Given two points: (x1, y1) and (x2, y2)
I believe that the midpoint between the two points will be found by moving the average (mean) distance in the x-direction and the average (mean) distance in the y-direction.
The average (mean) between x1 and x2 is (x1 + x2)/2; the average (mean) between y1 and y2 is (y1 + y2)/2.
Therefore, I believe that the midpoint of the two points (x1, y1) and (x2, y2) is ( (x1 + x2)/2, (y1 + y2)/2 ).
To prove that this is correct, I rely upon a theorem from geometry that states that any line segment has only one midpoint.
Now, I plan to show that the point ( (x1 + x2)/2, (y1 + y2)/2 ) is equally distant from the two endpoints, making it a midpoint (the theorem from geometry guarantees that it is the only midpoint).
Distance from (x1, y1 ) to ( (x1 + x2)/2, (y1 + y2)/2 ) is
√[ (x1 - (x1 + x2)/2 )² + (y1 - (y1 + y2)/2 )² ] (using the distance formula)
= √[ (2x1/2 - (x1 + x2)/2 )² + (2y1/2 - (y1 + y2)/2 )² ] (common denominator of 2)
= √[ (2x1/2 - x1/2 - x2/2 )² + (2y1/2 - y1/2 - y2/2 )² ] (distributive property)
= √[ (x1/2 - x2/2)² + (y1/2 - y2/2 )² ] (simplifying)
Similarly, distance from (x2, y2 ) to ( (x1 + x2)/2, (y1 + y2)/2 ) is
√[ (x2 - (x1 + x2)/2 )² + (y2 - (y1 + y2)/2 )² ] (using the distance formula)
= √[ (2x2/2 - (x1 + x2)/2 )² + (2y2/2 - (y1 + y2)/2 )² ] (common denominator of 2)
= √[ (2x2/2 - x1/2 - x2/2 )² + (2y2/2 - y1/2 - y2/2 )² ] (distributive property)
= √[ (x2/2 - x1/2)² + (y2/2 - y1/2 )² ] (simplifying)
Since (x1/2 - x2/2)² = (x2/2 - x1/2)² and (y1/2 - y2/2 )² = (y2/2 - y1/2 )²,
(you can check that by multiplying them out -- FOIL)
( (x1 + x2)/2. (y1 + y2)/2 ) is a midpoint (and the only midpoint).
