Midpoint of a Line Segment

The Midpoint Formula

On a number line, the point halfway between $3$ and $11$ is their average, $\dfrac{3+11}{2} = 7$. In the plane you do the same thing twice: once for the $x$-coordinates and once for the $y$-coordinates.

Why does it work? Halfway along the $x$-direction means $x_1$ plus half the horizontal change, and $x_1 + \dfrac{x_2 - x_1}{2} = \dfrac{x_1 + x_2}{2}$. The same is true for $y$.

Back to Maya and Jayden: you can already compute their meeting point. We will do it at the end of the page.

Finding a missing endpoint

Sometimes you know the midpoint $M$ and one endpoint $A$, and you need the other endpoint $B$. Solve the midpoint formula for the unknown.

The vector view

Because $M$ is halfway, the step from $A$ to $M$ equals the step from $M$ to $B$. In vector form,

To show two vectors are equal, subtract coordinates (head minus tail) and compare the components.

Centers and Intercepts

Whenever a problem says "center," look for a midpoint.

Midpoint of the segment cut off by the axes

The intercepts of a line are where it crosses the axes. Set $y = 0$ for the $x$-intercept and $x = 0$ for the $y$-intercept. Then take the midpoint of those two points.

If a problem asks for the midpoint of two midpoints, just apply the formula twice, once for each segment and once for the two results.

Distance and Lines Through a Midpoint

Two more tools show up in almost every midpoint problem.

Since $M$ is halfway, $AM = MB = \dfrac{1}{2}AB$.

Does a point lie on a line?

To test whether a point is on a line, plug in its coordinates. If the equation is true, the point is on the line.

Parallelograms, Rhombuses and Quadrilaterals

Rhombuses

A rhombus is a parallelogram, so its diagonals bisect each other. They are also perpendicular, which creates four right triangles. If the diagonals have lengths $d_1$ and $d_2$, each side is the hypotenuse of a right triangle with legs $\dfrac{d_1}{2}$ and $\dfrac{d_2}{2}$:

To find the side you need the length (or half-length) of both diagonals. When the second diagonal is described by a line, the line tells you where it goes, and a vertex on it tells you how long it is.

What if the second diagonal is given only by its line, with no vertex or length on it? Then the center and the perpendicularity can still be checked, and the side can still be written exactly, but as a formula in one unknown: the half-length $h$ of the second diagonal. Any rhombus with the given first diagonal has side $\sqrt{(\text{half of } AC)^2 + h^2}$, and $h$ can be any positive number, so the side can be any value greater than half of $AC$.

Use the same steps whenever a rhombus is given by one diagonal's endpoints and the line of the other diagonal. For the practice problem with $A(-1, 3)$, $C(3, 1)$ and the line $y = 2x$: the center is $(1, 2)$, which is on $y = 2x$; the slopes $-\dfrac{1}{2}$ and $2$ are negative reciprocals; and half of $AC$ is $\sqrt{5}$, so the side is $\sqrt{5 + h^2}$, where $h$ is half of the other diagonal. If you can't tell $h$ from the problem, give the answer in this form and say what is missing, rather than guessing a number.

The midpoints of any quadrilateral's sides

To prove it with numbers, compute the four midpoints and show that two opposite sides are the same vector. To find an area, use two adjacent side vectors $\mathbf{u} = \langle a, b\rangle$ and $\mathbf{v} = \langle c, d\rangle$:

Check it on a rectangle: $\langle 3, 0\rangle$ and $\langle 0, 2\rangle$ give $|3 \cdot 2 - 0| = 6$.

Triangles: Medians and the Centroid

Rebuilding a triangle from the midpoints of its sides

Each midpoint gives one equation per coordinate. With three midpoints you get a system you can solve by elimination.

The equation of a median

A median is just a line through a vertex and the opposite midpoint. Find the midpoint, then use the slope and point-slope form $y - y_1 = m(x - x_1)$.

If both points have the same $x$-coordinate, the median is a vertical line $x = c$, and the slope formula would divide by zero. Just write $x = c$.

Reflections Through a Point and Over a Line

The reflection $A'$ ("$A$ prime") of $A$ through a point $P$ is the point so that $P$ is the midpoint of $\overline{AA'}$. Some problems call $A'$ the symmetric point of $A$. The key equation is again $A' = 2P - A$.

Reflecting through the origin is the special case $P = (0, 0)$: $(x, y) \to (-x, -y)$. If a problem asks for a second reflection, just use the answer of the first one as the new $A$.

Reflecting over a line

Unknown Coordinates and Parameters

When a coordinate is a letter, the midpoint formula gives an equation for each coordinate. Solve them one at a time.

A midpoint that must lie on a line

Find the midpoint (it will contain the parameter), then substitute into the line's equation.

Minimizing a distance

If the midpoint depends on a parameter $t$, the distance from the origin to it is a function of $t$. Its square $d^2$ is a quadratic in $t$. The smallest $d$ happens at the same $t$ as the smallest $d^2$, and a quadratic $at^2 + bt + c$ with $a > 0$ has its minimum at its vertex, $t = -\dfrac{b}{2a}$.

Endpoints on Lines and Loci

One endpoint on each of two lines

If a segment has midpoint $M$ and its endpoints lie on lines $r$ and $s$, call the endpoint on $r$ a point with a variable, say $P(a, \dots)$. The other endpoint is $Q = 2M - P$. Then require $Q$ to lie on $s$.

The answers may be fractions. That is fine; leave them exact.

Locus of midpoints

A locus is the set of all points that satisfy a condition. To find the locus of the midpoints of segments from a fixed point $F$ to points $P$ on a line:

1. Let $M(x, y)$ be a midpoint and $P(p, q)$ a point on the line.

2. Use the midpoint formula to write $p$ and $q$ in terms of $x$ and $y$.

3. Substitute into the equation of the line.

Summary

Back to the start: Maya at $(1, 7)$ and Jayden at $(9, 3)$ meet at $\left(\dfrac{1 + 9}{2}, \dfrac{7 + 3}{2}\right) = (5, 5)$, five blocks east and five blocks north of the origin of the map.