Points: (3, -4) and (3, 3)
Distance: 7 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
5 is.
Points: (2, 3) and (2, 7) Distance works out as: 4 units
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
√98, or 9.8995 (4dp)
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
Points: (-4, 5) and (3, 16) Distance: square root of 170 which is about 13
1
The distance between the points of (4, 3) and (0, 3) is 4 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
(-3-(-6))2 + (7-4)2 = 18 and the square root of this is the distance between the two points
5 is.
Points: (2, 3) and (2, 7) Distance works out as: 4 units
If you mean: (-8, 4) and (5, 4) Then the distance between the points works out as 13
50
Points: (-3, -4) and (-8, 1) Distance: square root of 50 or about 7.071 to three decimal places