Distance between two points calculator

• $(1, 2)$ and $(4, 6)$: changes of 3 and 4, so $d = \sqrt{9 + 16} = \sqrt{25} = 5$.
• $(0, 0)$ and $(2, 4)$: $d = \sqrt{4 + 16} = \sqrt{20} = 2\sqrt{5} \approx 4.47$.
• $(-3, 1)$ and $(2, -11)$: changes of 5 and $-12$, so $d = \sqrt{25 + 144} = 13$.

Draw the two points and complete a right-angled triangle with a horizontal and a vertical side. The horizontal side is the change in $x$, the vertical side is the change in $y$, and the distance between the points is the hypotenuse. Squaring removes any negative signs, so the order you subtract in doesn't matter.

When the distance isn't a whole number, exam questions often want it in surd form. $\sqrt{20} = \sqrt{4 \times 5} = 2\sqrt{5}$. Our square root simplifier shows how to simplify any surd.

A circle with centre $(0, 0)$ and radius $r$ has the equation $x^2 + y^2 = r^2$: every point on it is distance $r$ from the centre. To check whether a point lies on, inside or outside a circle, find its distance from the centre and compare with the radius. See our equation of a circle questions.

• Forgetting to square-root at the end.
• Squaring a negative change wrongly: $(-12)^2 = 144$, not $-144$.
• Adding the changes before squaring.

Practise with our Pythagoras questions, and see the midpoint calculator.

• Sketch the triangle on the grid: it makes the method obvious and earns method marks.
• Give exact answers in simplified surd form if asked, otherwise round to 3 significant figures.
• Watch for 3-4-5 and 5-12-13 triangles: they give whole-number distances.

On a grid map where each square is 1 km, a walk from $(2, 1)$ to $(8, 9)$ is $\sqrt{36 + 64} = 10$ km in a straight line ("as the crow flies"), even though walking along the grid lines would be 14 km.

In three dimensions, add the third change too: $d = \sqrt{\Delta x^2 + \Delta y^2 + \Delta z^2}$. The space diagonal of a 3 × 4 × 12 box is $\sqrt{9 + 16 + 144} = 13$.