Midpoint calculator with working

• $(2, 3)$ and $(8, 11)$: $\left(\dfrac{2 + 8}{2}, \dfrac{3 + 11}{2}\right) = (5, 7)$.
• $(-4, 1)$ and $(6, -5)$: $\left(\dfrac{-4 + 6}{2}, \dfrac{1 + (-5)}{2}\right) = (1, -2)$.
• $(0, 0)$ and $(3, 7)$: $(1.5, 3.5)$.

If you know one end and the midpoint, work backwards: the other end is the same step again. With $A(2, 3)$ and midpoint $M(5, 7)$, the step from $A$ to $M$ is $+3$ in $x$ and $+4$ in $y$, so $B = (8, 11)$. As a formula: $x_2 = 2x_M - x_1$ and $y_2 = 2y_M - y_1$.

The perpendicular bisector of a line segment passes through its midpoint at right angles to it. To find its equation, take the midpoint and the negative reciprocal of the segment's gradient, then use our equation of a line calculator. The centre of a circle is the midpoint of any diameter.

Questions often ask for both the midpoint and the length of a segment. For the length, use Pythagoras on the change in $x$ and $y$, or our distance calculator. For $(2, 3)$ to $(8, 11)$, the changes are 6 and 8, so the length is 10.

• Subtracting instead of adding the coordinates.
• Forgetting to halve.
• Sign slips with negative coordinates: put negatives in brackets.

Practise with our midpoint questions.

• Sketch the two points: the midpoint should look halfway between them.
• Give answers as exact fractions or decimals ($1.5$ or $\tfrac{3}{2}$), not rounded.
• On the SAT, midpoint questions often appear with circles: the centre is the midpoint of a diameter.

The same averaging works in three dimensions: the midpoint of $(1, 2, 3)$ and $(5, 6, 7)$ is $(3, 4, 5)$.

On the SAT, midpoint questions are often combined with circles ("the endpoints of a diameter are...find the centre") or with the equation of a perpendicular bisector. Finding the midpoint is always the first step, so write it down before doing anything else.

You can also use midpoints to check whether a shape is a parallelogram: its two diagonals share the same midpoint.