2. Calculate the area of the triangle formed by the x-axis and the lines $p$ and $q$ whose equations are $p \ldots y=\frac{1}{2} x-3$ and $\quad q \ldots y=-3 x+4$.
2. Sketch:

![](https://cdn.mathpix.com/cropped/2024_05_30_4e3bea37fa3879b468d2g-2.jpg?height=711&width=870&top_left_y=267&top_left_x=593)

Determine the intersections of the lines with the x-axis:

$y=\frac{1}{2} x-3, y=0 \Rightarrow x=6 \rightarrow A(6,0)$

$y=-3 x+4, y=0 \Rightarrow x=\frac{4}{3} \rightarrow B\left(\frac{4}{3}, 0\right)$

Determine the length of side $\overline{A B}$:

$|A B|=6-\frac{4}{3}=\frac{18-4}{3}=\frac{14}{3}$

$1 \text{POINT}$

Determine the intersection of the lines $y=\frac{1}{2} x-3$ and $y=-3 x+4$ to find the height of triangle $A B C$ on side $\overline{A B}$:

$\frac{1}{2} x-3=-3 x+4$

$\frac{1}{2} x+3 x=3+4$

$\frac{7}{2} x=7 \Rightarrow x=2$

It follows that $y=-3 x+4=-3 \cdot 2+4=-6+4=-2$.

Point $C$ has coordinates $C(2,-2)$, and the height on side $\overline{A B}$ is $v=2$. Determine the area of triangle $A B C$:

$P_{A B C}=\frac{|A B| \cdot v}{2}=\frac{\frac{14}{3} \cdot 2}{2}=\frac{14}{3}$

The area of triangle $A B C$ is $P_{A B C}=\frac{14}{3}$ (square units).