A line with slope equal to 1 and a line with slope equal to 2 intersect at the point $P(1,6)$, as shown. The area of $\triangle P Q R$ is
(A) 6
(B) 9
(C) 12
(D) 15
(E) 18

![](https://cdn.mathpix.com/cropped/2024_04_20_e85ef69091c57cb633f3g-088.jpg?height=363&width=444&top_left_y=1019&top_left_x=1293)
The slope of line segment $Q P$ is 1 . Since the "rise" of $Q P$ is 6 units, then the "run" of $Q P$ should also be 6 units. Therefore, $Q$ is 6 units horizontally to the left of $P$, and so has coordinates $(-5,0)$.

The slope of line segment $R P$ is 2 . Since the rise of $R P$ is 6 units, then the run of $R P$ is $\frac{1}{2}(6)=3$ units. Therefore, $R$ is 3 units horizontally to the left of $P$, and so has coordinates $(-2,0)$.

(We could have used the coordinates of $P$ and the slopes of the lines to find that the equations of the lines are $y=x+5$ and $y=2 x+4$ and used them to find the coordinates of $Q$ and $R$.) Therefore, $Q R=-2-(-5)=3$ and $P$ is 6 units above the $x$-axis.

Thus, treating $Q R$ as the base of $\triangle P Q R$, we find that its area is $\frac{1}{2}(3)(6)=9$.

ANSwER: (B)