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 $\left(x_{2},\right.$ $y $$_{2}$ $)$ is given by $m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}, x_{2} \neq x_{1}$

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

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

Slope of $CA \left( m _{3}\right)=\frac{4+1}{4+1}=\frac{5}{5}=1$

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.