Respuesta :
Step-by-step explanation:
always Pythagoras with the coordinate differences as sides and the distance the Hypotenuse.
c² = (2/3 - 5/3)² + (-10/9 - -7/9)² = (-3/3)² + (-10/9 + 7/9)² =
= (-1)² + (-3/9)² = 1 + (-1/3)² = 1 + 1/9 = 10/9
c = sqrt(10)/3
Answer:
Step-by-step explanation:
Point 1 ([tex]\frac{2}{3}[/tex] , [tex]\frac{-10}{9}[/tex]) in the form (x1,y1)
Point 2 ( [tex]\frac{5}{3}[/tex] , [tex]\frac{-7}{9}[/tex]) in the form (x2,y2)
use the distance formula
dist = sqrt[ (x2-x1)^2 + (y2-y1)^2 ]
dist = sqrt [ [tex]\frac{5}{3}[/tex] -[tex]\frac{2}{3}[/tex])^2 + ( [tex]\frac{-7}{9}[/tex] - ( [tex]\frac{-10}{9}[/tex] ) )^2 ]
dist = sqrt [ ([tex]\frac{3}{3}[/tex])^2 + ([tex]\frac{3}{9}[/tex])^2 ]
dist = sqrt [ 1 + ([tex]\frac{1}{3}[/tex])^2 ]
dist = sqrt [ [tex]\frac{9}{9}[/tex] + [tex]\frac{1}{9}[/tex] ]
dist = [tex]\sqrt{\frac{10}{9} }[/tex]
dist = [tex]\sqrt{10}[/tex] *[tex]\sqrt{\frac{1}{9} }[/tex]
dist = [tex]\sqrt{10}[/tex] * [tex]\frac{1}{3}[/tex]
dist = [tex]\frac{\sqrt{10} }{3}[/tex]