Distance between two points is Pythagoras

The gap between two coordinates is always a right triangle in disguise.

The last lesson gave every point an address. This one asks how far apart two addresses are.

Two points, A and B, on the coordinate plane, joined by a segment. Dashed lines mark the horizontal and vertical gap between them. A live reading gives both gaps, and the distance AB by Pythagoras.

A labelled x and y axis with two draggable points, A and B, joined by a segment. Dashed lines mark the horizontal and vertical legs between them. A live reading gives both points' coordinates, the horizontal and vertical gap, and the distance AB by Pythagoras. This needs JavaScript switched on.

Drag A or B. The distance between them is always the hypotenuse of the gap.

Drag A or B until the two dashed legs are wildly different lengths from each other — the diagonal distance still comes out of the same Pythagorean sum every time.

The straight-line gap between two points, computed this way, is their distance — and the whole method is worth its own name, the distance formula.

Where the right triangle is hiding

The horizontal gap between A and B is just their x-coordinates subtracted; the vertical gap is their y-coordinates subtracted. Those two gaps are exactly the two legs of a right triangle — the same right triangle the dashed lines draw — with AB itself as the hypotenuse. G6.6 already proved a hypotenuse’s square equals the sum of the legs’ squares, so:

AB = \sqrt{(x_B - x_A)^2 + (y_B - y_A)^2}

Nothing new is being proved here. This is Pythagoras, applied to a triangle whose legs happen to be readable straight off two points’ coordinates, without ever drawing the triangle by hand.

One to watch out for.

Notice the sign never survives: $(x_B-x_A)^2$ and $(x_A-x_B)^2$ are the same number. Distance does not care which point is “first” — AB and BA are always equal.

This is also the one fact every later coordinate lesson leans on for comparing two mapped locations — whether two shapes really are the same size, whether a point sits on a given circle — because a circle is itself defined as every point the same distance from a centre. Coordinates only pay off once distance is something you can calculate from them directly, rather than something you still have to measure by hand.

Trying it on numbers

Find the distance between $(0, 0)$ and $(3, 4)$ .

$\sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5$ .

Find the distance between $(1, 1)$ and $(4, 5)$ .

$\sqrt{(4-1)^2 + (5-1)^2} = \sqrt{9 + 16} = 5$ .

Find the distance between $(-2, 3)$ and $(2, 0)$ .

$\sqrt{(2-(-2))^2 + (0-3)^2} = \sqrt{4^2 + (-3)^2} = \sqrt{16+9} = \sqrt{25} = 5$ .

Do the exercise

This lesson counts as done once you get its exercise right. Nothing to tick off by hand.