The Midpoint of a Segment

The point halfway along

A segment is the part of a straight line between two endpoints. Its midpoint is the point on it that is exactly halfway from one end to the other, so it is the same distance from both ends.

Take the segment from (1, 2) to (5, 8). Which point is its midpoint?

Halfway on a number line

Start with one number line. Halfway from 1 to 5: the gap is 5 − 1 = 4, half the gap is 2, and 1 + 2 = 3. The number 3 is also the average of 1 and 5, because (1 + 5) ÷ 2 = 6 ÷ 2 = 3.

That is always so. Halfway from a to b is $a+\frac{\left(b-a\right)}{2}$a + (b − a)/2. Write a as $\frac{2a}{2}$2a/2 and add: $\frac{\left(2a+b-a\right)}{2}=\frac{\left(a+b\right)}{2}$(2a + b − a)/2 = (a + b)/2, which is the average of a and b. So the number halfway between two numbers is their average.

Half the run and half the rise

From (1, 2) to (5, 8), the run is 5 − 1 = 4 and the rise is 8 − 2 = 6. Halfway along the segment you have gone half of each: 2 across and 3 up, which reaches (1 + 2, 2 + 3) = (3, 5).

Half of each is right because the segment is straight. The slope triangle from (1, 2) to the halfway point has the same shape as the whole one, at half the size, so both its run and its rise are halved.

So each coordinate of the midpoint is halfway between the two ends, on its own axis. Across: (1 + 5) ÷ 2 = 3. Up: (2 + 8) ÷ 2 = 5. The midpoint is (3, 5).

From (1, 2) to the midpoint (3, 5) is 2 across and 3 up, drawn below the line. From (3, 5) to (5, 8) is 3 up and 2 across again, drawn above it. The two halves of the segment are equal.

The midpoint formula

For the segment from $\left(x_1,y_1\right)$(x₁, y₁) to $\left(x_2,y_2\right)$(x₂, y₂), the midpoint is $M=\left(\frac{\left(x_1+x_2\right)}{2},\frac{\left(y_1+y_2\right)}{2}\right)$M = ((x₁ + x₂)/2, (y₁ + y₂)/2). Add the two x-coordinates and halve; then add the two y-coordinates and halve. Each coordinate is averaged separately.

Negative coordinates go in the same way. The midpoint of (−3, 4) and (5, −2) is $\left(\frac{\left(-3+5\right)}{2},\frac{\left(4+\left(-2\right)\right)}{2}\right)=\left(\frac{2}{2},\frac{2}{2}\right)=\left(1,1\right)$((−3 + 5)/2, (4 + (−2))/2) = (2/2, 2/2) = (1, 1).

A and B sit at opposite corners of the dashed box. M is the middle of the box: its x-coordinate, marked on the x-axis, is halfway between the x-coordinates of A and B, and its y-coordinate, marked on the y-axis, is halfway between their y-coordinates. Move either point and M follows. The figure also prints the distance AB, the length of the box’s diagonal.

Finding a missing end

The formula also works backward. M(5, 4) is the midpoint of A(2, 1) and an unknown point B(p, q). Then $\frac{\left(2+p\right)}{2}=5$(2 + p)/2 = 5 and $\frac{\left(1+q\right)}{2}=4$(1 + q)/2 = 4.

Multiply both sides of each equation by 2: 2 + p = 10, so p = 8, and 1 + q = 8, so q = 7. B is (8, 7).

There is a second way to see it. From A(2, 1) to M(5, 4) is 3 across and 3 up. M is halfway, so B is the same step again beyond M: (5 + 3, 4 + 3) = (8, 7).

The usual mistakes

Subtracting instead of adding. For (1, 2) and (5, 8), halving the differences gives (2, 3), which is half the step from one end to the other, not a point on the segment. Add that step to (1, 2) and you reach the midpoint, (3, 5).

Adding without halving. (1 + 5, 2 + 8) = (6, 10) is not between the ends at all. The average needs the division by 2.

Averaging only one coordinate. (3, 8) averages the x-coordinates but keeps the y-coordinate of one end. Both coordinates need averaging.

Looking ahead: the distance between two points

Going across and then up makes a right angle, so the run, the rise and the straight line between two points form a right triangle, and the straight line is its hypotenuse.

So Pythagoras’ theorem gives the distance. From (1, 2) to (4, 6), the run is 3 and the rise is 4, and the distance is $\sqrt{3^2+4^2}=\sqrt{9+16}=\sqrt{25}=5$√(3² + 4²) = √(9 + 16) = √25 = 5.