Two lines with slopes $\frac{1}{4}$ and $\frac{5}{4}$ intersect at $(1,1)$. What is the area of the triangle formed by these two lines and the vertical line $x=5$ ?
(A) 5
(B) 10
(C) 8
(D) 12
(E) 15
We start by drawing a graph that includes the point $(1,1)$, the lines with slopes $\frac{1}{4}$ and $\frac{5}{4}$ that pass through this point, the vertical line with equation $x=5$, and the horizontal line $y=1$ (which passes through $(1,1)$ and is perpendicular to the vertical line with equation $x=5$ ). We label the various points of intersection $(A, B, C)$ as shown.

![](https://cdn.mathpix.com/cropped/2024_04_20_688a3b42f69739f2e244g-081.jpg?height=366&width=423&top_left_y=397&top_left_x=905)

We want to determine the area of $\triangle P B C$.

Since $P$ has coordinates $(1,1)$ and $A$ has coordinates $(5,1)$, then $P A=4$.

Since the slope of $P B$ is $\frac{1}{4}$ and $P A=4$, then thinking about slope as "rise over run", we see that $A B=1$.

Since the slope of $P C$ is $\frac{5}{4}$ and $P A=4$, then $A C=5$.

Since $A C=5$ and $A B=1$, then $B C=A C-A B=5-1=4$.

We can view $\triangle P B C$ as having base $B C$ and perpendicular height $P A$. (This is because the length of $P A$ is the perpendicular distance from the line through $B$ and $C$ to the point $P$.)

Therefore, the area of this triangle is $\frac{1}{2}(4)(4)$ which equals 8.

Alternatively, we could note that the area of $\triangle P B C$ equals the area of $\triangle P A C$ minus the area of $\triangle P A B$.

$\triangle P A C$ has base $P A=4$ and height $A C=5$ and so has area $\frac{1}{2}(4)(5)=10$.

$\triangle P A B$ has base $P A=4$ and height $A B=1$ and so has area $\frac{1}{2}(4)(1)=2$.

Thus, the area of $\triangle P B C$ is $10-2=8$.

ANswer: (C)