26. Find out if there is an obtuse angle among the interior angles of the triangle with vertices $A(1 ; 2), B(-3 ; 4), C(0 ;-2)$.
Solution. Using formula (1), we find the lengths of the sides of the triangle:

$$
\begin{aligned}
& AB=\sqrt{(-3-1)^{2}+(4-2)^{2}}=\sqrt{16+4}=2 \sqrt{5} \\
& BC=\sqrt{3^{2}+(-2-4)^{2}}=\sqrt{9+36}=3 \sqrt{5} \\
& AC=\sqrt{1^{2}+[2-(-2)]^{2}}=\sqrt{1+16}=\sqrt{17}
\end{aligned}
$$

We find the squares of the lengths of the sides of triangle $ABC: AB^{2}=20$, $BC^{2}=45$, $AC^{2}=17$. From this, it is clear that $BC^{2}>AB^{2}+AC^{2}$, therefore, $\angle BAC$, which lies opposite side $BC$, is obtuse, since it lies opposite the greater side, the square of which exceeds the sum of the squares of the other two sides of the triangle.