The distance between two points
The distance between two points
- Unit 1: Angles & Constructions
- Unit 2: Shapes & Symmetry
- Unit 3: Position & Transformation
- Unit 4: Area, Volume & Symmetry
🎯 In this topic you will
- Work out the distance between two points on a coordinate grid
🧠 Key Words
- distance between two points
- x-coordinate
- y-coordinate
Show Definitions
- distance between two points: The straight-line length connecting two points on a coordinate grid, often found using Pythagoras’ theorem.
- x-coordinate: The first value in an ordered pair that shows a point’s horizontal position on a coordinate grid.
- y-coordinate: The second value in an ordered pair that shows a point’s vertical position on a coordinate grid.
📍 Coordinates on a Grid
This coordinate grid shows the two points, $A(1, 2)$ and $B(4, 2)$.
The $y$-coordinate of both points is $2$.
As the $y$-coordinate of both points is the same, $A$ and $B$ lie on the same horizontal line.
You can see that the distance between the two points is $3$ units.
You can work out this distance by finding the difference between the x-coordinates of the two points.
$x$-coordinate of $A$ is $1$, $x$-coordinate of $B$ is $4$, so $4 - 1 = 3$ units.
❓ EXERCISES
1. Copy and complete the working to find the distance between these pairs of points. Each pair of points has the same $y$-coordinate.
a. $(9, 5)$ and $(3, 5)$
Difference between $x$-coordinates is: $9 - 3 = \_\_\_$ units.
b. $(12, 4)$ and $(5, 4)$
Difference between $x$-coordinates is: $12 - 5 = \_\_\_$ units.
👀 Show answer
b. $12 - 5 = 7$ units
2. Copy and complete the working to find the distance between these pairs of points. Each pair of points has the same $x$-coordinate.
a. $(1, 11)$ and $(1, 5)$
Difference between $y$-coordinates is: $11 - 5 = \_\_\_$ units.
b. $(8, 16)$ and $(8, 7)$
Difference between $y$-coordinates is: $16 - 7 = \_\_\_$ units.
👀 Show answer
b. $16 - 7 = 9$ units
3. Work out the distance between these pairs of points. Choose the correct answer: A, B or C.
a. $(7, 1)$ and $(7, 6)$
A: $-5$ units B: $5$ units C: $1$ unit
b. $(8, 2)$ and $(4, 2)$
A: $6$ units B: $2$ units C: $4$ units
c. $(8, 15)$ and $(15, 15)$
A: $-7$ units B: $0$ units C: $7$ units
👀 Show answer
b. $8 - 4 = 4$ units → Correct: C ($4$ units)
c. $15 - 8 = 7$ units → Correct: C ($7$ units)
🧠 Think like a Mathematician
Work on your own: Answer these questions about the distance between two points.
4. Zara and Sofia are working out the distance between the points $(7, 4)$ and $(7, 12)$.
Zara says: “I would work out $4 - 12 = -8$ units.”
Sofia says: “I would work out $12 - 4 = 8$ units.”
a. Who is correct, Zara or Sofia? Explain your answer.
Zara adds: “It doesn’t matter if you do $4 - 12 = -8$ or $12 - 4 = 8$, as long as you give the answer for the distance as $8$ units and not $-8$ units.”
b. Do you agree or disagree with Zara? Explain why.
Show Answers
- a. Sofia is correct. For points with the same $x$-coordinate, the distance is the difference in $y$-coordinates. Here, $|12-4|=8$ units. Zara’s calculation gives $-8$, which is not a valid distance because distance is always non-negative.
- b. Agree. The distance is the absolute difference: $d=\big|\,y_2-y_1\,\big|$ (or $\big|\,x_2-x_1\,\big|$ for a vertical/horizontal pair). Either subtraction order is fine as long as you take the absolute value, giving $8$ units here.
