Answer:
Isosceles.
Step-by-step explanation:
We are going to check it with the distance formula
d(R,T) = [tex]\sqrt[2]{(3-4)^{2}+(-3-1)^{2}}[/tex]
= [tex]\sqrt[2]{(-1)^{2}+(-4)^{2}}[/tex]
= [tex]\sqrt[2]{1+16}[/tex]
= [tex]\sqrt[2]{17}[/tex].
d(T,S) = [tex]\sqrt[2]{(4-8)^{2}+(1-0)^{2}}[/tex]
= [tex]\sqrt[2]{(-4)^{2}+(1)^{2}}[/tex]
= [tex]\sqrt[2]{16+1}[/tex]
= [tex]\sqrt[2]{17}[/tex].
d(S,R) = [tex]\sqrt[2]{(8-3)^{2}+(0-(-3))^{2}}[/tex]
= [tex]\sqrt[2]{(5)^{2}+(3)^{2}}[/tex]
= [tex]\sqrt[2]{25+9}[/tex]
= [tex]\sqrt[2]{34}[/tex].
Then, as the triangle has two sides with the same length, the triangle is isosceles.
