The midpoint $M$ of a segment joining $A(x_1,y_1)$ and $B(x_2,y_2)$ is simply the average of the two $x$-coordinates and the average of the two $y$-coordinates:

Thinking of it as "the mean of the coordinates" is the best way to avoid confusing it with gradient or distance, which both use subtraction rather than addition.

Always substitute negative values in brackets to avoid sign slips. For $A(-4,3)$ and $B(8,-12)$:

Midpoint: $(2,-4.5)$. Non-integer answers are common and perfectly acceptable as a fraction or decimal.

Method (vector shift): work out the "shift" from the known endpoint to the midpoint, then apply the same shift again to reach the other endpoint.

Example: Given $A(4,8)$ and midpoint $M(6,9)$, find $B$.

1. Shift from $A$ to $M$: $x$: $6-4=+2$; $y$: $9-8=+1$.
2. Apply the same shift from $M$ to $B$: $x_B = 6+2=8$; $y_B=9+1=10$.

$B=(8,10)$.

To show that the diagonals of a quadrilateral bisect each other (a defining property of a parallelogram), calculate the midpoint of both diagonals — if they're identical, the diagonals cross at the same central point, proving the bisection property. This is often combined with the distance formula to additionally confirm side lengths or symmetry.

Worked Example

$P(5,1)$ and $Q(8,6)$ are two vertices of a parallelogram, with $M$ the midpoint of one diagonal. Find $M$, and state it as a coordinate pair with any fractions simplified.

1. $x_M = \dfrac{5+8}{2}=\dfrac{13}{2}=6.5$
2. $y_M = \dfrac{1+6}{2}=\dfrac{7}{2}=3.5$

$M=(6.5, 3.5)$.

All three coordinate-geometry formulas use the same two points $(x_1,y_1)$ and $(x_2,y_2)$, but combine them differently:

Keeping this table in mind is the fastest way to avoid reaching for the wrong formula under exam pressure — if the question wants a "centre" or "average" point, it's midpoint; if it wants "steepness" or "length," it's one of the other two.