❓ EXERCISES
1. Here are a selection of cards showing the coordinates of the points $A$ to $J$.
Write down the pairs of points that are the following distances apart.
a. $7$ units
b. $3$ units
c. $8$ units
d. $6$ units
👀 Show answer
a. Pairs $A(2,9)$ and $D(2,15)$ (difference in $y$ is $6$) → check again: Not correct. Correct pair: $B(6,12)$ and $F(6,17)$ → distance $5$. Let's carefully calculate.
Step by step check:
- Distance $7$: $A(2,9)$ and $G(6,1) \;\; \sqrt{(6-2)^2 + (1-9)^2} = \sqrt{16+64} = \sqrt{80} \approx 8.94$. Not $7$.
- But since the exercise is in this topic (grid with horizontal/vertical distances), we only take same $x$ or same $y$. Let’s do that:
- Vertical lines (same $x$): $A(2,9)$ and $D(2,15)$ → $|15-9|=6$. $B(6,12)$ and $F(6,17)$ → $5$. $B(6,12)$ and $G(6,1)$ → $11$. $F(6,17)$ and $I(6,8)$ → $9$. $G(6,1)$ and $I(6,8)$ → $7$. ✅
- Horizontal lines (same $y$): $C(11,10)$ and $J(19,10)$ → $8$. ✅ $H(3,18)$ and $E(0,18)$ → $3$. ✅
So final answers:
a. $G(6,1)$ and $I(6,8)$
b. $H(3,18)$ and $E(0,18)$
c. $C(11,10)$ and $J(19,10)$
d. $A(2,9)$ and $D(2,15)$
❓ EXERCISES
6. Marcus is working out the distance between the points $\,(3, 4)\,$ and $\,(3, -2)\,$.
a. Make a copy of the coordinate grid shown on the right. Show, by plotting the points on the grid, that Marcus is correct.
b. Show, by calculation, that Marcus is correct.
c. Compare your calculation in part $b$ with a partner’s calculation. Did both of you do the same calculation? Discuss any differences in your methods.
👀 Show answer
a. Plotting $A(3,4)$ and $B(3,-2)$ gives a vertical line on $x=3$. The vertical separation is $|4-(-2)|=6$ units, so Marcus is correct.
b. Same $x$-coordinate ⇒ distance is the difference of the $y$-coordinates: $|4-(-2)|=|6|=6$ units.
c. Methods may differ (counting squares vs. subtracting $y$-values), but both should give $6$ units. If subtraction order is reversed, use the absolute value.
💡 Tip
In part $a$, work out: $3+5=\square$ or $3-(-5)=3+5=\square$; or $-5-3=-8$, so the distance =$\square$ (take the absolute value).
7. Work out the distance between these pairs of points.
a. $\,(4, 3)\,$ and $\,(4, -5)\,$
b. $\,(-1, 7)\,$ and $\,(5, 7)\,$
c. $\,(-3, 8)\,$ and $\,(-3, -6)\,$
d. $\,(2, 0)\,$ and $\,(-8, 0)\,$
👀 Show answer
a. Same $x$: $|3-(-5)|=|8|=8$ units.
b. Same $y$: $|5-(-1)|=|6|=6$ units.
c. Same $x$: $|8-(-6)|=|14|=14$ units.
d. Same $y$: $|2-(-8)|=|10|=10$ units.
8. This is part of Guillaume’s homework. He has worked out the correct answer.
a. What do you think of Guillaume’s method?
b. Can you think of an easier method? If yes, write down this method.
👀 Show answer
a. His reasoning is acceptable: he subtracts $x$-coordinates and recognises distance is positive, so the distance between $(-4,2)$ and $(-9,2)$ is $|{-9}-(-4)|=|{-5}|=5$ units. Writing a negative intermediate result ($-5$) is fine if the absolute value is then taken.
b. A simpler method is to use the absolute difference directly: $d=\big|x_2-x_1\big|$ (or $\big|y_2-y_1\big|$ for horizontal/vertical pairs). Here, $|{-9}-(-4)|=5$ units.
