The required distance is the hypotenuse of a right triangle with sides of lengths [2 - (-1)] and [4 - (-3)] = 3 and 7 respectively. From the Pythagorean theorem, this distance is the square root of the sum of the squares of the lengths of each side, in this instance the square root of (9 + 49) or the square root of 58. There is no integral square root, but the decimal approximation is about 7.616 units.
8.54
Answer: 1
If you mean points of (1, -2) and (-9, 3) then the distance is about 11 units using the distance formula
Points: (-1, -9) and (4, -2) Distance: (-1-4)2+(-9--2)2 = 74 and the square root of this is the distance which is about 8.602 to 3 decimal places
since you know of one points and the halfway point between the other point. just multiply the halfway point by 2 and this is the total distance between the two points.
8.54
11 points
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
Answer: 1
Points: (-6, 1) and (-2, -2) Distance: 5 units
(3-1)2 + (5-8)2 = 13 and the square root of this is the distance between the points
The distance between the points is two times the square root of 3.
Distance = (9-5)2+(-6-1)2 = 65 and the square root of this is the distance between the points which is about 8.062257748
Points: (2, 1) and (14, 6) Distance: 13
The shortest distance between the two points is zero
If you mean points of (2, 4) and (-1, 8) then the distance works as 5