Part A
By the distance formula,
[tex]rs=\sqrt{(4-2)^2 + (4-3)^2}=\sqrt{5}\\\\st=\sqrt{(4-5)^2 + (4-0)^2}=\sqrt{17}\\\\rt=\sqrt{(2-5)^2 + (3-0)^2}=3\sqrt{2}[/tex]
Part B
[tex]m_{rs}=\frac{4-3}{4-2}=\frac{1}{2}\\\\m_{rt}=\frac{3-0}{2-5}=-1\\\\m_{st}=\frac{4-0}{4-5}=-4[/tex]
Part C
The triangle is scalene because each side has a different length.