Linear Equations II

Hong Kong

Stage 1 - Stage 3

Lesson

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

Pythagoras' Theorem

$a^2+b^2=c^2$`a`2+`b`2=`c`2

Did you know we can also use Pythagoras' theorem to find the distance between two points on a number plane? Let's see how by looking at an example.

Let's say we wanted to find the distance between $\left(-3,6\right)$(−3,6) and $\left(5,4\right)$(5,4).

Firstly, we can plot the points of a number 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:

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.

The good news is that we don't have to graph the points to be able to find the distance between two points. We can use the distance formula which is derived from Pythagoras' theorem.

The Distance Formula

$d=\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}$`d`=√(`x`2−`x`1)2+(`y`2−`y`1)2

Now let's look at the following written and video examples and see this process in action!

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 |

The points $P$`P` $\left(-3,-1\right)$(−3,−1), $Q$`Q` $\left(-3,-4\right)$(−3,−4) and $R$`R` $\left(2,-4\right)$(2,−4) are the vertices of a right-angled triangle, as shown on the number plane.

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 value of $c$`c`to three decimal places.

Consider the interval joining points $P$`P`$\left(16,-10\right)$(16,−10) and $Q$`Q`$\left(4,6\right)$(4,6).

Loading Graph...

Evaluate $PQ^2$

`P``Q`2, the square of the length of the interval $PQ$`P``Q`.Point $N$

`N`is the midpoint of interval $PQ$`P``Q`. What is the distance from $P$`P`to $N$`N`?