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
7.2111 (rounded)
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
Distance between (1, 3) and (-1, 0.5) = sqrt[(1 - -1)2 + (3 - 0.5)2] = sqrt(22 + 2.52) = sqrt(10.25) = 3.20 approx.
(-3-5)2+(-1--1)2 = 64 and the square root of this is the distance which is 8
4
3-4 = -1 -21 the distance between -1 and -21 is -20 -20 -1 = -21
3 and 1/2 miles
3 and 1/2 miles
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
7.2111 (rounded)
(3-1)2 + (5-8)2 = 13 and the square root of this is the distance between the points
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
Distance between (1, 3) and (-1, 0.5) = sqrt[(1 - -1)2 + (3 - 0.5)2] = sqrt(22 + 2.52) = sqrt(10.25) = 3.20 approx.
The distance between the points is two times the square root of 3.
Using the distance formula from (3, 1) to (7, 1) is 4 units
(-3-5)2+(-1--1)2 = 64 and the square root of this is the distance which is 8