Midpoint formula: Worked examples — Maths deep dive

This is the drill floor for the Midpoint formula. Here we hunt down every kind of situation the exam (or life) can throw at you, so that when you meet a new problem you already recognise its species.

Notation reminder (read this before anything else)

Before we write a single formula, let us agree on what every symbol means. Nothing below is assumed.

Now that $x_{1}, y_{1}, x_{2}, y_{2}, A, B$ and $M$ all have meaning, the parent note's formula reads cleanly:

Why the formula is true (the halfway picture)

Do not take the formula on faith — let us see why averaging lands you exactly in the middle.

Deriving the x-coordinate, step by step:

Step A — the horizontal gap is $x_{2} - x_{1}$ .

Why? The sideways distance between the two points is just the difference of their x-values.

Step B — half of that gap is $\frac{x _{2} - x _{1}}{2}$ .

Why? "Middle" literally means you have travelled half the sideways distance.

Step C — start at $x_{1}$ and add that half: $x_{M} = x_{1} + \frac{x _{2} - x _{1}}{2} .$

Why start at $x_{1}$ ? A midpoint is a position, and we measure position starting from point $A$ .

Step D — simplify over a common denominator of $2$ : $x_{M} = \frac{2 x _{1} + x _{2} - x _{1}}{2} = \frac{x _{1} + x _{2}}{2} .$

Why? $2 x_{1} - x_{1} = x_{1}$ , leaving the clean average.

The vertical coordinate follows by the identical argument, so $y_{M} = \frac{y _{1} + y _{2}}{2}$ . That is the whole formula — two averages, each proven, not memorised.

Figure 1 makes this visible: the black segment from $A$ to $B$ , the horizontal gap and vertical gap drawn as black brackets, and the red midpoint sitting after exactly half of each gap.

Alt text: segment AB with horizontal gap and vertical gap bracketed; red midpoint after half of each gap.

The scenario matrix

Every midpoint problem you will ever see belongs to one of these classes:

Cell	Case class	What makes it tricky	Example
C1	Both points in Quadrant I (all positive)	Warm-up, no sign traps	Ex 1
C2	Both points in the same non-I quadrant (e.g. both negative, Quadrant III)	Two negatives averaged	Ex 2
C3	Points across different quadrants (mixed signs)	Negative $+$ positive averaging	Ex 3
C4	Horizontal or vertical segment ( $y_{1} = y_{2}$ or $x_{1} = x_{2}$ ), both coords nonzero	One coordinate stays fixed	Ex 4
C5	One point on an axis (a single zero coordinate)	One point has a $0$ , the other does not	Ex 5
C6	Symmetric about the origin (double zero midpoint)	Equal-and-opposite coordinates cancel	Ex 6
C7	Degenerate: the two points are identical	"Segment" has length $0$	Ex 7
C8	Reverse solve: find a missing endpoint	Must double, not halve	Ex 8
C9	Fractions / odd sums giving a non-integer midpoint	Answer isn't a whole number	Ex 9
C10	Geometry proof (parallelogram via diagonals)	Two midpoints must match	Ex 10
C11	Real-world word problem	Translate words → coordinates	Ex 11
C12	Exam twist: midpoint + a parameter to solve for	Two unknowns, use both averages	Ex 12

Everything below is tagged with the cell it fills. Together they cover the whole table.

C1 — Both points in Quadrant I

Figure 2 below draws segment $A B$ in black with $M$ in red; the two red double-arrows show the equal halves you just proved. Look at how the red dot splits the black line into two visually identical pieces.

Alt text: segment from A(2,4) to B(8,10) with midpoint M(5,7) in red, equal-length arrows on each side.

C2 — Both points in the SAME non-I quadrant (both negative)

The matrix must also cover two points sitting together in a quadrant other than the first — e.g. both in Quadrant III, where both coordinates are negative.

Worked example Example 2 (Cell C2)

Find the midpoint of $A (- 6, - 2)$ and $B (- 2, - 8)$ . (Both points are in Quadrant III.)

Forecast: Everything is negative, so the middle must also be negative — down and to the left. Halfway between $- 6$ and $- 2$ feels like $- 4$ ; between $- 2$ and $- 8$ feels like $- 5$ . Guess $(- 4, - 5)$ .

