Distance between points using Pythagoras | Year 9

01 / LEARN

Understand the idea.

Make a right triangle

Draw a horizontal step and a vertical step between the two points. These form perpendicular legs of a right triangle; the direct segment is its hypotenuse. Their lengths come from the coordinate differences.

Pythagoras' theorem adds the squares of the legs. Signed differences may be negative, but squaring removes the sign. Take the non-negative square root at the end because a distance cannot be negative.

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

Equal-unit Cartesian grid from negative six to eight on both axes. Segment AB joins A at negative two, negative one to B at four, seven. Dashed horizontal and vertical legs join A to four, negative one and then B. No distance is labelled.

Exact or approximate?

Keep a square-root answer exact unless a decimal approximation is requested. A perfect-square total gives an integer distance; otherwise retain the radical while working and round the final decimal only once.

On horizontal or vertical segments, one coordinate difference is zero, so the formula reduces to the size of the other difference. Include units: the result measures direct separation, rather than a route along grid lines.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

A coordinate triangle

Find the distance from $A (- 2, - 1)$ to $B (4, 7)$ .

Worked solution

1
Find the two perpendicular changes.
$\begin{aligned} Δ x & = 4 - (- 2) = 6 \\ Δ y & = 7 - (- 1) = 8 \end{aligned}$
The coordinate differences give the horizontal and vertical legs.
2
Apply Pythagoras.
$d^{2} = 6^{2} + 8^{2} = 36 + 64 = 100$
The direct segment is the hypotenuse of this right triangle.
3
Take the non-negative square root.
$d = \sqrt{100} = 10 units$
Distance is a non-negative length.

Answer

10 units

EXAMPLE 02

A non-integer distance

Find the distance between $(1, 2)$ and $(5, 7)$ , first exactly and then to 2 decimal places.

Worked solution

1
Calculate the coordinate changes.
$\begin{aligned} Δ x & = 5 - 1 = 4 \\ Δ y & = 7 - 2 = 5 \end{aligned}$
The two perpendicular legs have lengths 4 and 5.
2
Find the exact distance.
$d = \sqrt{4^{2} + 5^{2}} = \sqrt{41} units$
Pythagoras gives an exact radical because 41 is not a perfect square.
3
Round the decimal at the end.
$\sqrt{41} = 6.403124 \dots \approx 6.40 units$
The third decimal digit is 3, so the hundredths digit stays unchanged.

Answer

$\sqrt{41}$ units; approximately 6.40 units

03 / YOUR TURN

Try it yourself.

Check your understanding

Find the distance between $(0, 0)$ and $(6, 8)$ .

Show answer & reasoning

10 units

The legs are $6$ and $8$ , so $d = \sqrt{6^{2} + 8^{2}} = \sqrt{100} = 10 units$ .

Keep practising

6 worksheet questions · 16 marks

Open worksheet Worked solutions

Read the questions online

Find the distance between $(1, 2)$ and $(4, 6)$ .
2 marks
Find the distance between $(- 3, - 2)$ and $(5, 4)$ .
2 marks
Find the exact distance between $(0, 1)$ and $(3, 5)$ .
2 marks
Find the exact distance between $(2, - 1)$ and $(7, 1)$ , then round it to 1 decimal place.
3 marks
A map grid uses 100 m per coordinate unit. A gate is at $(1, 1)$ and a shelter at $(4, 5)$ . Find their straight-line separation.
3 marks
Points $A (- 2, 3)$ , $B (4, 3)$ and $C (4, - 5)$ form a triangle. Find its perimeter.
4 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
The changes are 3 and 4.
$d = \sqrt{9 + 16} = 5$ .
5 units
Find the distance between $(- 3, - 2)$ and $(5, 4)$ .
2 marks
The changes are 8 and 6.
$d = \sqrt{64 + 36} = 10$ .
10 units
Find the exact distance between $(0, 1)$ and $(3, 5)$ .
2 marks
The changes are 3 and 4.
$d = \sqrt{3^{2} + 4^{2}} = 5$ .
5 units
Find the exact distance between $(2, - 1)$ and $(7, 1)$ , then round it to 1 decimal place.
3 marks
The changes are 5 and 2.
$d = \sqrt{25 + 4} = \sqrt{29} \approx 5.385$ .
To 1 decimal place this is 5.4.
$\sqrt{29}$ units; 5.4 units
A map grid uses 100 m per coordinate unit. A gate is at $(1, 1)$ and a shelter at $(4, 5)$ . Find their straight-line separation.
3 marks
The grid distance is $\sqrt{3^{2} + 4^{2}} = 5$ units.
Multiply by 100 m per unit.
500 m
Points $A (- 2, 3)$ , $B (4, 3)$ and $C (4, - 5)$ form a triangle. Find its perimeter.
4 marks
$A B = 6$ and $B C = 8$ .
$A C = \sqrt{6^{2} + 8^{2}} = 10$ .
The perimeter is $6 + 8 + 10 = 24$ .
24 units

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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.

Make a right triangle

Exact or approximate?

See the method in action.

A coordinate triangle

Find the two perpendicular changes.

Apply Pythagoras.

Take the non-negative square root.

A non-integer distance

Calculate the coordinate changes.

Find the exact distance.

Round the decimal at the end.

Try it yourself.

Check your understanding

Keep practising