Answer:
1. 8
2 . 11
[tex]3 . \sqrt{128} \ or \ 8 \sqrt2[/tex]
Step-by-step explanation:
[tex]Distance = \sqrt{(x_2 -x_1)^2 + (y_2 - y_1)^2 }[/tex]
1 . ( 5 , - 2 ) and ( 5 , 6 )
[tex]Distance = \sqrt{(5 -5)^2 + ( 6-(-2))^2} = \sqrt{ 0 + 8^2 } = \sqrt{64} = 8[/tex]
2 . ( 8 , 6 ) and ( - 3 , 6 )
[tex]Distance = \sqrt{( -3 -8)^2 + ( 6 - 6)^2 } = \sqrt{ (-11)^2 + 0 } = \sqrt { 121 } = 11[/tex]
3. ( 5 , - 2 ) and ( -3 , 6 )
[tex]Distance = \sqrt{(-3 - 5)^2 + ( 6 --2)^2} = \sqrt{(-8)^2 + ( 8)^2} = \sqrt{ 64 + 64 } = \sqrt{128}[/tex]
[tex][ \ \sqrt{128} = \sqrt{ 2 \times 64} = \sqrt{ 2 \times 8^2 } = 8 \sqrt{2} \ ][/tex]