If you mean points of (-5, 1) and (-2, 3) then using the distance formula it is the square root of 13 or about 3.61 rounded to 2 decimal places
Points: (2, 4) and (5, 0) Distance: 5
If you mean points of (-5, 1) and (-2, 3) then using the distance formula it is the square root of 13 or about 3.6
You can use Pythagoras to solve this to get the co-ordinates. from Pythagoras we have a^2 + b^2 = c^2 For this a and b will be the x and y distance and c the distance between the 2 points so we get (7-2) ^2 + (8-y1)^2 = 13^2 So we need so solve this to get a correct solution which is (8-y1)^2 = 144 so 8 - y1 = + or - 12 so y1 = -4 or y1 = 20
0 0
Points: (2, 1) and (14, 6) Distance: 13
(3-1)2 + (5-8)2 = 13 and the square root of this is the distance between the points
If you mean points of: (2, 1) and (14, 6) then the distance is 13
the distance between two points is length
(2-5)2+(5-7)2 = 13 and the square root of this is the distance which is about 3.60555 to 5 decimal places
If you mean points of (-5, 1) and (-2, 3) then using the distance formula it is the square root of 13 or about 3.61 rounded to 2 decimal places
11 points
Points: (-6, 1) and (-2, -2) Distance: 5 units
Distance2 = (21-9)2 + (16-11)2 = 169 and the square root of this is the distance which is 13 units
Points: (2, 2) and (8, -6) Distance: 10
length
the distance between 2 points