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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:

Pythagorean Theorem



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.


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 $\left(-3,-2\right)$(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...

  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 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).

Loading Graph...

  1. Evaluate $PQ^2$PQ2, the square of the length of the interval $PQ$PQ.

  2. Point $N$N is the midpoint of interval $PQ$PQ. 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.

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