Midpoint | Year 9 | Ascent Library

01 / LEARN

The idea, step by step

Halfway in two directions

The midpoint lies halfway along a line segment. Its $x$ -coordinate is halfway between the endpoints' $x$ -coordinates, and its $y$ -coordinate is halfway between their $y$ -coordinates. Average each pair separately.

This works for horizontal, vertical and sloping segments. A midpoint can have fractional coordinates even if both endpoints lie on grid intersections. It need not be the origin or the point where the segment crosses an axis.

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

Work backwards from the middle

If the midpoint and one endpoint are given, the other endpoint is the same distance beyond the midpoint in each coordinate. Equivalently, double the midpoint coordinate and subtract the known endpoint coordinate.

Check that your two endpoints really average to the stated midpoint. This is often safer than judging a sketch, especially if the axes are not drawn to equal scales.

A common trap

Averaging all four numbers together.

Instead: Keep horizontal and vertical positions separate: average the two $x$ -values, then the two $y$ -values.

02 / WORKED EXAMPLES

See the thinking

Example 1

Across the axes

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

Average the $x$ -coordinates: $(- 4 + 6) / 2 = 1$ .
Average the $y$ -coordinates: $(- 2 + 4) / 2 = 1$ .
The midpoint is $(1, 1)$ .

(1, 1)

Example 2

Recover an endpoint

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

The missing $x$ -coordinate is $2 (5) - 2 = 8$ .
The missing $y$ -coordinate is $2 (1) - (- 3) = 5$ .
Check: the averages of 2 and 8, and of $- 3$ and 5, are 5 and 1.

B = (8, 5)

Check your understanding

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

Show answer

$(1, 1)$

$(- 3 + 5) / 2 = 1$ and $(4 - 2) / 2 = 1$ .

03 / WORKSHEET

Your turn

6 questions · 14 marks · Work at your own pace.

Worksheet PDF 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: $(2 + 8) / 2 = 5$ and $(4 + 10) / 2 = 7$ .
$(5, 7)$
Find the midpoint of $(- 6, 3)$ and $(2, - 5)$ .
2 marks
$(- 6 + 2) / 2 = - 2$ and $(3 - 5) / 2 = - 1$ .
$(- 2, - 1)$
Find the midpoint of $(- 1, - 4)$ and $(4, 5)$ .
2 marks
$(- 1 + 4) / 2 = 1.5$ and $(- 4 + 5) / 2 = 0.5$ .
$(1.5, 0.5)$
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.
$((- 4 + 8) / 2, (1 + 5) / 2) = (2, 3)$ .
$(2, 3)$
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 ↗

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