Question

2/3 - 10/9and5/3 and 7/9

Answer 1

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 2

Answer:

Step-by-step explanation:

Point 1  ($\frac{2}{3}$ , $\frac{-10}{9}$)   in the form (x1,y1)

Point 2 ( $\frac{5}{3}$ , $\frac{-7}{9}$)  in the form (x2,y2)

use the distance formula

dist = sqrt[ (x2-x1)^2 + (y2-y1)^2 ]

dist = sqrt [ $\frac{5}{3}$ -$\frac{2}{3}$)^2 + (  $\frac{-7}{9}$ - ( $\frac{-10}{9}$ ) )^2 ]

dist = sqrt [ ($\frac{3}{3}$)^2 + ($\frac{3}{9}$)^2 ]

dist = sqrt [  1 + ($\frac{1}{3}$)^2 ]

dist = sqrt [  $\frac{9}{9}$ + $\frac{1}{9}$ ]

dist = $\sqrt{\frac{10}{9} }$

dist = $\sqrt{10}$ *$\sqrt{\frac{1}{9} }$

dist = $\sqrt{10}$  * $\frac{1}{3}$

dist = $\frac{\sqrt{10} }{3}$