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Distance between two points

Covers AS91256

The formula

The distance between $(x_1, y_1)$ and $(x_2, y_2)$ is:

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

Each part:
- $(x_2 - x_1)$ — the horizontal separation of the two points
- $(y_2 - y_1)$ — the vertical separation
- $d$ — the straight-line distance between them, always positive

Where it comes from

The formula is Pythagoras' theorem in disguise.
Joining the two points and drawing a horizontal and a vertical line makes a right-angled triangle, with the distance as the hypotenuse.

The horizontal side has length $|x_2 - x_1|$ and the vertical side $|y_2 - y_1|$ , so:

$d^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2$

Taking the square root gives the formula. Understanding this means you never need to memorise it — you can rebuild it from a sketch.

Why the order does not matter

$(x_2 - x_1)$ is squared, so subtracting the other way round gives the same result:

$(3 - 7)^2 = (-4)^2 = 16 \qquad (7 - 3)^2 = 4^2 = 16$

You can take either point as "point 1", as long as you are consistent within one calculation.
The distance is never negative. If your answer is negative, you have taken the square root incorrectly.

Working through an example

The distance between $A(2, 3)$ and $B(8, 11)$ :
- Horizontal difference: $8 - 2 = 6$
- Vertical difference: $11 - 3 = 8$
- Apply the formula: $d = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10$
Negative coordinates need care. For $C(-3, 5)$ and $D(4, -1)$ :
- Horizontal: $4 - (-3) = 7$
- Vertical: $-1 - 5 = -6$
- $d = \sqrt{49 + 36} = \sqrt{85}$
Use brackets when subtracting negatives. $4 - (-3) = 7$ , not 1.

Leaving the answer exact

Most distances are surds, and the exact form is usually preferred.
Simplify by taking out the largest perfect square:
- $\sqrt{50} = \sqrt{25 \times 2} = 5\sqrt{2}$
- $\sqrt{72} = 6\sqrt{2}$
- $\sqrt{85}$ has no square factor, so it cannot be simplified
Give a decimal only if the question asks for one, and round only at the very end.

What distance is used for

Showing two sides are equal — an isosceles triangle, a rhombus, a square
Showing a point is equidistant from two others
Finding a perimeter — add the side lengths
Testing for a right angle with the converse of Pythagoras: if $a^2 + b^2 = c^2$ , the angle opposite $c$ is $90°$
Note that comparing distances often does not need the square root at all. If you only want to know whether two lengths are equal, comparing $d^2$ values is enough and avoids surds entirely.

Tips and Tricks

Square the differences, do not subtract the squares. $\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$ is not $\sqrt{x_2^2 - x_1^2 + \ldots}$ . Work out each difference first, then square it.

Brackets around negative coordinates. $y_2 - y_1$ where $y_1 = -4$ becomes $y_2 + 4$ . Writing the subtraction out in full prevents the sign error.

Leave surds exact and simplified. $2\sqrt{13}$ rather than $\sqrt{52}$ , and rather than 7.21 unless a decimal is requested. In a multi-step problem, decimals introduce rounding error that compounds.

A distance cannot be negative. If you get one, you have made a sign or arithmetic error — go back rather than dropping the minus.

To show two lengths are equal, you can compare $d^2$ instead of $d$ and skip the surds altogether. State that you are doing so, since it is a genuine simplification worth communicating.

Sketch the points. A rough plot tells you whether a distance of 10 is plausible between points that look 3 units apart, and it catches transcription errors before they propagate.

Worked ExampleUsing distances to classify a triangle

The points are $A(1, 2)$ , $B(5, 6)$ and $C(9, 2)$ . Show that triangle $ABC$ is isosceles, and determine whether it contains a right angle.

Step 1 — Sketch the points

$A$ and $C$ both have $y = 2$ , so they lie on a horizontal line, 8 units apart. $B$ sits above and between them. The triangle looks symmetric — which is worth confirming.

