Midpoint Calculator

Find the midpoint of a segment, a missing endpoint, or midpoint and distance together โ€” with every step shown
Midpoint of (-3, 7) and (5, -1)
Find endpoint B if A = (2, -5) and the midpoint is (6, 1)
Midpoint and distance between (1, 2, 3) and (7, -4, 9)
Midpoint of (0, 0) and (9, 5)

The Midpoint Formula

The midpoint of a segment is the point exactly halfway between its endpoints โ€” the average of the coordinates.

2D, for P1=(x1,y1)P_1 = (x_1, y_1) and P2=(x2,y2)P_2 = (x_2, y_2):

M=(x1+x22,โ€…โ€Šy1+y22)M = \left(\frac{x_1 + x_2}{2},\; \frac{y_1 + y_2}{2}\right)

3D simply averages a third coordinate too:

M=(x1+x22,โ€…โ€Šy1+y22,โ€…โ€Šz1+z22)M = \left(\frac{x_1 + x_2}{2},\; \frac{y_1 + y_2}{2},\; \frac{z_1 + z_2}{2}\right)

The formula is symmetric: swapping the two points changes nothing, because addition is commutative. It assumes a straight segment in a Cartesian coordinate system, and it works for any real coordinates โ€” negatives, fractions, decimals โ€” with no special cases.

A useful check: the midpoint must lie between the endpoints in each coordinate. If your xx-value is outside [minโก(x1,x2),maxโก(x1,x2)][\min(x_1,x_2), \max(x_1,x_2)], you have made a sign error.

Missing Endpoint, and Midpoint with Distance

Finding a missing endpoint

Given one endpoint AA and the midpoint MM, solve the midpoint equation for the other endpoint BB:

xA+xB2=xMโ€…โ€ŠโŸนโ€…โ€ŠxB=2xMโˆ’xA\frac{x_A + x_B}{2} = x_M \;\Longrightarrow\; x_B = 2x_M - x_A

So the compact rule is

B=2Mโˆ’A=(2xMโˆ’xA,โ€…โ€Š2yMโˆ’yA)B = 2M - A = (2x_M - x_A,\; 2y_M - y_A)

Double, then subtract โ€” not "add the difference once". The most common error here is computing Mโˆ’AM - A and stopping.

Midpoint and distance together

The two formulas use the same coordinates but different operations โ€” averages versus differences:

M=(x1+x22,y1+y22)d=(x2โˆ’x1)2+(y2โˆ’y1)2M = \left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right) \qquad d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}

A handy consequence: the distance from either endpoint to the midpoint is exactly d/2d/2, which is a fast way to verify your answer.

Where it gets used

  • Centre of a circle from the endpoints of a diameter.
  • Perpendicular bisector: passes through MM with slope โˆ’1/m-1/m.
  • Centroid of a triangle, and proving a quadrilateral is a parallelogram (its diagonals share a midpoint).

Common Mistakes to Avoid

  • Subtracting instead of averaging. The midpoint uses x1+x2x_1 + x_2; the distance formula uses x2โˆ’x1x_2 - x_1. Mixing them up is the single most frequent error, and it produces a point nowhere near the segment.
  • Dividing only one coordinate by 2. Both (or all three) coordinates must be halved.
  • Sign slips with negatives. For (โˆ’3,7)(-3, 7) and (5,โˆ’1)(5, -1): (โˆ’3+5)/2=1(-3 + 5)/2 = 1, not โˆ’4-4 or 44. Write the addition out before simplifying.
  • Using Mโˆ’AM - A for a missing endpoint. That gives the half-vector, not the endpoint. The correct rule is B=2Mโˆ’AB = 2M - A.
  • Reporting the midpoint as a single number. It is an ordered pair (or triple), not a length.
  • Assuming the midpoint splits the segment into equal slopes. Slope is constant along a straight segment; what the midpoint halves is length.

Examples

Step 1: xx-coordinate: โˆ’3+52=22=1\dfrac{-3 + 5}{2} = \dfrac{2}{2} = 1
Step 2: yy-coordinate: 7+(โˆ’1)2=62=3\dfrac{7 + (-1)}{2} = \dfrac{6}{2} = 3
Step 3: Check: 11 lies between โˆ’3-3 and 55; 33 lies between โˆ’1-1 and 77 โœ“
Answer: M=(1,3)M = (1, 3)

Step 1: Use B=2Mโˆ’AB = 2M - A
Step 2: xB=2(6)โˆ’2=12โˆ’2=10x_B = 2(6) - 2 = 12 - 2 = 10
Step 3: yB=2(1)โˆ’(โˆ’5)=2+5=7y_B = 2(1) - (-5) = 2 + 5 = 7
Step 4: Verify: midpoint of (2,โˆ’5)(2,-5) and (10,7)(10,7) is (122,22)=(6,1)\left(\dfrac{12}{2}, \dfrac{2}{2}\right) = (6, 1) โœ“
Answer: B=(10,7)B = (10, 7)

Step 1: Midpoint: (1+72,โ€…โ€Š2+(โˆ’4)2,โ€…โ€Š3+92)=(4,โˆ’1,6)\left(\dfrac{1+7}{2},\; \dfrac{2+(-4)}{2},\; \dfrac{3+9}{2}\right) = (4, -1, 6)
Step 2: Differences: ฮ”x=6\Delta x = 6, ฮ”y=โˆ’6\Delta y = -6, ฮ”z=6\Delta z = 6
Step 3: Distance: 36+36+36=108=63โ‰ˆ10.39\sqrt{36 + 36 + 36} = \sqrt{108} = 6\sqrt{3} \approx 10.39
Step 4: Check: from (1,2,3)(1,2,3) to (4,โˆ’1,6)(4,-1,6) is 9+9+9=27=33\sqrt{9+9+9} = \sqrt{27} = 3\sqrt{3}, exactly half โœ“
Answer: M=(4,โˆ’1,6)M = (4, -1, 6) and d=63โ‰ˆ10.39d = 6\sqrt{3} \approx 10.39

Frequently Asked Questions

M = ((xโ‚ + xโ‚‚)/2, (yโ‚ + yโ‚‚)/2) โ€” the average of the x-coordinates paired with the average of the y-coordinates. In three dimensions you average the z-coordinates as well. The order of the two points does not matter.

Use B = 2M โˆ’ A: double each coordinate of the midpoint, then subtract the corresponding coordinate of the known endpoint. For A = (2, โˆ’5) and M = (6, 1) this gives B = (10, 7). Subtracting A from M without doubling first is the usual mistake.

The midpoint formula adds and halves the coordinates to locate a point; the distance formula subtracts, squares, sums, and takes a root to produce a length. One returns an ordered pair, the other a single number. They are useful together โ€” the distance from an endpoint to the midpoint is always half the total length.

Yes, and in any number of dimensions. Just average each coordinate separately, so a 3D midpoint is ((xโ‚+xโ‚‚)/2, (yโ‚+yโ‚‚)/2, (zโ‚+zโ‚‚)/2). Nothing about the reasoning changes โ€” the midpoint is still the componentwise average of the endpoints.

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