The answer will be the diagonal (hypotenuse) for a horizontal distance x2-x1 (4) and a vertical distance y2-y1 (4). The square root of the squares is sqrt [42 + 42] = sqrt [32] = approx 5.66
0 0
8.54
Points: (2, 4) and (5, 0) Distance: 5
Points: (-5, -2) and (3, 13)Distance works out as 17 units
The distance between two points is Square root of [ (difference in their 'x' coordinates)2 + (difference in their 'y' coordinates)2 ]
To find the distance between the points (-2, 5) and (-2, 0), we can use the distance formula. Since both points have the same x-coordinate (-2), the distance is simply the difference in their y-coordinates: |5 - 0| = 5. Therefore, the distance between the two points is 5 units.
0 0
10
8.54
If the points are (3, 2) and (9, 10) then the distance works out as 10
To find the distance between two points on a graph, you can use the distance formula: √((x₂ - x₁)² + (y₂ - y₁)²). Plug in the coordinates of the two points to calculate the distance.
the distance between two points is length
The distance between the points of (4, 3) and (0, 3) is 4 units
To find the distance between the points (3, -8) and (3, -19), you can use the distance formula, which is (d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}). Substituting the points into the formula: (d = \sqrt{(3 - 3)^2 + (-19 - (-8))^2} = \sqrt{0 + (-11)^2} = \sqrt{121} = 11). Thus, the distance between the points is 11.
11 points
Points: (-6, 1) and (-2, -2) Distance: 5 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