Because the same unit of length is used for each dimension of the Cartesian plane, it is possible to express the distance between two points in terms of their coordinates. The distance formula is based on the Pythagorean Theorem:
The square of the length of the hypotenuse of a right triangle is equal to the sum of the squares of the lengths of the sides.When the lengths of the sides are the differences between corresponding coordinates of each point, the square root of the hypotenuse is the distance between the two points. That is, the distance between any two points,
{x1, y1}, {x2, y2}
in the Cartesian plane is given by:
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d = √(x2 - x1)2 + (y2 - y1)2
Example, find the distance between {-3, 1} and {1, 4}.
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d = √(1 + 3)2 + (4 - 1)2
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= √ 42 + 32 = 5
The distance formula can be extended to higher dimensions. In a three dimensional Cartesian
coordinate system, the distance between any two points,
{x1, y1, z1}, {x2, y2, z2}
would be given by:
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d = √(x2 - x1)2 + (y2 - y1)2 + (z2 - z1)2
Example, find the distance between {-2, 2, -5} and {2, 5, 7}.
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d = √(2 + 2)2 + (5 - 2)2 + (7 + 5)2
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= √ 42 + 32 + 122 = 13
{(x1 + x2)/2, (y1 + y2)/2}
For example, find the midpoint between {2, 1} and {-2, 5}.
{(2 - 2)/2, (1 + 5)/2} -> {0, 3}
The midpoint in three dimensions is similar; given two end points, {x1, y1,
z1} and {x2, y2, z2},
the midpoint is:
{(x1 + x2)/2, (y1 + y2)/2, (z1 + z2)/2}
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r = √(x - xo)2 + (y - yo)2
Squaring both sides, we get the standard form of the equation of a circle:
Example, find the center and radius of the circle given by:r2 = (x - xo)2 + (y - yo)2
Convert to standard form by completing the squares in x, y.0 = x2 + y2 + 2x - 4y + 1
Then, r = 2, xo = -1, yo = 2.0 = (x2 + 2x + 1) - 1 + (y2 - 4y + 4) - 4 + 1 0 = (x + 1)2 + (y - 2)2 - 4 4 = (x + 1)2 + (y - 2)2
r2 = (x - xo)2 + (y - yo)2 + (z - zo)2