7.02 Distances in the plane using the Pythagorean theorem
We already learned how to use Pythagorean theorem to calculate the side lengths in a right triangle. Pythagorean' theorem states:
$a^2+b^2=c^2$a2+b2=c2
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.
Worked example
question 1
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 .
Question 2
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
Practice questions
Question 2
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 triangle, as shown on the number plane.
Find the length of interval $PQ$PQ.
Find the length of interval $QR$QR.
If the length of $PR$PR is denoted by $c$c, use Pythagoras’ theorem to find the value of $c$c to three decimal places.
Question 3
Consider the interval joining points $P$P$\left(16,-10\right)$(16,−10) and $Q$Q$\left(4,6\right)$(4,6).
Evaluate $PQ^2$PQ2, the square of the length of the interval $PQ$PQ.
Point $N$N is the midpoint of interval $PQ$PQ. What is the distance from $P$P to $N$N?