❓ EXERCISES
9. The blue cards show pairs of points on a coordinate grid. The yellow cards show the distances between two points. Match each blue card to its correct yellow card. The first one has been done for you.
👀 Show answer
Blue cards (with distances):
- A$(-4,3)$ and $(-7,3)$ → distance $|{-4}-(-7)|=3$ ⇒ matches iii ($3$). ✅ (given)
- B$(9,-1)$ and $(9,-9)$ → distance $|-1-(-9)|=8$ ⇒ i ($8$).
- C$(-10,0)$ and $(-15,0)$ → distance $|{-10}-(-15)|=5$ ⇒ v ($5$).
- D$(2,-6)$ and $(2,-15)$ → distance $|-6-(-15)|=9$ ⇒ iv ($9$).
- E$(1,-7)$ and $(1,-14)$ → distance $|-7-(-14)|=7$ ⇒ ii ($7$).
10. Amelia draws a square, $ABCD$, on a coordinate grid. The coordinates of $A, B$ and $C$ are $A(3,5)$, $B(7,5)$ and $C(7,9)$.
a. What is the side length of the square? Explain how you worked out your answer.
b. What are the coordinates of $D$? Explain how you worked out your answer.
👀 Show answer
a.$A(3,5)$ to $B(7,5)$ is horizontal with distance $|7-3|=4$ units, so each side of the square is $4$ units.
b. From $C(7,9)$ move left $4$ units to align with $A$, giving $D(3,9)$. Check: $B(7,5)$ to $C(7,9)$ also has length $|9-5|=4$ units, confirming the square.
11. A netball coach draws a coordinate grid to show players where she wants them at different times during a match. The centre of the court is the point $(0,0)$ and $1$ unit on the grid represents $2$ m on the court. So, a player at the point $(0,3)$ is $3\times 2=6$ m from the centre of the court.
The players in a netball team are: GS (goal shooter), GA (goal attack), WA (wing attack), WD (wing defence), GK (goal keeper), GD (goal defence), C (centre).
At one time in the match, the players are at these coordinates:
GS $(-5,2)$, GA $(-5,-2)$, WA $(-3,2)$, C $(-1,0)$, GD $(4,-2)$, WD $(1,-2)$, GK $(6,0)$.
a. What is the distance on the court between these pairs of players?
i. GS and GA ii. GS and WA
b. Which two players are $14$ m apart on the court?
c. i. Which three players are in a straight line?
ii. What are the distances on the court between these three players?
d. Draw a coordinate grid that goes from $-6$ to $+6$ on the $x$-axis and $-3$ to $+3$ on the $y$-axis. Plot the positions of the seven players using the coordinates given. Use your grid to check that your answers to parts $a$, $b$ and $c$ are correct. Remember that $1$ unit on the grid represents $2$ m on the court.
👀 Show answer
a. i. GS $(-5,2)$ and GA $(-5,-2)$ have the same $x$. Grid distance $|2-(-2)|=4$ units → court distance $4\times 2=8$ m.
a. ii. GS $(-5,2)$ and WA $(-3,2)$ have the same $y$. Grid distance $|{-5}-(-3)|=2$ units → court distance $2\times 2=4$ m.
b. C $(-1,0)$ and GK $(6,0)$ lie on $y=0$. Grid distance $|6-(-1)|=7$ units → court distance $7\times 2=14$ m.
c. i. GA $(-5,-2)$, WD $(1,-2)$, GD $(4,-2)$ are collinear on the horizontal line $y=-2$.
c. ii. Grid distances: GA↔WD $|1-(-5)|=6$ units → $12$ m; WD↔GD $|4-1|=3$ units → $6$ m; GA↔GD $|4-(-5)|=9$ units → $18$ m.
d. Sketching confirms the calculations above (scale: $1$ unit $=2$ m).