The distance formula

The first big formula of coordinate geometry tells us how far apart two points are.

Statement

The distance between $P = (x_1, y_1)$ and $Q = (x_2, y_2)$ is $\boxed{PQ = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}.}$

A special case: distance from the origin $O = (0, 0)$ to $P = (x, y)$ is $\sqrt{x^2 + y^2}$.

Derivation

Drop perpendiculars from $P$ and $Q$ to the $x$-axis, meeting it at $P' = (x_1, 0)$ and $Q' = (x_2, 0)$. From $P$, drop a perpendicular to $QQ'$, meeting at $R = (x_2, y_1)$. Then $PRQ$ forms a right triangle with $\angle R = 90^\circ$ (legs parallel to the axes).

• $PR = |x_2 - x_1|$ (horizontal distance).
• $QR = |y_2 - y_1|$ (vertical distance).
• $PQ$ is the hypotenuse.

By Pythagoras: $PQ^2 = PR^2 + QR^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2$. Take square roots. $\blacksquare$

(We use $(x_2 - x_1)^2 = (x_1 - x_2)^2$, so the formula is symmetric in $P$ and $Q$.)

Applications

The formula lets us test several geometric properties from coordinates alone.

• Equality of distances (e.g., a point equidistant from two given points lies on the perpendicular bisector).
• Identifying types of triangles: compute all three sides; if two are equal, isoceles; if all three, equilateral; if the Pythagorean relation holds, right.
• Identifying quadrilaterals: compute all four sides and both diagonals; the combinations identify squares, rectangles, rhombi, parallelograms.
• Verifying a point lies on a circle of given centre and radius (distance from centre equals radius).

The distance formula is robust against permutations: it doesn't matter which point you call "first".

Worked examples

Example 1. Find the distance between $(3, -2)$ and $(-1, 4)$.

$d = \sqrt{(-1 - 3)^2 + (4 - (-2))^2} = \sqrt{16 + 36} = \sqrt{52} = 2\sqrt{13}$.

Example 2. Determine if the points $(1, 5), (2, 3), (-2, -11)$ are collinear (using distances).

$d(A, B) = \sqrt{1 + 4} = \sqrt{5}$. $d(B, C) = \sqrt{16 + 196} = \sqrt{212}$. $d(A, C) = \sqrt{9 + 256} = \sqrt{265}$.

If collinear: $d(A, B) + d(B, C) = d(A, C)$. Check: $\sqrt{5} + \sqrt{212} \approx 2.236 + 14.56 = 16.80$. $\sqrt{265} \approx 16.28$. Not equal, so the three points are not collinear.

Example 3. Show that $(1, -1), (-1, -3), (3, -3), (-1, 1)$ form an isoceles triangle.

Wait, that's four points. Take the three non-collinear: $A = (1, -1), B = (-1, -3), C = (3, -3)$.

$AB = \sqrt{4 + 4} = 2\sqrt{2}$. $BC = \sqrt{16 + 0} = 4$. $CA = \sqrt{4 + 4} = 2\sqrt{2}$.

$AB = CA$ , isoceles.

Example 4. Find a point on the $x$-axis equidistant from $A = (-2, 5)$ and $B = (2, -3)$.

Let $P = (x, 0)$. $PA^2 = (x + 2)^2 + 25$. $PB^2 = (x - 2)^2 + 9$. Set equal: $(x + 2)^2 + 25 = (x - 2)^2 + 9 \Rightarrow 8 x + 16 = -16 \Rightarrow x = -2$.

So $P = (-2, 0)$.

Example 5. Verify that $(1, 7), (4, 2), (-1, -1), (-4, 4)$ are the vertices of a square.

Label $A, B, C, D$ in order.

$AB = \sqrt{9 + 25} = \sqrt{34}$. $BC = \sqrt{25 + 9} = \sqrt{34}$. $CD = \sqrt{9 + 25} = \sqrt{34}$. $DA = \sqrt{25 + 9} = \sqrt{34}$.

All sides equal , at least a rhombus. Check diagonals:

$AC = \sqrt{4 + 64} = \sqrt{68}$. $BD = \sqrt{64 + 4} = \sqrt{68}$.

Equal diagonals , together with equal sides, this is a square. ✓

Try it yourself

1. Distance between $(-3, 4)$ and $(4, -3)$.
2. Point $(a, b)$ from origin?
3. Find the point on the $y$-axis equidistant from $(5, -2)$ and $(-3, 2)$.
4. Show $(2, -2), (-2, 1), (5, 2)$ are vertices of a right triangle.
5. The point $(0, 0)$ is equidistant from $(a, 0)$ and $(0, a)$. Show distances equal.
6. Find $k$ if the distance between $(2, 3)$ and $(4, k)$ is $5$.
7. Show $A(0, 5), B(-2, -3), C(2, -3)$ form an isoceles triangle.
8. Are $(1, 1), (-1, -1), (-\sqrt{3}, \sqrt{3})$ vertices of an equilateral triangle?
9. Find the centre of a circle passing through $(6, -6), (3, -7), (3, 3)$.
10. The point $P = (x, y)$ is equidistant from $A = (5, 1)$ and $B = (-1, 5)$. Show $3 x = 2 y$.

Pitfalls / Insight

• Always square first, then add, then square-root. Order of operations matters under a radical.
• Distance is always non-negative. A negative answer is a sign of arithmetic error.
• For "equidistant from $X$ and $Y$", set $PX^2 = PY^2$ , the squares are easier than the roots.

Insight. The distance formula is Pythagoras dressed in coordinates. Every metric question in the plane reduces to it.