Midpoint formula: Exercises — Maths deep dive

This page is your practice ladder for the midpoint formula. Climb from simple recognition to full synthesis. Try each problem before opening the solution — every solution is hidden in a collapsible callout so the page tests you honestly.

Reminders of what each symbol means, so nothing is assumed:

$x_{1}, y_{1}$ ::: the horizontal and vertical positions of the first point $A$ .
$x_{2}, y_{2}$ ::: the horizontal and vertical positions of the second point $B$ .
$M$ ::: the midpoint — the single point exactly halfway along the segment $A B$ .

Level 1 — Recognition

Goal: read off coordinates and average them. No traps yet, just fluency.

Recall Solution L1.1

Average the x's, then the y's — that is the whole procedure. $x_{M} = \frac{2 + 10}{2} = \frac{12}{2} = 6$ What/Why: we added the two horizontal positions and split them in half to land halfway across. $y_{M} = \frac{6 + 2}{2} = \frac{8}{2} = 4$ Midpoint $= (6, 4)$ .

Recall Solution L1.2

Negatives change nothing about the method — only about the arithmetic. Keep signs attached. $x_{M} = \frac{- 3 + ( - 1 )}{2} = \frac{- 4}{2} = - 2$ $y_{M} = \frac{- 5 + 3}{2} = \frac{- 2}{2} = - 1$ Midpoint $= (- 2, - 1)$ .

Recall Solution L1.3

A zero is just another number to average — do not skip it. $x_{M} = \frac{0 + 6}{2} = 3, y_{M} = \frac{8 + 0}{2} = 4$ Midpoint $= (3, 4)$ .

Level 2 — Application

Goal: run the formula backwards, and use it inside small procedures.

Recall Solution L2.1

We know the average and one ingredient; we want the other ingredient. Undo the average by multiplying by $2$ . $\frac{2 + x}{2} = 4 \Rightarrow 2 + x = 8 \Rightarrow x = 6$ Why $\times 2$ first? To cancel the " $\div 2$ " that hides inside the average and free $x$ . $\frac{3 + y}{2} = - 1 \Rightarrow 3 + y = - 2 \Rightarrow y = - 5$ $B = (6, - 5)$ . Check via the shortcut: $2 x_{M} - x_{1} = 8 - 2 = 6$ , $2 y_{M} - y_{1} = - 2 - 3 = - 5$ . ✔

Recall Solution L2.2

Why the midpoint? The centre of a circle is exactly halfway along any diameter — that is the definition of a midpoint. So we average $P$ and $Q$ . $x_{C} = \frac{- 2 + 6}{2} = 2, y_{C} = \frac{4 + ( - 2 )}{2} = 1$ Centre $= (2, 1)$ .

Recall Solution L2.3

Set up one equation per axis; each has exactly one unknown. $\frac{a + 9}{2} = 7 \Rightarrow a + 9 = 14 \Rightarrow a = 5$ $\frac{5 + b}{2} = 2 \Rightarrow 5 + b = 4 \Rightarrow b = - 1$ $a = 5, b = - 1$ .

Level 3 — Analysis

Goal: use midpoints to test geometric claims — parallelograms, medians, collinearity.

The figure below sets up L3.1: the purple outline is the quadrilateral $P QR S$ , the two dashed lines (coral and mint) are its diagonals, and the single slate dot where they cross is the shared midpoint. Watch that both dashed lines pass through the same dot — that visual coincidence is exactly the "diagonals bisect each other" test you are about to compute.

Recall Solution L3.1

Why midpoints? Diagonals bisect each other exactly when they share the same midpoint (they cross at the middle of both) — that is the crossing dot in the figure above. So compute the midpoint of each diagonal and compare. The diagonals of $P QR S$ are $P R$ and $QS$ . $M_{P R} = (\frac{1 + 6}{2}, \frac{0 + 5}{2}) = (3.5, 2.5)$ $M_{QS} = (\frac{4 + 3}{2}, \frac{1 + 4}{2}) = (3.5, 2.5)$ Same point ⇒ diagonals bisect each other. Yes — and this makes $P QR S$ a parallelogram (see Properties of parallelograms).

Recall Solution L3.2

Key insight: in a parallelogram the diagonals bisect each other, so $A C$ and $B D$ share a midpoint. First find the midpoint of the known diagonal $A C$ . $M = (\frac{1 + 6}{2}, \frac{2 + 7}{2}) = (3.5, 4.5)$ This must also be the midpoint of $B D$ . Use the reverse formula with $B (4, 3)$ : $D = (2 x_{M} - x_{B}, 2 y_{M} - y_{B}) = (2 (3.5) - 4, 2 (4.5) - 3) = (3, 6)$ $D = (3, 6)$ .

Recall Solution L3.3

Compute the midpoint of $A$ and $C$ and see if it equals $B$ . $M_{A C} = (\frac{- 2 + 2}{2}, \frac{- 1 + 5}{2}) = (0, 2)$ This equals $B$ . Yes, $B$ is the midpoint — so $A$ , $B$ , $C$ are collinear and evenly spaced.

Level 4 — Synthesis

Goal: combine midpoint with distance, section, or centroid ideas.

The next figure carries both L4.1 and L4.2: the purple triangle is $A B C$ , the coral segment is the median from $A$ down to $M$ (the mint dot, the midpoint of side $B C$ ), and the slate dot on that coral segment is the centroid $G$ . Notice $G$ sits closer to $M$ than to $A$ — that is the $2 : 1$ split you will prove numerically below.

