The distance is 5. The x distance is 3, the y distance is 4, and the diagonal is
sqrt(32 + 42) = sqrt (9 + 16) = sqrt 25 = 5
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The distance is 4
It is the square root of (-6-4)2+(1-3)2 = 2 times sq rt of 26 or about 10.198 to 3 decimal places
If you mean (4, 5) and (10, 13) then the distance is 10
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