Without using the Pythagoras theorem,

The vertices of the given triangle are A (4, 4), B (3, 5), and C (–1, –1).

It is known that the slope $(m)$ of a non-vertical line passing through the points $\left(x_{1}, y_{1}\right)$ and $\left(x_{2}, y_{2}\right)$ is given by $m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}, x_{2} \neq x_{1}$.

$\therefore$ Slope of $\mathrm{AB}\left(m_{1}\right)=\frac{5-4}{3-4}=-1$

Slope of BC $\left(m_{2}\right)=\frac{-1-5}{-1-3}=\frac{-6}{-4}=\frac{3}{2}$

Slope of $C A\left(m_{3}\right)=\frac{4+1}{4+1}=\frac{5}{5}=1$

It is observed that $m_{1} m_{3}=-1$

This shows that line segments AB and CA are perpendicular to each other

i.e., the given triangle is right-angled at A (4, 4).

Thus, the points (4, 4), (3, 5), and (–1, –1) are the vertices of a right-angled triangle.