The coordinates of the vertices of a triangle are: $A(2; -1), B(3; 1), C\left(2^{1999}; 2^{2000}\right)$. Calculate the area of triangle $ABC$.


Notice that $2^{2000}=2 \cdot 2^{1999}$, so the $x, y$ coordinates of point $C$ satisfy $y=2 x$, meaning $C$ lies on the line $y=2 x$.

The slope of line $A B$ is: $m=\frac{1-(-1)}{3-2}=2$, so it is parallel to the line $y=2 x$.

![](https://cdn.mathpix.com/cropped/2024_05_02_f81b1bd4ef8e5489ac76g-1.jpg?height=556&width=429&top_left_y=310&top_left_x=856)

To determine the area of triangle $A B C$, we need to know the height of the triangle, which is equal to the distance between the two parallel lines. Drop a perpendicular from the origin to the line $A B$. The equation of this perpendicular is $y=-\frac{1}{2} x$, so it passes through point $A$. Thus, the distance between the two lines is

$$
\overline{A O}=\sqrt{(-1)^{2}+2^{2}}=\sqrt{5}
$$

The base of the triangle: $\overline{A B}=\sqrt{(1-(-1))^{2}+(3-2)^{2}}=\sqrt{5}$. The area of triangle $A B C$ is:

$$
T=\frac{\sqrt{5} \cdot \sqrt{5}}{2}=2.5 \text { square units. }
$$

Miklós Németh (Budapest, K. Könyves Gymnasium, 8th grade)