Step 1. $x_{M} = \frac{- 6 + ( - 2 )}{2} = \frac{- 8}{2} = - 4$ . Why this step? Two negatives add to a bigger negative: $- 6 + (- 2) = - 8$ . We still average, never subtract.

Step 2. $y_{M} = \frac{- 2 + ( - 8 )}{2} = \frac{- 10}{2} = - 5$ . Why this step? Same — $- 2 + (- 8) = - 10$ , halved gives $- 5$ ; the middle stays in Quadrant III.

Answer: $M = (- 4, - 5)$ .

Verify (distance formula): $A M = (- 4 - (- 6))^{2} + (- 5 - (- 2))^{2} = 2^{2} + (- 3)^{2} = 13$ ; $M B = (- 2 - (- 4))^{2} + (- 8 - (- 5))^{2} = 2^{2} + (- 3)^{2} = 13$ . Equal. ✔

Alt text: both points in the lower-left (Quadrant III) with the red midpoint also in Quadrant III.

Figure 3 shows both endpoints and the red midpoint all sitting in the lower-left region — proof that a "same-quadrant, all-negative" pair behaves exactly like Quadrant I, only mirrored.

C3 — Across different quadrants (mixed signs)

Alt text: slanted segment crossing both axes, red midpoint near the origin.

Figure 4 draws the axes; watch the black segment cross from the third quadrant up into the first, with the red midpoint sitting just left of the y-axis.

C4 — Horizontal or vertical segment (one coordinate stays fixed)

A very common edge case: the two points share the same $y$ (a flat, horizontal segment) or the same $x$ (an upright, vertical segment). Because one coordinate never changes, its average is just that same value — nothing moves in that direction.

Figure 5 draws this flat black segment with the red midpoint sitting exactly halfway along it; notice the height line stays perfectly level.

Alt text: horizontal segment from (2,4) to (8,4) with red midpoint (5,4) at the same height.

C5 — One point on an axis (a single zero coordinate)

The common exam subcase: one point sits on an axis (has a zero coordinate) while the other point is fully off the axes. Nothing special — the $0$ is just a number to average.

Figure 6 shows point $A$ pinned on the y-axis, point $B$ off it, and the red midpoint floating between them — off both axes.

Alt text: point A on the y-axis, point B off it, red midpoint between them off both axes.

C6 — Symmetric about the origin (double-zero midpoint)

Figure 7 shows the two points equal-and-opposite on the y-axis, with the red midpoint pinned exactly at the origin.

Alt text: points (0,6) and (0,-6) on the y-axis with red midpoint at the origin (0,0).

C7 — Degenerate: identical points

What if $A$ and $B$ are the same point? The "segment" shrinks to a dot. The formula still works — and tells you the truth.

C8 — Reverse: find the missing endpoint

Now the midpoint is given and one endpoint is hidden. You must run the machine backwards.

Worked example Example 8 (Cell C8)

$M (2, - 1)$ is the midpoint of $A (- 3, 4)$ and $B (x, y)$ . Find $B$ .

Forecast: $M$ is to the right of $A$ (from $x = - 3$ to $x = 2$ ) and below it. So $B$ must be even further right and further down. Rough guess: positive $x$ , negative $y$ .

Step 1. Set up the x-average: $\frac{- 3 + x}{2} = 2$ . Why this step? The x of the midpoint is the average of the two endpoints' x's — that's the definition, written as an equation.

Step 2. Multiply both sides by $2$ : $- 3 + x = 4$ , so $x = 7$ . Why multiply first? To undo the " $\div 2$ " trapping $x$ . We double, we do not halve — that's the classic trap.

Step 3. Set up the y-average: $\frac{4 + y}{2} = - 1 \Rightarrow 4 + y = - 2 \Rightarrow y = - 6$ . Why this step? Same reverse process for the vertical coordinate.

Answer: $B = (7, - 6)$ .

Verify: Midpoint of $(- 3, 4)$ and $(7, - 6)$ : $x_{M} = \frac{- 3 + 7}{2} = 2$ , $y_{M} = \frac{4 - 6}{2} = - 1$ . Gives back $M (2, - 1)$ . ✔

Shortcut check: $B = (2 x_{M} - x_{1}, 2 y_{M} - y_{1}) = (2 (2) - (- 3), 2 (- 1) - 4) = (7, - 6)$ . Same answer. ✔

C9 — Non-integer midpoint (odd sums)

The midpoint is not required to have whole-number coordinates. Don't round — keep the fraction.

