6. Find the area of triangle $ABC$, the vertices of which have coordinates $A(0,0), B(1424233,2848467), C(1424234,2848469)$. Round your answer to the hundredths.
Solution. Note that points $B$ and $C$ lie on the line $y=2x+1$. Their abscissas differ by 1, hence $BC=\sqrt{5}$. The length of the height of triangle $ABC$, drawn from vertex $A$, is equal to the distance $h$ from point $A$ to the line $y=2x+1$, which, in turn, is equal to $1 / \sqrt{5}$. The desired area

$$
S_{ABC}=\frac{1}{2} \cdot h \cdot BC=\frac{1}{2}
$$

Answer: 0.50.