Answer:
[tex]since \: the \: three \: sides \: are \: not\: equal : \\ its \: a \to \:\underline{\boxed{scalene}} \\ line \: \overline{AB} = 2 \\ line \: \overline{AC}= 6 \\ line \: \overline{BC} = 2 \sqrt{10} [/tex]
Step-by-step explanation:
[tex]to \: prove \: this \to \: we \: will \: need \: to \: find \\ \: the \: lenghts \: of \: the \: sides \: of \: \\ this \: triangle \to \\ using \: the \: shortest \: distance \: between \: \\ two \: points \to \: d \: = \sqrt{( {x_2 -x_1) }^{2} + (y_2 - y_1)^{2} } \\ \\ \underline{ \boxed{\overline{AB}}} \\ line \: \overline{AB} \: has \: points \to: ( - 3, - 3) \: and \:( 3, 5) \\ d \: = \sqrt{ (( - 3) - ( - 3))^{2} +( 5 - 3))^{2} } \\ d = \sqrt{(0) + (4)} \\ d = \sqrt{4} \\ \boxed{ d = 2 = AB} \\ \\ \underline{ \boxed{\overline{AC}}} \\ line \: \overline{AC} \: has \: points \to: ( - 3, - 3) \: and \:( 5, - 1) \\ d \: = \sqrt{ (( - 3) - ( - 3))^{2} +(( - 1) - 5)^{2} } \\ d = \sqrt{(0) + ( - 6) {}^{2} } \\ d = \sqrt{36} \\ \boxed{ d = 6 = AC} \\ \\ \underline{ \boxed{\overline{BC}}} \\ line \: \overline{BC} \: has \: points \to: ( 3, 5) \: and \:( 5, - 1) \\ d \: = \sqrt{ (5 - 3)^{2} +(( - 1) - 5)^{2} } \\ d = \sqrt{(2) {}^{2} + ( - 6) {}^{2} } \\ d = \sqrt{40} \\ \boxed{ d =2 \sqrt{10} = BC} \\ its \: a \: scalene : since \: the \: three \: sides \: are \: not\: equal[/tex]