C10 — Geometry proof (do diagonals bisect?)

Worked example Example 10 (Cell C10)

A quadrilateral has vertices $P (- 1, 0), Q (3, 1), R (4, 5), S (0, 4)$ . Do its diagonals $P R$ and $QS$ bisect each other?

Forecast: If the two diagonals' midpoints are the same point, they cross at each other's centre → parallelogram. Let's test.

Step 1. Midpoint of diagonal $P R$ : $(\frac{- 1 + 4}{2}, \frac{0 + 5}{2}) = (\frac{3}{2}, \frac{5}{2}) = (1.5, 2.5)$ . Why this step? A diagonal is just a segment; its midpoint is where it would be cut in half.

Step 2. Midpoint of diagonal $QS$ : $(\frac{3 + 0}{2}, \frac{1 + 4}{2}) = (\frac{3}{2}, \frac{5}{2}) = (1.5, 2.5)$ . Why this step? Compute the other diagonal's centre to compare.

Step 3. Compare: both midpoints are $(1.5, 2.5)$ — identical. Why this matters? Diagonals sharing a midpoint means each cuts the other in half. That is exactly the parallelogram property.

Answer: Yes — the diagonals bisect each other, so $P QR S$ is a parallelogram. This is the coordinate-geometry way to prove a shape is a parallelogram: show the diagonals share a midpoint.

Verify: Both computed midpoints equal $(1.5, 2.5)$ . Match confirmed. ✔

Figure 8 draws the quadrilateral in black with both diagonals dashed; the single red dot is where they cross — the shared midpoint you just computed. Notice there is only one red dot: that is the whole point.

Alt text: quadrilateral PQRS with both diagonals dashed meeting at one red midpoint (1.5, 2.5).

C11 — Real-world word problem

Worked example Example 11 (Cell C11)

Two friends are hiking. Aisha is at map-grid point $(2, 3)$ and Ben is at $(10, 7)$ (grid units = km east and km north of the car park). They agree to meet at the exact midpoint of their positions and pitch a tent. Where is the tent, and how does Aisha check she's halfway?

Forecast: The tent is east of Aisha (higher x) and north of her (higher y), roughly between $2$ and $10$ , and $3$ and $7$ . Guess $(6, 5)$ .

Step 1. East coordinate: $x_{M} = \frac{2 + 10}{2} = 6$ km east. Why this step? Their east–west positions average to the meeting east coordinate.

Step 2. North coordinate: $y_{M} = \frac{3 + 7}{2} = 5$ km north. Why this step? Average of the north–south positions.

Answer: Tent at $(6, 5)$ — i.e. $6$ km east, $5$ km north of the car park.

Verify (distance formula, with units): Aisha's walk $= (6 - 2)^{2} + (5 - 3)^{2} = 16 + 4 = 20$ km. Ben's walk $= (10 - 6)^{2} + (7 - 5)^{2} = 16 + 4 = 20$ km. Both walk the identical distance → truly halfway. ✔

C12 — Exam twist: solve for a parameter

Recall Did we cover every cell?

C1 Quadrant I (Ex1), C2 same non-I quadrant / both negative (Ex2), C3 mixed signs (Ex3), C4 horizontal/vertical segment (Ex4), C5 one zero coordinate (Ex5), C6 double-zero symmetric (Ex6), C7 degenerate (Ex7), C8 reverse-solve (Ex8), C9 non-integer (Ex9), C10 geometry proof (Ex10), C11 word problem (Ex11), C12 parameter twist (Ex12). Every row of the matrix is filled. ✅

Active Recall

Connections

Midpoint formula — the parent formula these examples drill.
Section formula — the general ratio version; midpoint is the $1 : 1$ case.
Distance formula — used in every "Verify" step to confirm equal distances $A M = M B$ .
Properties of parallelograms — the logic behind Example 10.
Centroid of a triangle — averaging idea extended to three points.
Coordinate Geometry basics — plotting the points in each figure.

Worked examples — Midpoint formula