Midpoint | Year 9 | Ascent Library

01 / LEARN

Understand the idea.

Halfway in two directions

The midpoint splits a segment into two equal lengths. Halfway horizontally is the average of the two $x$ -coordinates; halfway vertically is the average of the two $y$ -coordinates. Keep these two calculations separate.

Averaging works even when a segment slopes or crosses an axis. Include negative signs in each sum. An answer between grid lines is allowed: whole-number endpoints can have a midpoint with fractional coordinates.

M = (\frac{x_{1} + x_{2}}{2}, \frac{y_{1} + y_{2}}{2})

Equal-unit Cartesian grid from negative six to eight on both axes. A gold segment joins the sage points A at negative four, negative two and B at six, four. Both coordinates are labelled; no midpoint is marked.

Work backwards from the middle

When one endpoint is missing, write the midpoint formula with that coordinate unknown. Multiplying by $2$ gives the sum of the endpoints, so subtract the known endpoint to recover the missing one.

You can also continue the same horizontal and vertical steps beyond the midpoint. Check by averaging your recovered endpoint with the known endpoint; the result must be the given midpoint in both coordinates.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Across the axes

Find the midpoint of $A (- 4, - 2)$ and $B (6, 4)$ .

Worked solution

1
Average the horizontal coordinates.
$x_{M} = \frac{- 4 + 6}{2} = \frac{2}{2} = 1$
The midpoint is halfway between the endpoint x-values.
2
Average the vertical coordinates.
$y_{M} = \frac{- 2 + 4}{2} = 1$
Use the y-values only for the vertical position.
3
Write the ordered pair.
$M = (1, 1)$
Coordinates are written in horizontal, then vertical order.

Answer

$(1, 1)$

EXAMPLE 02

Recover an endpoint

$A = (2, - 3)$ and the midpoint of $A B$ is $(5, 1)$ . Find $B$ .

Worked solution

1
Use the horizontal midpoint equation.
$\frac{2 + x_{B}}{2} = 5$
The average of the two endpoint x-values must be 5.
2
Solve for the missing horizontal coordinate.
$\begin{aligned} 2 + x_{B} & = 10 \\ x_{B} & = 8 \end{aligned}$
Double the midpoint coordinate, then subtract the known endpoint coordinate.
3
Recover the vertical coordinate.
$\begin{aligned} \frac{- 3 + y_{B}}{2} & = 1 \\ - 3 + y_{B} & = 2 \\ y_{B} & = 5 \end{aligned}$
The same averaging relationship holds vertically.
4
State and check the endpoint.
$B = (8, 5), (\frac{2 + 8}{2}, \frac{- 3 + 5}{2}) = (5, 1)$
Both averages reproduce the stated midpoint.

Answer

$B = (8, 5)$

03 / YOUR TURN

Try it yourself.

Check your understanding

Find the midpoint of $(- 3, 4)$ and $(5, - 2)$ .

Show answer & reasoning

$(1, 1)$

The averages are $\frac{- 3 + 5}{2} = 1$ and $\frac{4 + (- 2)}{2} = 1$ , so the midpoint is $(1, 1)$ .

Keep practising

6 worksheet questions · 14 marks

Open worksheet Worked solutions

Read the questions online

Find the midpoint of $(2, 4)$ and $(8, 10)$ .
2 marks
Find the midpoint of $(- 6, 3)$ and $(2, - 5)$ .
2 marks
Find the midpoint of $(- 1, - 4)$ and $(4, 5)$ .
2 marks
One endpoint is $(1, 6)$ and the midpoint is $(4, 2)$ . Find the other endpoint.
3 marks
The endpoints of a diameter are $(- 4, 1)$ and $(8, 5)$ . Find the circle's centre.
2 marks
A student says the midpoint of $(2, 1)$ and $(10, 7)$ is $(6, 4)$ . Verify this using coordinate differences.
3 marks

Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

Find the midpoint of $(2, 4)$ and $(8, 10)$ .
2 marks
Average coordinates: $\frac{2 + 8}{2} = 5$ and $\frac{4 + 10}{2} = 7$ .
$(5, 7)$
Find the midpoint of $(- 6, 3)$ and $(2, - 5)$ .
2 marks
$\frac{- 6 + 2}{2} = - 2$ and $\frac{3 - 5}{2} = - 1$ .
$(- 2, - 1)$
Find the midpoint of $(- 1, - 4)$ and $(4, 5)$ .
2 marks
$\frac{- 1 + 4}{2} = 1.5$ and $\frac{- 4 + 5}{2} = 0.5$ .
$(1.5, 0.5)$
One endpoint is $(1, 6)$ and the midpoint is $(4, 2)$ . Find the other endpoint.
3 marks
Double the midpoint: $(8, 4)$ .
Subtract $(1, 6)$ to obtain $(7, - 2)$ .
$(7, - 2)$
The endpoints of a diameter are $(- 4, 1)$ and $(8, 5)$ . Find the circle's centre.
2 marks
The centre is the midpoint of the diameter.
$(\frac{- 4 + 8}{2}, \frac{1 + 5}{2}) = (2, 3)$ .
$(2, 3)$
A student says the midpoint of $(2, 1)$ and $(10, 7)$ is $(6, 4)$ . Verify this using coordinate differences.
3 marks
From $(2, 1)$ to $(6, 4)$ the changes are $(4, 3)$ .
From $(6, 4)$ to $(10, 7)$ the changes are also $(4, 3)$ .
Equal steps on the same segment place the point halfway.
The student is correct.

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Curriculum reference & lesson scope

Australian Curriculum v9 and SA Curriculum Mathematics

Gradient, midpoint and distance; algebraic and graphical straight-line relationships; interpreting models and testing parameters. Parallel and perpendicular lines are included as a natural school-course extension. All explanations and questions are original.

Official curriculum source ↗

In this lesson

Understand the idea.

Halfway in two directions

Work backwards from the middle

See the method in action.

Across the axes

Average the horizontal coordinates.

Average the vertical coordinates.

Write the ordered pair.

Recover an endpoint

Use the horizontal midpoint equation.

Solve for the missing horizontal coordinate.

Recover the vertical coordinate.

State and check the endpoint.

Try it yourself.

Check your understanding

Keep practising