7. Geometry

Lesson

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

Pythagorean Theorem

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

Did you know we can also use the Pythagorean theorem to find the distance between two points on a coordinate 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).

**Think:** Firstly, we can plot the points of a number plane like so:

Then we can draw a right 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 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 Pythagorean 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 Pythagorean 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 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 Pythagorean 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 |

Let's try a couple of practice questions

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

Loading Graph...

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`?

Apply the pythagorean theorem to find the distance between two points in a coordinate plane.