Why it is exact
Distance and midpoint use the same two points and behave completely differently.
The distance squares, adds and roots. That last step almost always produces an irrational number, and every decimal you write is an approximation.
The midpoint adds and halves. Both operations keep rational numbers rational, so if the endpoints are exact so is the answer — a midpoint of 1/6 stays 1/6.
It is worth noticing because it decides where rounding can safely happen. Chain midpoints together and nothing degrades; chain distances and error accumulates.
The reverse problem
“One end is at (1,2) and the midpoint is at (4,6). Where is the other end?” This is asked as often as the forward question and rarely has a calculator for it.
Rearrange: if M = (A + B)/2 then B = 2M − A. Double the midpoint, subtract the known end. Here that is (8−1, 12−2) = (7,10).
A quick check: the midpoint of (1,2) and (7,10) is (4,6). Substituting back is worth the five seconds, because the doubling step is easy to apply to the wrong point.
The page reports this alongside the midpoint, so whichever question you arrived with is answered.
Dividing in a ratio
The midpoint is the 1 : 1 case. For any other split, the section formula applies — and it contains one detail that catches nearly everyone.
To divide from A to B in the ratio m : n, the point is (n·A + m·B) / (m + n). The m weights B, not A. The part nearer your starting point multiplies the FAR endpoint.
A sanity check that settles it: a 1 : 3 split from A should land close to A. With the formula that is (3A + B)/4, which is indeed mostly A. With the weights swapped it would be (A + 3B)/4, which is mostly B and clearly wrong.
Setting m = n = 1 recovers (A + B)/2, which is a useful check that you have the formula the right way round.
Two segments worked by hand: negative coordinates, then three dimensions
Every figure below is the calculator’s own output, and each one is pinned as a test in its engine suite, so this working cannot drift from what the tool computes.
From (−3, 5) to (7, −1)
Take the x-coordinates first: −3 + 7 = 4, and half of 4 is 2. Then the y-coordinates: 5 + (−1) = 4, and half of that is 2. The midpoint is (2, 2).
The brackets around the negative are not decoration. Written as 5 + −1 it is easy to read as 5 + 1, which gives y = 3 and a point that is not on the segment at all.
The check costs one line. The squared distance from (−3, 5) to (2, 2) is 5² + 3² = 34, and from (2, 2) to (7, −1) it is also 5² + 3² = 34. The whole segment squares to 136 = 4 × 34, and a length that doubles has a square that quadruples, so each half is exactly half the segment.
The reverse reading comes with the result: if (7, −1) were the midpoint instead, the far end would be 2 × (7, −1) − (−3, 5) = (17, −7).
From (2, −4, 6) to (−8, 10, 1), and a quarter of the way along
A third coordinate changes nothing about the method; it is one more average. x: (2 + (−8)) ÷ 2 = −3. y: (−4 + 10) ÷ 2 = 3. z: (6 + 1) ÷ 2 = 3.5. The midpoint is (−3, 3, 3.5), and 3.5 is exactly 7/2, not a rounding.
Now the point a quarter of the way from the first end, which is the ratio 1 : 3. The section formula weights each end by the opposite part: (3 × A + 1 × B) ÷ 4. For x that is (6 − 8) ÷ 4 = −0.5; for y, (−12 + 10) ÷ 4 = −0.5; for z, (18 + 1) ÷ 4 = 4.75. The point is (−0.5, −0.5, 4.75).
Swap the weights by mistake and the answer is (−5.5, 6.5, 2.25), three quarters of the way along and nearer the far end. The z-values catch it: the first end sits at z = 6 and the far end at z = 1, so a point near the start needs a z near 6. 4.75 qualifies; 2.25 does not.
The slips that move a midpoint
- Dropping a sign. (−3 + 7) ÷ 2 done as (3 + 7) ÷ 2 gives 5 instead of 2.
- Crossing the pairs. Averaging the x of one point with the y of the other gives a point that lies on the segment only by coincidence.
- Halving some axes and not others. In three dimensions the z-coordinate is the one most often left un-halved, giving 7 instead of 3.5.
- Rounding early. Writing 3.5 as 4 moves the point. Nothing in a midpoint needs rounding, because adding and halving never produce an irrational number.
Where it gets used
Construction and layout. Marking the centre of a wall, a beam or an opening.
Geometry proofs. The midpoint theorem, medians of a triangle, diagonals of a parallelogram bisecting each other — all of them start by computing midpoints.
Graphics and animation. Interpolating halfway between two positions is a midpoint; interpolating a fraction of the way is the section formula, which is what a linear tween does on every frame.
Statistics, carefully. The midpoint of a class interval is used in grouped-data calculations, and it is a midpoint rather than a mean of the underlying values.
The centroid, and what it is not
For two points, the midpoint is the centre. For three or more, the equivalent is the centroid — the average of all the coordinates — and it is not the same as the centre of the shape.
The centroid of a triangle’s three VERTICES sits where the medians meet, two thirds of the way along each from the vertex. That is the balance point of three equal weights at the corners.
The centroid of the triangle as a solid lamina happens to be the same point, which is a genuine coincidence of triangles and does not hold for other shapes. For a general polygon the vertex average and the area centroid are different points.
A common slip
The midpoint formula ADDS and the distance formula SUBTRACTS, and under exam pressure they get swapped.
A midpoint computed with subtraction gives half the gap rather than the centre — a number that looks plausible and is not a point on the segment at all.
The check that catches it immediately: the midpoint must lie BETWEEN the two endpoints on every axis. If it does not, the wrong operation was used.
Sources and methodology
The midpoint and section formulas are curriculum standards; these are the references.
Method. Coordinates are carried as exact rationals and averaged, so nothing is rounded at any point — a midpoint of 1/6 is reported as 1/6 rather than 0.1666666667. The reverse problem and the section formula run on the same exact arithmetic. The suite checks that the returned midpoint is genuinely equidistant from both endpoints across three thousand generated segments, which is the definition rather than the formula. That engine is verified on every change against 73 hand-written assertions, including that the midpoint is equidistant from both endpoints across three thousand generated segments — a check against the definition rather than against the formula that produced it. The count and the per-case breakdown are published on the formula verification page.
Read the guide
The distance between the same two points does involve a square root, which is what makes the midpoint the easier of the two: averaging coordinates is exact in whatever arithmetic you are already using, while a distance is usually irrational and has to be rounded somewhere. Keeping the exact surd and its decimal apart is worth doing whenever the result feeds another step, because rounding early is how a coordinate-geometry answer drifts without any single step looking wrong.