The Midpoint of a Segment
Two friends live on opposite sides of town and want to meet up. To be fair, they agree to meet at
the point exactly halfway between their houses. If one house is at
(2, 1) on the map and the other at (8, 5),
where's the fair meeting spot?
Finding it is wonderfully easy — you don't measure anything, you just average the
coordinates. The midpoint's x is the average of the two
x-values, and its y is the average of the two
y-values. Halfway in every direction at once.
M = \left(\tfrac{x_1+x_2}{2},\ \tfrac{y_1+y_2}{2}\right)
For the two houses: x = \tfrac{2+8}{2} = 5 and
y = \tfrac{1+5}{2} = 3, so they meet at
(5, 3). Done.
Why averaging lands you in the middle
Think about just the x-direction for a second. If you start at
x = 2 and finish at x = 8, the number
sitting dead centre between them is their average, \tfrac{2+8}{2} = 5 —
it's exactly 3 away from each end. The same logic handles the
y-direction. Average both, and you've pinned the point that's halfway
along the whole segment.
For a segment from (x_1, y_1) to
(x_2, y_2):
-
the midpoint is
\left(\tfrac{x_1+x_2}{2},\ \tfrac{y_1+y_2}{2}\right) —
average the x's, average the
y's;
-
it lies exactly halfway along the segment, splitting it into two equal halves.
Worked example 1: see the averaging
Step through the figure for the two houses, A = (2, 1) and
B = (8, 5). The midpoint M lands at the
average of the endpoints' coordinates, (5, 3), so it sits right in the
middle of the segment. The dashed guides show 5 is halfway between
2 and 8, and 3 is
halfway between 1 and 5.
Worked example 2: working backward to a missing endpoint
Here's the trickier — and more interesting — version. Suppose you know one endpoint
A = (3, 4) and you know the midpoint
M = (7, 6). Where's the other endpoint
B?
The key idea: the midpoint is halfway, so B is the same
distance beyond M as A is before it.
To get from A to M you go
+4 in x (from 3 to 7) and
+2 in y (from 4 to 6). Take that same
step again from M:
B = (7 + 4,\ 6 + 2) = (11, 8)
Prefer a formula? Since M is the average,
x_M = \tfrac{x_A + x_B}{2}, which rearranges to
x_B = 2x_M - x_A = 2(7) - 3 = 11, and likewise
y_B = 2(6) - 4 = 8. Same answer: (11, 8).
Worked example 3: the halfway meeting point
Priya lives at grid reference (4, 2) and Sam at
(10, 12), where each unit is one kilometre. They want a café that's a
fair, equal walk for both. Where should they look?
- Average the x's: \tfrac{4+10}{2} = 7.
- Average the y's: \tfrac{2+12}{2} = 7.
The fair meeting point is (7, 7) — and because it's the midpoint, each
of them walks the same straight-line distance to get there.
Two mix-ups trip people up every time — here's how to stay clear of both:
-
Midpoint means average, not difference. You add the
coordinates and divide by 2. Some students subtract them — but subtracting gives you the
gaps between the points (the raw material for
distance), not the
middle. For (2, 1) and (8, 5), the midpoint
is (5, 3), not (6, 4).
-
Working backward, don't just halve. Given an endpoint and the midpoint, the
missing endpoint is the same distance beyond the midpoint as the known one is before it —
use x_B = 2x_M - x_A. A common slip is to average
A and M again, which lands you at the
quarter-point, not the far end.
"Average the coordinates" turns out to be the simplest case of a genuinely deep idea: the
centre of mass — the balance point — of a set of points is just the average of all
their coordinates. Two equal weights? Their balance point is the midpoint between them, exactly what
you've been computing. Hang a mobile from that spot and it hangs level.
Push it further and the same trick keeps working. Average the three corners of a triangle and you
get its centroid — balance a cardboard triangle on a pin there and it won't tip.
Average a whole cloud of data points and you get their mean, the "centre" of the
cluster that statisticians live by. One little averaging move, scaling all the way from a school
segment to the balance point of anything.
See it explained