## Task B-2.3.

The graphs of two linear functions have slopes of 3 and $\frac{1}{3}$, and intersect at the point $(3,3)$. Determine the area of the triangle bounded by these graphs and the $x$-axis.
## Solution.

The graphs of the given linear functions are lines with the equation $y=k x+l$.

The slope $k$ is given for both lines, as well as a point that belongs to these lines. Therefore, we can determine the equations of the given lines.

$p_{1} \ldots y=3 x+l$, or $3=9+l$ so $l=-6$, and the equation of the first line is $y=3 x-6$.

Similarly for the second line: $p_{2} \ldots y=\frac{1}{3} x+l$, or $3=1+l$ so $l=2$, and the equation of the second line is $y=\frac{1}{3} x+2$.

The sketch of the desired area is:

![](https://cdn.mathpix.com/cropped/2024_05_30_9b9f9f32e16538559f3ag-09.jpg?height=628&width=1176&top_left_y=968&top_left_x=363)

The intersection of the line $p_{1}$ with the $x$-axis is the point $B(2,0)$, and for the line $p_{2}$, the point $A(-6,0)$.

The area of the triangle $A B T$ is given by the expression:

$$
P=\frac{d(A, B) \cdot y_{T}}{2}=\frac{8 \cdot 3}{2}=12
$$

Note: The area of the triangle can also be calculated using the coordinate method.