Recall Solution L4.1

Step 1 — why a midpoint? A median goes from a vertex to the midpoint of the opposite side (the mint dot $M$ in the figure). So first find $M$ , the midpoint of $B C$ . $M = (\frac{8 + 4}{2}, \frac{3 + 9}{2}) = (6, 6)$ Step 2 — why the distance formula now? We want the length of the coral segment, and the Distance formula measures the straight-line gap between two known points $A (2, 1)$ and $M (6, 6)$ . $A M = (6 - 2)^{2} + (6 - 1)^{2} = 4^{2} + 5^{2} = 16 + 25 = 41$ Median length $= 41 \approx 6.40$ .

Recall Solution L4.2

Why the section formula? $G$ divides $A M$ in the ratio $2 : 1$ (not the middle), so a simple average won't do — we need the weighted version. With $A (2, 1)$ , $M (6, 6)$ and ratio $m : n = 2 : 1$ (from $A$ toward $M$ ): $G = (\frac{m \cdot x _{M} + n \cdot x _{A}}{m + n}, \frac{m \cdot y _{M} + n \cdot y _{A}}{m + n}) = (\frac{2 ( 6 ) + 1 ( 2 )}{3}, \frac{2 ( 6 ) + 1 ( 1 )}{3})$ $G = (\frac{14}{3}, \frac{13}{3})$ Cross-check with the Centroid of a triangle shortcut (average all three vertices): $G = (\frac{2 + 8 + 4}{3}, \frac{1 + 3 + 9}{3}) = (\frac{14}{3}, \frac{13}{3}) ✓$ $G = (\frac{14}{3}, \frac{13}{3})$ .

Recall Solution L4.3

Step 1: find $B$ using the reverse formula from $M$ and $A$ . $B = (2 (3) - 1, 2 (5) - 2) = (5, 8)$ Step 2: now average $B (5, 8)$ with $C (11, 0)$ . $N = (\frac{5 + 11}{2}, \frac{8 + 0}{2}) = (8, 4)$ $N = (8, 4)$ .

Level 5 — Mastery

Goal: unknowns, conditions, and proof-style reasoning.

Recall Solution L5.1

Build one equation per axis. Start with the x-equation (it isolates $k$ cleanly): $\frac{2 k + 4}{2} = 5 \Rightarrow 2 k + 4 = 10 \Rightarrow 2 k = 6 \Rightarrow k = 3$ Check consistency with the y-equation — a good problem must agree: $\frac{3 + ( k - 1 )}{2} = 3 \Rightarrow 3 + (k - 1) = 6 \Rightarrow k = 4 ?$ Substituting $k = 3$ : $\frac{3 + 2}{2} = 2.5 \neq = 3$ . The two axes disagree, so no single $k$ satisfies both. Conclusion: no value of $k$ works — the stated midpoint is impossible. Lesson: always verify both coordinates; a parameter must satisfy the x-and y-condition simultaneously.

Recall Solution L5.2

Why diagonals share a midpoint ⇒ parallelogram? Because that is exactly the bisection condition proved in L3.1. First the four side-midpoints: $E = (\frac{0 + 4}{2}, \frac{0 + 0}{2}) = (2, 0), F = (\frac{4 + 6}{2}, \frac{0 + 4}{2}) = (5, 2)$ $G = (\frac{6 + 2}{2}, \frac{4 + 6}{2}) = (4, 5), H = (\frac{2 + 0}{2}, \frac{6 + 0}{2}) = (1, 3)$ Now the midpoints of diagonals $E G$ and $F H$ : $M_{E G} = (\frac{2 + 4}{2}, \frac{0 + 5}{2}) = (3, 2.5)$ $M_{F H} = (\frac{5 + 1}{2}, \frac{2 + 3}{2}) = (3, 2.5)$ Equal ⇒ diagonals bisect each other ⇒ $E F G H$ is a parallelogram. (This is Varignon's theorem — it holds for every quadrilateral.)

Recall Solution L5.3

Translate the words: " $A$ on the x-axis" means $A = (a, 0)$ ; " $B$ on the y-axis" means $B = (0, b)$ . Now enforce the midpoint condition: $\frac{a + 0}{2} = 3 \Rightarrow a = 6$ $\frac{0 + b}{2} = - 2 \Rightarrow b = - 4$ $A = (6, 0)$ and $B = (0, - 4)$ .

Score yourself

Recall Answer key (all levels)

L1.1 $(6, 4)$ · L1.2 $(- 2, - 1)$ · L1.3 $(3, 4)$ L2.1 $B = (6, - 5)$ · L2.2 centre $(2, 1)$ · L2.3 $a = 5, b = - 1$ L3.1 yes, parallelogram, shared midpoint $(3.5, 2.5)$ · L3.2 $D = (3, 6)$ · L3.3 yes, $B$ is the midpoint L4.1 $41 \approx 6.40$ · L4.2 $G = (14/3, 13/3)$ · L4.3 $N = (8, 4)$ L5.1 no valid $k$ (axes disagree) · L5.2 parallelogram, shared midpoint $(3, 2.5)$ · L5.3 $A = (6, 0), B = (0, - 4)$

Connections

Midpoint formula — the parent topic these exercises drill.
Section formula — used at L4 when the split is not $1 : 1$ .
Distance formula — combined with midpoint at L4.1 (median length).
Centroid of a triangle — the averaging idea extended to three points (L4.2).
Properties of parallelograms — the bisecting-diagonals test (L3, L5.2).
Coordinate Geometry basics — plotting and reading coordinates.