It is the square root of (-3-0)2+(-4-0)2 = 5
The distance is 4
Points: (3, -4) and (3, 3) Distance: 7 units
60
Points: (-4, 3) and (3, -1) Distance: (3--4)2+(-1-3)2 = 65 and the square root if this is the distance which is just over 8
Use Pythagoras to find the distance between two points (x0,.y0) and (x1, y1): distance = √(change_in_x² + change_in_y²) → distance = √((x1 - x0)² + (y1 - y0)²) → distance = √((4 - 1)² + (-1 -2)²) → distance = √(3² + (-2)²) → distance = √(9 + 9) → distance = √18 = 3 √2
1
The distance between the points of (4, 3) and (0, 3) is 4 units
The distance is 4
Points: (3, -4) and (3, 3) Distance: 7 units
The distance between the points of (4, 3) and (0, 3) is 4 units
The distance between points: (9, 4) and (3, 4) is 6
4
60
3-4 = -1 -21 the distance between -1 and -21 is -20 -20 -1 = -21
Just subtract the lowest number from the greatest number. For example, the distance between 3 and 8, is 8 - 3 = 5 units, the distance between -2 and 3, is 3 - (-2) = 3 + 2 = 5 units, the distance between -4 and -2, is -2 - (-4) = -2 + 4 = 2 units.
Points: (-4, 3) and (3, -1) Distance: (3--4)2+(-1-3)2 = 65 and the square root if this is the distance which is just over 8
-4