Lesson

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.

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).

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.

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$29−5=24

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 $xy$`x``y`-plane. Let's see how by looking at an example.

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 $xy$`x``y`-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` .

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 $\left(x_1,y_1\right)$(`x`1,`y`1) and $\left(x_2,y_2\right)$(`x`2,`y`2) 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.

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

Round your answer to two decimal places.

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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.

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Find the length of interval $PQ$

`P``Q`.Find the length of interval $QR$

`Q``R`.If the length of $PR$

`P``R`is denoted by $c$`c`, use Pythagoras’ theorem to find the exact value of $c$`c`.

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

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

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determines the midpoint, gradient and length of an interval, and graphs linear relationships