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
∴
S
x
1
+
x
2
2
;
y
1
+
y
2
2
X
=
x
1
+
x
2
2
Y
=
y
1
+
y
2
2
∴
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
∴
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
∴
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.
