The distance formula

When two points lie on the same hor­i­zon­tal or ver­ti­cal line, the dis­tance between them is just a dif­fer­ence of coor­di­nates. But most pairs of points are not lined up like that. The dis­tance for­mula finds the dis­tance between any two points from their coor­di­nates, and it comes straight from Pythago­ras' the­o­rem. With it we can test tri­an­gles for being isosce­les or right-angled, check whether points lie on a line, and find unknown coor­di­nates.

Deriv­ing the for­mula#

Let $A(x_1, y_1)$ and $B(x_2, y_2)$ be two points. Through $A$ draw a hor­i­zon­tal line and through $B$ a ver­ti­cal line; they meet at $C(x_2, y_1)$.

• $AC$ is hor­i­zon­tal, so its length is $|x_2 - x_1|$.
• $BC$ is ver­ti­cal, so its length is $|y_2 - y_1|$.
• The angle at $C$ is a right angle.

By Pythago­ras' the­o­rem, $AB^2 = AC^2 + BC^2$, so

$$AB = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$

Because the dif­fer­ences are squared, it does not mat­ter which point you call the first. In par­tic­u­lar, the dis­tance of a point $(x, y)$ from the ori­gin is $\sqrt{x^2 + y^2}$.

Exam­ple 1#

Find the dis­tance between $(2, 3)$ and $(8, 11)$.

$$d = \sqrt{(8 - 2)^2 + (11 - 3)^2} = \sqrt{36 + 64} = \sqrt{100} = 10$$

Exam­ple 2#

Find the dis­tance between $(-1, -2)$ and $(4, 10)$.

$$d = \sqrt{(4 - (-1))^2 + (10 - (-2))^2} = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13$$

Put brack­ets round neg­a­tive coor­di­nates when you sub­tract them: $4 - (-1) = 5$, not $3$.

Exam­ple 3#

How far is the point $(-5, 12)$ from the ori­gin?

$$\sqrt{(-5)^2 + 12^2} = \sqrt{25 + 144} = 13$$

Test­ing tri­an­gles#

Find all three sides, then com­pare them.

Exam­ple 4#

Show that $A(1, 1)$, $B(4, 5)$ and $C(8, 2)$ are the ver­tices of an isosce­les right-angled tri­an­gle.

$$AB^2 = (4 - 1)^2 + (5 - 1)^2 = 9 + 16 = 25$$

$$BC^2 = (8 - 4)^2 + (2 - 5)^2 = 16 + 9 = 25$$

$$AC^2 = (8 - 1)^2 + (2 - 1)^2 = 49 + 1 = 50$$

$AB = BC = 5$, so the tri­an­gle is isosce­les. Also $AB^2 + BC^2 = 25 + 25 = 50 = AC^2$, so by the con­verse of Pythago­ras' the­o­rem the angle at $B$ is a right angle.

Work­ing with squared dis­tances avoids square roots until the very end.

Test­ing for collinear points#

Points are collinear if they lie on one straight line. Three points $A$, $B$, $C$ are collinear exactly when the longest of the three dis­tances equals the sum of the other two.

Exam­ple 5#

Are $A(1, 2)$, $B(3, 6)$ and $C(4, 8)$ collinear?

$$AB = \sqrt{4 + 16} = \sqrt{20} = 2\sqrt{5}, \quad BC = \sqrt{1 + 4} = \sqrt{5}, \quad AC = \sqrt{9 + 36} = \sqrt{45} = 3\sqrt{5}$$

$AB + BC = 2\sqrt{5} + \sqrt{5} = 3\sqrt{5} = AC$, so the points are collinear.

Find­ing unknown coor­di­nates#

Exam­ple 6#

The dis­tance between $(x, 3)$ and $(2, -1)$ is $5$. Find $x$.

$$\begin{aligned} (x - 2)^2 + (3 - (-1))^2 &= 25 \\ (x - 2)^2 + 16 &= 25 \\ (x - 2)^2 &= 9 \\ x - 2 &= \pm 3 \end{aligned}$$

So $x = 5$ or $x = -1$. Both answers are cor­rect: there are two such points.

Exam­ple 7#

Find the point on the x-axis that is equidis­tant from $A(2, 5)$ and $B(-2, 9)$.

A point on the x-axis has the form $P(x, 0)$. Set $PA^2 = PB^2$:

$$\begin{aligned} (x - 2)^2 + 25 &= (x + 2)^2 + 81 \\ x^2 - 4x + 4 + 25 &= x^2 + 4x + 4 + 81 \\ -8x &= 56 \\ x &= -7 \end{aligned}$$

The point is $(-7, 0)$. Check: $PA^2 = 81 + 25 = 106$ and $PB^2 = 25 + 81 = 106$.

Prac­tice#

1. Find the dis­tance between $(3, 4)$ and $(6, 8)$.
2. Find the dis­tance between $(-2, 5)$ and $(4, -3)$.
3. Find the dis­tance of $(7, -24)$ from the ori­gin.
4. Find the dis­tance between $(a, b)$ and $(-a, -b)$.
5. The point $(2, y)$ is at a dis­tance $10$ from $(8, 4)$. Find $y$.
6. Find the point on the y-axis that is equidis­tant from $(6, 5)$ and $(-4, 3)$.
7. Show that $(0, 0)$, $(3, 0)$ and $(3, 4)$ form a right-angled tri­an­gle, and find its perime­ter.
8. Are the points $(-1, -1)$, $(2, 3)$ and $(8, 11)$ collinear?
9. Show that $(0, 0)$, $(2, 0)$ and $(1, \sqrt{3})$ are the ver­tices of an equi­lat­eral tri­an­gle.
10. A cir­cle has cen­tre $(2, -3)$ and passes through $(5, 1)$. Find its radius.

Answers#