Distance and midpoint | Year 11

01 / LEARN

Understand the idea.

Distance comes from a right triangle

Horizontal and vertical coordinate changes form the perpendicular sides of a right triangle. The distance between the points is its hypotenuse, so Pythagoras’ theorem gives $d = \sqrt{{(x_{2} - x_{1})}^{2} + {(y_{2} - y_{1})}^{2}}$ . Reversing the point order leaves the squared differences unchanged.

Distance is non-negative. Keep an exact square root unless rounding is requested, and use compatible length units on the axes. A diagram mixing hours and dollars does not usually give a useful physical distance between its plotted points.

d = \sqrt{{(x_{2} - x_{1})}^{2} + {(y_{2} - y_{1})}^{2}}

A midpoint averages matching coordinates

The midpoint lies halfway along a segment. Average the two horizontal coordinates together and the two vertical coordinates together. The result is a point, and its coordinates can be fractional even when both endpoints have integer coordinates.

If a midpoint and one endpoint are known, reverse those averages or repeat the displacement from the endpoint to the midpoint. Equal lengths and common diagonal midpoints can support coordinate geometry arguments, but each conclusion needs the relevant geometric condition.

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

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Length and midpoint

Find the distance and midpoint between A(-2,1) and B(4,9).

Worked solution

1
Find the horizontal and vertical changes.
$\begin{aligned} Δ x & = 4 - (- 2) = 6 \\ Δ y & = 9 - 1 = 8 \end{aligned}$
These are the lengths, with signs, of the right triangle’s coordinate changes.
2
Apply Pythagoras to find the distance.
$d = \sqrt{6^{2} + 8^{2}} = \sqrt{100} = 10$
The segment joining the two points is the hypotenuse.
3
Average matching coordinates for the midpoint.
$M = (\frac{- 2 + 4}{2}, \frac{1 + 9}{2}) = (1, 5)$
Each coordinate is halfway between its two endpoint values.

Answer

Distance 10; midpoint (1,5).

EXAMPLE 02

Recover an endpoint

A segment has endpoint P(3,-1) and midpoint M(5,2). Find the other endpoint Q.

Worked solution

1
Name the unknown endpoint and use the midpoint.
$Q = (a, b), (\frac{3 + a}{2}, \frac{- 1 + b}{2}) = (5, 2)$
The average of corresponding endpoint coordinates must equal each midpoint coordinate.
2
Solve the two coordinate equations.
$\begin{aligned} 3 + a & = 10 & a & = 7 \\ - 1 + b & = 4 & b & = 5 \end{aligned}$
Multiplying each midpoint equation by two reverses the averaging operation.
3
Check equal displacements on both halves.
$\begin{aligned} M - P & = (5 - 3, 2 - (- 1)) = (2, 3) \\ Q - M & = (7 - 5, 5 - 2) = (2, 3) \end{aligned}$
The same movement reaches the midpoint and then the recovered endpoint, so Q has coordinates seven and five.

Answer

Q(7,5).

03 / YOUR TURN

Try it yourself.

Check your understanding

Find the midpoint of (-4,7) and (2,-1).

Show answer & reasoning

(-1,3).

The midpoint is $(\frac{- 4 + 2}{2}, \frac{7 + (- 1)}{2}) = (- 1, 3)$ . Averaging is done separately for each coordinate.

Keep practising

6 worksheet questions · 12 marks

Open worksheet Worked solutions

Read the questions online

Find the distance between $(1, 2)$ and $(4, 6)$ .
2 marks
Find the exact distance between $(- 1, 3)$ and $(4, 5)$ .
2 marks
Find the midpoint of $(- 5, - 2)$ and $(3, 8)$ .
2 marks
A segment has endpoint $(2, 7)$ and midpoint $(- 1, 4)$ . Find its other endpoint.
2 marks
Show that the triangle with vertices $(0, 0)$ , $(4, 0)$ and $(2, 3)$ is isosceles.
2 marks
A map uses kilometres on both axes. Two stations are at $(- 3, 2)$ and $(5, - 4)$ . Find their straight-line separation and explain why this may not be the road distance.
2 marks

Read the worked solutions

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

Find the distance between $(1, 2)$ and $(4, 6)$ .
2 marks
$d = \sqrt{3^{2} + 4^{2}} = 5$ .
$5$
Find the exact distance between $(- 1, 3)$ and $(4, 5)$ .
2 marks
$d = \sqrt{5^{2} + 2^{2}} = \sqrt{29}$ .
$\sqrt{29}$
Find the midpoint of $(- 5, - 2)$ and $(3, 8)$ .
2 marks
$(\frac{- 5 + 3}{2}, \frac{- 2 + 8}{2}) = (- 1, 3)$ .
$(- 1, 3)$ .
A segment has endpoint $(2, 7)$ and midpoint $(- 1, 4)$ . Find its other endpoint.
2 marks
Let $Q = (a, b)$ . Then $2 + a = - 2$ and $7 + b = 8$ , so $a = - 4$ and $b = 1$ .
$(- 4, 1)$ .
Show that the triangle with vertices $(0, 0)$ , $(4, 0)$ and $(2, 3)$ is isosceles.
2 marks
Each sloping side has horizontal change of magnitude 2 and vertical change of magnitude 3.
Their equal lengths establish the isosceles claim.
The two sloping sides both have length $\sqrt{13}$ .
A map uses kilometres on both axes. Two stations are at $(- 3, 2)$ and $(5, - 4)$ . Find their straight-line separation and explain why this may not be the road distance.
2 marks
$d = \sqrt{8^{2} + {(- 6)}^{2}} = 10$ km.
The formula measures the direct geometric distance, not a route following roads.
10 km; roads need not follow the straight segment.

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Curriculum reference & lesson scope

SACE Board of South Australia

Stage 1 Mathematics, Topic 1: Functions and graphs. Prerequisite algebra and coordinate geometry are embedded as supporting lessons.

Official curriculum source ↗

In this lesson

Understand the idea.

Distance comes from a right triangle

A midpoint averages matching coordinates

See the method in action.

Length and midpoint

Find the horizontal and vertical changes.

Apply Pythagoras to find the distance.

Average matching coordinates for the midpoint.

Recover an endpoint

Name the unknown endpoint and use the midpoint.

Solve the two coordinate equations.

Check equal displacements on both halves.

Try it yourself.

Check your understanding

Keep practising