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# Midpoint of a line

## Analytical Geometry; Midpoint of a line

Sometimes, knowing the co-ordinates of the middle point or midpoint of a line is useful. For example, what is the midpoint of the line between point PP with co-ordinates (2;1)(2;1) and point QQ with co-ordinates (-2;-2)(-2;-2).

The co-ordinates of the midpoint of any line between any two points AA and BB with co-ordinates (x1;y1)(x1;y1) and (x2;y2)(x2;y2), is generally calculated as follows. Let the midpoint of ABAB be at point SS with co-ordinates (X;Y)(X;Y). The aim is to calculate XX and YY in terms of (x1;y1)(x1;y1) and (x2;y2)(x2;y2).

X = x 1 + x 2 2 Y = y 1 + y 2 2 âˆ´ S x 1 + x 2 2 ; y 1 + y 2 2 X = x 1 + x 2 2 Y = y 1 + y 2 2 âˆ´ S x 1 + x 2 2 ; y 1 + y 2 2
(1)

Then the co-ordinates of the midpoint (SS) of the line between point PP with co-ordinates (2;1)(2;1) and point QQ with co-ordinates (-2;-2)(-2;-2) is:

X = x 1 + x 2 2 = - 2 + 2 2 = 0 Y = y 1 + y 2 2 = - 2 + 1 2 = - 1 2 âˆ´ S is at ( 0 ; - 1 2 ) X = x 1 + x 2 2 = - 2 + 2 2 = 0 Y = y 1 + y 2 2 = - 2 + 1 2 = - 1 2 âˆ´ S is at ( 0 ; - 1 2 )
(2)

It can be confirmed that the distance from each end point to the midpoint is equal. The co-ordinate of the midpoint SS is (0;-0,5)(0;-0,5).

P S = ( x 1 - x 2 ) 2 + ( y 1 - y 2 ) 2 = ( 0 - 2 ) 2 + ( - 0 . 5 - 1 ) 2 = ( - 2 ) 2 + ( - 1 . 5 ) 2 = 4 + 2 . 25 = 6 . 25 P S = ( x 1 - x 2 ) 2 + ( y 1 - y 2 ) 2 = ( 0 - 2 ) 2 + ( - 0 . 5 - 1 ) 2 = ( - 2 ) 2 + ( - 1 . 5 ) 2 = 4 + 2 . 25 = 6 . 25
(3)

and

Q S = ( x 1 - x 2 ) 2 + ( y 1 - y 2 ) 2 = ( 0 - ( - 2 ) ) 2 + ( - 0 . 5 - ( - 2 ) ) 2 = ( 0 + 2 ) ) 2 + ( - 0 . 5 + 2 ) ) 2 = ( 2 ) ) 2 + ( - 1 . 5 ) ) 2 = 4 + 2 . 25 = 6 . 25 Q S = ( x 1 - x 2 ) 2 + ( y 1 - y 2 ) 2 = ( 0 - ( - 2 ) ) 2 + ( - 0 . 5 - ( - 2 ) ) 2 = ( 0 + 2 ) ) 2 + ( - 0 . 5 + 2 ) ) 2 = ( 2 ) ) 2 + ( - 1 . 5 ) ) 2 = 4 + 2 . 25 = 6 . 25
(4)

It can be seen that PS=QSPS=QS as expected.

The following video provides a summary of the midpoint of a line.

Figure 3
Khan academy video on midpoint of a line

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