2} x-2$ meet at $(4,0)$, as shown. The area of the triangle formed by these two lines and the line $x=-2$ is
(A) 15
(B) 27
(C) 30
(D) 36
(E) 45

![](https://cdn.mathpix.com/cropped/2024_04_20_e85ef69091c57cb633f3g-111.jpg?height=626&width=635&top_left_y=1281&top_left_x=1081)
We first determine where the lines $y=-2 x+8$ and $y=\frac{1}{2} x-2$ cross the line $x=-2$.

For the line $y=-2 x+8$, when $x=-2, y=-2(-2)+8=12$, so the point of intersection is $(-2,12)$.

For the line $y=\frac{1}{2} x-2$, when $x=-2, y=\frac{1}{2}(-2)-2=-3$, so the point of intersection is $(-2,-3)$.

![](https://cdn.mathpix.com/cropped/2024_04_20_688a3b42f69739f2e244g-203.jpg?height=637&width=740&top_left_y=1563&top_left_x=752)

Therefore, we can think of $\triangle A B C$ as having base $A B$ of length $12-(-3)=15$ and height being the distance from $C$ to the line segment $A B$, or $4-(-2)=6$.

Therefore, the area of $\triangle A B C$ is $\frac{1}{2}(15)(6)=45$.

ANSWER: (E)