Problem 4. The vertex $A$ of triangle $A B C$ is the intersection of the linear function $y=\frac{3}{4} x+12$ with the $x$-axis, and vertex $B$ is the intersection of the linear function $y=-\frac{4}{3} x+12$ with the $x$-axis. Vertex $C$ is the intersection point of the graphs of these two linear functions.

a) Prove that triangle $A B C$ is a right triangle

b) Calculate the perimeter and area of triangle $A B C$.
Solution. From $\frac{3}{4} x+12=0$ it follows that $x=-16$, so $A(-16,0)$. (3) From $-\frac{4}{3} x+12=0$ it follows that $x=9$, so $B(9,0)$. From $\frac{3}{4} x+12=-\frac{4}{3} x+12$ we get $x=0$, so $y=12$. Therefore, $C(0,12)$. The lengths of the sides of triangle $ABC$ are

$$
a=\sqrt{9^{2}+12^{2}}=15, b=\sqrt{12^{2}+16^{2}}=20, c=9+16=25
$$

Now, from $a^{2}+b^{2}=15^{2}+20^{2}=625=25^{2}=c^{2}$ it follows that triangle $ABC$ is a right triangle.

b) The perimeter of triangle $ABC$ is

$$
L=15+20+25=60
$$

and the area is

$$
P=\frac{15 \cdot 20}{2}=150
$$