\(\overline{AB}=\sqrt{(5-3)^2+(-1-(-4))^2}=\sqrt{4+9}=\sqrt{13}\\
\overline{AC}=\sqrt{(6-3)^2+(-2-(-4))^2}=\sqrt{9+4}=\sqrt{13}\\
\overline{BC}=\sqrt{(6-5)^2+(-2-(-1))^2}=\sqrt{1+1}=\sqrt{2}\\\)
The triangle has exactly two sides of the same length, so it's isosceles triangle.
