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Q: What is the distance between 5 -2 and -3 4?

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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

It is the square root of (-3-0)2+(-4-0)2 = 5

Points: (2, 3) and (2, 7) Distance works out as: 4 units

-8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 There is a distance of 12.

The distance is 4

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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.

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

It is the square root of (-3-0)2+(-4-0)2 = 5

What is the distance between (4, -2) and (-1,6)?

It is the square root of (-3-0)2+(-4-0)2 = 5

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

What is the distance between (4, -2) and (-1,6)?

Using Pythagoras: distance = √(difference_in_x^2 + difference_in_y^2) = √((6 - 2)^2 + (3 - 4)^2) = √(16 + 1) = √17 ≈ 4.12

4

Points: (2, 3) and (2, 7) Distance works out as: 4 units

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