1. Given are the line $p$ with the equation $-x+2 y=1$ and the line $q$ with the equation $-4 x+3 y=16$. Let point $A$ be the intersection of line $p$ with the $x$-axis, let point $B$ be the intersection of line $q$ with the $x$-axis, let point $C$ be the intersection of lines $p$ and $q$, and let point $D$ be the orthogonal projection of point $C$ onto the $x$-axis.

a) Calculate and write down the coordinates of points $A, B, C$, and $D$.

b) Calculate the measure of the interior angle $\beta$ in triangle $A B C$.

c) Calculate the area of triangle $A B C$.
1. The intersection of the line $\mathrm{z}$ with the $x$-axis has a $y$ coordinate of 0. Thus, we get $A(-1,0)$ and $B(4,0)$. Solving the system of equations for $p$ and $q$, we obtain the coordinates of point $C\left(-\frac{29}{5},-\frac{12}{5}\right)$. Point $D$ has coordinates $D\left(-\frac{29}{5}, 0\right)$. The angle $\beta$ in triangle $A B C$ is also the exterior angle at vertex $B$ in triangle $B D C$. Segment $D C$ is $\frac{12}{5}$ long, and segment $B D$ has a length of $\frac{29}{5}-4=\frac{9}{5}$. Triangle $D B C$ is a right triangle, and $\tan \beta^{\prime}=\frac{\frac{12}{5}}{\frac{9}{5}}=\frac{4}{3}$, giving us $\beta \doteq 53.13^{\circ}$ and $\beta \doteq 126.87^{\circ}=126^{\circ} 52^{\prime}$.

Side $c$ in triangle $A B C$ has a length of 5. The area of triangle $A B C$ can be calculated in several ways:

(a) Side $c$ in triangle $A B C$ has a length of 3. Segment $D C$ is perpendicular to the $x$-axis and thus to side $c$ in triangle $A B C$. Segment $D C$ is the height to side $c$ in triangle $A B C$, and its length is $\frac{12}{5}$. The area of the triangle can be calculated as $S=\frac{c \cdot v_{c}}{2}=\frac{3 \cdot \frac{12}{5}}{2}=\frac{18}{5}=$ 3.6.
(b) Side $a$ in triangle $A B C$ is also the hypotenuse in the right triangle $B D C$. By the Pythagorean theorem, we get its length as 3. The same result is obtained by calculating the distance of segment $B C$. The area of the triangle can be calculated as $S=\frac{c \cdot a \cdot \sin \beta}{2}=\frac{3 \cdot 3 \cdot \sin 126.87^{\circ}}{2}=3.6$.
(c) In addition to $a=3$ and $c=3$, we also calculate side $b$, which is the length of segment $A C$:

$|A C|=\sqrt{\left(\frac{24}{5}\right)^{2}+\left(\frac{12}{5}\right)^{2}}=\sqrt{\frac{144}{5}}=\frac{12 \sqrt{5}}{5} \doteq 5.37$. Now we can use Heron's formula to calculate the area of the triangle. $s=\frac{a+b+c}{2}=\frac{15+6 \sqrt{5}}{5} \doteq 5.68$ and

$S=\sqrt{s \cdot(s-a) \cdot(s-b) \cdot(s-c)}=$

$\sqrt{\frac{15+6 \sqrt{5}}{5} \cdot \frac{6 \sqrt{5}}{5} \cdot \frac{6 \sqrt{5}}{5} \cdot \frac{15-6 \sqrt{5}}{5}}=\frac{18}{5}=3.6$.

(d) The area of triangle $A B C$ can be calculated as the difference between the areas of triangle $A D C$ and triangle $B D C: S=\frac{\frac{12}{5} \cdot \frac{24}{5}}{2}-\frac{\frac{12}{5} \cdot \frac{9}{5}}{2}=3.6$.

Calculation and writing of $A(-1,0)$ .1 point

Calculation and writing of $B(-4,0)$ 1 point

Calculation and writing of $C\left(-\frac{29}{5},-\frac{12}{5}\right)$ 1 point

Determination and writing of $D\left(-\frac{29}{5}, 0\right)$ 1 point

Calculation or writing of the lengths of sides $|C D|=\frac{12}{5}$ and $|B D|=\frac{9}{5}$ 1 point

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Writing or using the appropriate formula for the area of the triangle $\left(\frac{c \cdot v_{0}}{2}, \frac{c \cdot b \cdot \sin \beta}{2}\right.$, Heron, ... . . 1 point Calculation of necessary quantities..................................................................................................................................

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