The Midpoint Formula

The numbers $4$ and $10$ have a halfway point of $7$. How did you get it? Now do the same for $-3$ and $5$. Does the method still work when one of the numbers is negative?

A midpoint is just an average, done twice. Average the two x-coordinates to get the midpoint's x, and average the two y-coordinates to get its y. The two axes never interfere with each other.

Before the plane, the line. What is halfway between $4$ and $10$?

Most people answer $7$ instantly, and there are two ways of explaining it. You can say the gap is $6$, so half the gap is $3$, and $4 + 3 = 7$. Or you can say the halfway point is the average, $\dfrac{4 + 10}{2} = 7$.

Both work, but the second is the one worth keeping. It survives negatives without any thought about which number is bigger. Halfway between $-3$ and $5$ is $\dfrac{-3 + 5}{2} = 1$, and no subtraction was needed anywhere.

A midpoint is an average. Everything else in this lesson is that one sentence applied twice.

Now take an interval from $A(x_1, y_1)$ to $B(x_2, y_2)$ and drop a dashed line from each endpoint straight down to the x-axis, as in the diagram.

Those two dashed lines are parallel, and so is the one from the midpoint $M$. Three parallel lines cutting the interval $AB$ into two equal parts must cut the x-axis into two equal parts as well. So the foot of $M$ is halfway between the feet of $A$ and $B$, which makes the x-coordinate of $M$ the average of $x_1$ and $x_2$.

Repeat the argument with horizontal dashed lines onto the y-axis and the same conclusion follows for the y-coordinate. Putting the two together:

$M = \left( \dfrac{x_1 + x_2}{2},\ \dfrac{y_1 + y_2}{2} \right)$

Notice that the two coordinates never mix. The x-coordinates only ever meet other x-coordinates. That is what makes the formula easy to remember and easy to misuse: halving only one of them leaves a point floating well off the interval.

A common version of the question gives you one endpoint and the midpoint, and asks for the other endpoint. There are two ways to handle it.

The formal way is to write the formula and solve. If $A(2, -4)$ and $M(-1, 3)$, then $\dfrac{2 + x_2}{2} = -1$, so $2 + x_2 = -2$ and $x_2 = -4$. The same on the y-coordinates gives $y_2 = 10$.

The quick way is to think in steps. Going from $A$ to $M$ you moved $-3$ across and $+7$ up. To reach $B$ you take exactly the same step again, landing at $(-4, 10)$. The midpoint is halfway, so the second half of the journey is identical to the first.

Use the quick way to get the answer and the formal way to check it, or the other way round. They are the same idea wearing different clothes.

Here is where the formula stops being arithmetic and starts being useful.

In a parallelogram the diagonals bisect each other. That means each diagonal's midpoint is the point where they cross, so both diagonals have the same midpoint.

The converse is also true, and it is the version you use. If the two diagonals of a quadrilateral have the same midpoint, they bisect each other, and the quadrilateral must be a parallelogram.

So to test whether $ABCD$ is a parallelogram, find the midpoint of $AC$ and the midpoint of $BD$. Equal means yes; different means no. Two applications of one formula settle a question about shape.

Be careful with the labelling. The diagonals of $ABCD$ are $AC$ and $BD$, joining opposite vertices. Testing $AB$ and $CD$ tests nothing, because those are sides.

Three errors account for almost every lost mark here.

Subtracting instead of adding. Writing $\dfrac{x_2 - x_1}{2}$ gives half the horizontal gap, which is a length, not a position. It is the right quantity for a different question.

Halving one coordinate only. The midpoint of $(2, 6)$ and $(8, 10)$ is $(5, 8)$, not $(5, 6)$ or $(5, 10)$. Both coordinates get averaged.

Sign slips with negatives. The midpoint of $(-2, -7)$ and $(-6, 1)$ is $\left( \dfrac{-2 + (-6)}{2}, \dfrac{-7 + 1}{2} \right) = (-4, -3)$. Write the brackets in before you simplify.

A quick check catches all three: the midpoint must lie between the endpoints on both axes. If your answer's x is not between $x_1$ and $x_2$, it is wrong, and you know it without redoing the working.

Five midpoint items. Check each answer lies between the two endpoints before you move on.

1. 1 Find the midpoint of $(0, 0)$ and $(8, 6)$.

   Show answer
   Average each coordinate.$(4, 3)$

2. 2 Find the midpoint of $(-5, 2)$ and $(3, 2)$.

   Show answer
   The interval is horizontal, so the y-coordinate is unchanged.$(-1, 2)$

3. 3 Find the midpoint of $(-2, -7)$ and $(-6, 1)$.

   Show answer
   $\left( \dfrac{-8}{2}, \dfrac{-6}{2} \right)$$(-4, -3)$

4. 4 Find the midpoint of $(1.5, -2)$ and $(4.5, 5)$.

   Show answer
   Decimals cause no trouble; add and halve as usual.$(3, 1.5)$

5. 5 $M(0, 0)$ is the midpoint of $AB$ and $A$ is $(-3, 4)$. Find $B$.

   Show answer
   The step from $A$ to $M$ is $+3$ across and $-4$ down; repeat it.$B(3, -4)$