The lines with equations $x=4$ and $y=3 x$ form a triangle with the positive $x$-axis, as shown. The area of the triangle is
(A) 12
(B) 24
(C) 36
(D) 48
(E) 60

![](https://cdn.mathpix.com/cropped/2024_04_20_e85ef69091c57cb633f3g-063.jpg?height=499&width=610&top_left_y=1309&top_left_x=1145)
Let $O$ be the origin (where the line with equation $y=3 x$ intersects the $x$-axis).

Let $P$ be the point where the line with equation $x=4$ intersects the $x$-axis, and let $Q$ be the point where the lines with equations $x=4$ and $y=3 x$ intersect.

![](https://cdn.mathpix.com/cropped/2024_04_20_688a3b42f69739f2e244g-122.jpg?height=518&width=616&top_left_y=359&top_left_x=814)

The line $x=4$ is perpendicular to the $x$-axis, so the given triangle is right-angled at $P$.

Therefore, the area of the triangle equals $\frac{1}{2}(O P)(P Q)$.

Now $P$ lies on the $x$-axis and on the line $x=4$, so has coordinates $(4,0)$.

Thus, $O P=4$.

Point $Q$ also has $x$-coordinate 4 . Since $Q$ lies on $y=3 x$, then its $y$-coordinate is $y=3(4)=12$. Since $P$ has coordinates $(4,0)$ and $Q$ has coordinates $(4,12)$, then $P Q=12$.

Therefore, the area of the triangle is $\frac{1}{2}(4)(12)=24$.

ANSwer: (B)