Let's think of the graphs:
|
|
| 1------( y = 1 )-----
|
--------------------- (x axis)
|
| -1
|
| -2
|
| -3 -------( y = -3 ) -----------
You can see the distance between these two lines is 4.
You can just subtract the two lines, since they are parallel. 1 - (-3) = 4
The distance is 4
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
7.2111 (rounded)
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
Distance between (1, 3) and (-1, 0.5) = sqrt[(1 - -1)2 + (3 - 0.5)2] = sqrt(22 + 2.52) = sqrt(10.25) = 3.20 approx.
4
It is sqrt(3)/2 = 0.8660 approx.
The distance is 4
Abs(x-y) is the distance between x and y So it's the distance between any point and 1+i that must be equal to 3, so it's a circle centered on 1+i point
Using the map scale, 1 inch represents 5 miles. If the map distance between Barnesville and Carlton is 3 inches, the actual distance between them would be 3 inches x 5 miles = 15 miles.
3-4 = -1 -21 the distance between -1 and -21 is -20 -20 -1 = -21
3 and 1/2 miles
3 and 1/2 miles
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
7.2111 (rounded)
(3-1)2 + (5-8)2 = 13 and the square root of this is the distance between the 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