AustraliaNSW
Stage 5.1-2

# 3.03 The distance between two points

Lesson

## Horizontal and Vertical Distances

### Horizontal

Points that lie on a horizontal line share the same $y$y-value.  They would have coordinates that look something like this: $\left(2,5\right)$(2,5) and $\left(-4,5\right)$(4,5). More generally, two points that lie on a horizontal line could have coordinates $\left(a,b\right)$(a,b) and $\left(c,b\right)$(c,b).

If you can recognise that the points lie on a horizontal line then the distance between them is the distance between the $x$x-values: the largest $x$x-value minus the smallest $x$x-value.

#### Worked example 1

##### Question 1

Find the distance between $\left(2,5\right)$(2,5) and $\left(-4,5\right)$(4,5).

Think: Notice that because the $y$y-values are the same, these points lie on a horizontal line.

Do: The distance between them will be the largest $x$x-value ($2$2) minus the smallest $x$x-value ($-4$4)

$2-\left(-4\right)=2+4$2(4)=2+4 = $6$6 (remember subtracting a negative is the same as addition).

### Vertical

Points that lie on a vertical line share the same $x$x-value.  They would have coordinates that look something like this: $\left(2,5\right)$(2,5) and $\left(2,29\right)$(2,29). More generally, two points that lie on a vertical line could have coordinates $\left(a,b\right)$(a,b) and $\left(a,c\right)$(a,c).

If you can recognise that the points lie on a vertical line then the distance between them is the distance between the $y$y-values: the largest $y$y-value minus the smallest $y$y-value.

#### Question 2

Find the distance between $\left(2,5\right)$(2,5) and $\left(2,29\right)$(2,29).

Think:  Notice that because the $x$x-values are the same, these points lie on a vertical line.

Do: The distance between them will be the largest $y$y-value ($29$29) minus the smallest $y$y-value ($5$5)

$29-5=24$295=24

## Distance between two points

What if we want to find the distance between two points that are not on a horizontal or vertical line?

We already learned how to use Pythagoras' theorem to calculate the side lengths in a right triangle. Pythagoras' theorem states:

Pythagoras' theorem
 $a^2+b^2$a2+b2 $=$= $c^2$c2 shorter side lengths hypotenuse

The value $c$c is used to represent the hypotenuse which is the longest side of the triangle. The other two lengths are represented by $a$a and $b$b

We can also use Pythagoras' theorem to find the distance between two points on an $xy$xy-plane. Let's see how by looking at an example.

#### Worked example

##### question 3

Find the distance between $\left(-3,6\right)$(3,6) and $\left(5,4\right)$(5,4). Give your answer rounded to two decimal places.

Think:  Firstly, we can plot the points of an $xy$xy-plane like so:

Then we can draw a right-angled triangle by drawing a line between the two points, as well as a vertical and a horizontal line. The picture below shows one way to do it but there are others:

Once we've created the right-angled triangle, we can calculate the distances of the vertical and horizontal sides by counting the number of units:

Do:  On the $y$y-axis, the distance from $4$4 to $6$6  is $2$2 units and, on the $x$x-axis, the distance from $-3$3 to $5$5 is $8$8 units.

Then, we can use these values to calculate the length of the hypotenuse using Pythagoras' theorem.  The length of the hypotenuse will be the distance between our two points.

Reflect:  The good news is that we don't have to graph the points to be able to find the distance between two points. Since we know that a slanted line on the coordinate plane will always represent the hypotenuse of a right triangle, we can use the a variation of Pythagoras' theorem which is already solved for $c$c .

##### Question 4

How far is the given point $P$P$\left(-15,8\right)$(15,8) from the origin?

Think: Let's plot the points then create a right-angled triangle so we can use Pythagoras' theorem to solve.

Do:

We can see that the distance from $P$P to the $x$x-axis is $8$8 units and the distance from $P$P to the $y$y-axis is $15$15 units.

So, using Pythagoras' theorem:

 $\text{Distance from origin }^2$Distance from origin 2 $=$= $8^2+15^2$82+152 $\text{Distance from origin }$Distance from origin $=$= $\sqrt{64+225}$√64+225 $=$= $\sqrt{289}$√289 $=$= $17$17 units

Distance Formula

The distance between two points $\left(x_1,y_1\right)$(x1,y1) and $\left(x_2,y_2\right)$(x2,y2) is given by:

$d=\sqrt{\text{run }^2+\text{rise }^2}$d=run 2+rise 2

• 'run' is the horizontal distance between the two $x$x-coordinates
• 'rise' is the vertical distance between the two $y$y-coordinates.

#### Practice questions

##### Question 1

How far is the given point $P$P$\left(-3,-7\right)$(3,7) from the origin?

Round your answer to two decimal places.

##### Question 2

The points $P$P $\left(-1,9\right)$(1,9), $Q$Q $\left(-1,6\right)$(1,6) and $R$R $\left(-5,6\right)$(5,6) are the vertices of a right-angled triangle, as shown on the number plane.

1. Find the length of interval $PQ$PQ.

2. Find the length of interval $QR$QR.

3. If the length of $PR$PR is denoted by $c$c, use Pythagoras’ theorem to find the exact value of $c$c.

##### Question 3

Consider the interval $AC$AC that has been graphed on the number plane.

Use Pythagoras’ theorem to find the length of $AC$AC rounded to two decimal places.