In order to find the distance between two coordinates, you first need to find the difference between the x and y coordinates. In this case, the difference between the x coordinates is 3-(-2) = 5. The difference between the y coordinates is -4-5 = -9. To find the distance you add up the squares of these differences then find the square root. 52 = 25. -92 = 81. 25+81 = 106.
Thus the distance is the square root of 106, or approximately 10.296
(-3-(-6))2 + (7-4)2 = 18 and the square root of this is the distance between the two points
The distance between the points of (4, 3) and (0, 3) is 4 units
If you mean points of: (-5, 1) and (-2, 3) then the distance is about 3.61 rounded to two decimal places
Distance formula: square root of (x1-x2)2+(y1-y2)2
We use the distance formula to find the distance between the points (2,3) and (3,0) The distance is Square root of ((3^2+(2-3)^2)= Square root of (9+1) Which is square root of 10. This is the distance. This works because if we draw a triangle with one side having length 3 and another side having length 1, we have a right triangle. THis is because the side of length 3 is vertical and the side of length 1 is horizontal. Now the hypotenuse of this triangle is the line between the two points in question. So the length of the hypotenuse is the distance between the points. However, the pythagorean theorem tells us this distance is the square root of 1^2 +3^2=Square root of 10
(3-1)2 + (5-8)2 = 13 and the square root of this is the distance between the points
Points: (-5, -2) and (3, 13)Distance works out as 17 units
Points: (2, 3) and (2, 7) Distance works out as: 4 units
(-3-(-6))2 + (7-4)2 = 18 and the square root of this is the distance between the two points
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
If the points are (3, 2) and (9, 10) then the distance works out as 10
The distance between the points of (2, 3) and (7, 0) is the square root of 34
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
Points: (1, -2) and (1, -5) Distance: 3 units by using the distance formula
If you mean points of: (-5, 1) and (-2, 3) then the distance is about 3.61 rounded to two decimal places
The distance between the points of (4, 3) and (0, 3) is 4 units
If you mean points of (1, -2) and (-9, 3) then the distance is about 11 units using the distance formula