Using Pythagoras:
distance = √(change_in_x2 + change_in_y2)
= √((5 - -8)2 + (4 - 4)2)
= √(132 + 02)
= √(132)
= 13 units.
If you mean: (-8, -3) and (4, 5) then the distance is 4 times square root 13
When y=4, x=8/5
4/5 - 1/2 = 8/10 - 5/10 =(8-5)/10 =3/10 =0.3 (point three)
Using the distance formula the length of ab is 5 units
8 times 4 is 32, but more to the point here, it is 4 eights. 8 times 5 is 40, but more to the point here, it is 5 eights. Obviously then, 5 eights is 1 eight more than 4 eights. Or 8. (and if you don't believe me, subtract 32 from 40 and see if I was right!)
If you mean points of (5, 4) and (5, -4) then the distance works out as 8
If you mean: (-8, 4) and (5, 4) Then the distance between the points works out as 13
If you mean: (-8, -3) and (4, 5) then the distance is 4 times square root 13
7.62
3
If you mean points of: (2, 5) and (-4, 8) Distance is the square root of (2--4)^2+(5-8)^2 = 6.708 rounded
-8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 There is a distance of 12.
5x/8+10y/13=-3/4
If you mean points of (2, 4) and (-1, 8) then the distance works as 5
When y=4, x=8/5
Fractions that are equal to .5 are 5/10, 1/2, 2/4, 4/8, etc.
4/5 - 1/2 = 8/10 - 5/10 =(8-5)/10 =3/10 =0.3 (point three)