Step 2 — Find $AB$

$AB = \sqrt{(5-1)^2 + (6-2)^2}$

$= \sqrt{4^2 + 4^2} = \sqrt{16 + 16} = \sqrt{32}$

Simplify: $\sqrt{32} = \sqrt{16 \times 2} = 4\sqrt{2}$

Step 3 — Find $BC$

$BC = \sqrt{(9-5)^2 + (2-6)^2}$

$= \sqrt{4^2 + (-4)^2} = \sqrt{16 + 16} = \sqrt{32} = 4\sqrt{2}$

Step 4 — Find $AC$

$AC = \sqrt{(9-1)^2 + (2-2)^2} = \sqrt{64 + 0} = 8$

Step 5 — Conclude that the triangle is isosceles

$AB = BC = 4\sqrt{2}$

Two sides are equal in length, so triangle $ABC$ is isosceles, with the equal sides meeting at $B$ .

Step 6 — Test for a right angle

The longest side is $AC = 8$ , since $4\sqrt{2} \approx 5.66$ . If there is a right angle, it must be opposite the longest side — that is, at $B$ .

Apply the converse of Pythagoras. Compare $AB^2 + BC^2$ with $AC^2$ :

$AB^2 + BC^2 = 32 + 32 = 64$

$AC^2 = 8^2 = 64$

Step 7 — Conclude

$AB^2 + BC^2 = AC^2$

Since the sum of the squares of the two shorter sides equals the square of the longest side, the converse of Pythagoras' theorem applies and the angle at $B$ is a right angle.

Note how working with $d^2$ made Step 6 trivial — no surds appeared at all, because squaring $4\sqrt{2}$ returns the 32 we already had.

Distance between two points

Covers AS91256

The formula

The distance between $(x_1, y_1)$ and $(x_2, y_2)$ is:

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

Each part:
- $(x_2 - x_1)$ — the horizontal separation of the two points
- $(y_2 - y_1)$ — the vertical separation
- $d$ — the straight-line distance between them, always positive

Where it comes from

The formula is Pythagoras' theorem in disguise.
Joining the two points and drawing a horizontal and a vertical line makes a right-angled triangle, with the distance as the hypotenuse.

The horizontal side has length $|x_2 - x_1|$ and the vertical side $|y_2 - y_1|$ , so:

$d^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2$

Taking the square root gives the formula. Understanding this means you never need to memorise it — you can rebuild it from a sketch.

Why the order does not matter

$(x_2 - x_1)$ is squared, so subtracting the other way round gives the same result:

$(3 - 7)^2 = (-4)^2 = 16 \qquad (7 - 3)^2 = 4^2 = 16$

You can take either point as "point 1", as long as you are consistent within one calculation.
The distance is never negative. If your answer is negative, you have taken the square root incorrectly.

Working through an example

The distance between $A(2, 3)$ and $B(8, 11)$ :
- Horizontal difference: $8 - 2 = 6$
- Vertical difference: $11 - 3 = 8$
- Apply the formula: $d = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10$
Negative coordinates need care. For $C(-3, 5)$ and $D(4, -1)$ :
- Horizontal: $4 - (-3) = 7$
- Vertical: $-1 - 5 = -6$
- $d = \sqrt{49 + 36} = \sqrt{85}$
Use brackets when subtracting negatives. $4 - (-3) = 7$ , not 1.

Leaving the answer exact

Most distances are surds, and the exact form is usually preferred.
Simplify by taking out the largest perfect square:
- $\sqrt{50} = \sqrt{25 \times 2} = 5\sqrt{2}$
- $\sqrt{72} = 6\sqrt{2}$
- $\sqrt{85}$ has no square factor, so it cannot be simplified
Give a decimal only if the question asks for one, and round only at the very end.

What distance is used for

Showing two sides are equal — an isosceles triangle, a rhombus, a square
Showing a point is equidistant from two others
Finding a perimeter — add the side lengths
Testing for a right angle with the converse of Pythagoras: if $a^2 + b^2 = c^2$ , the angle opposite $c$ is $90°$
Note that comparing distances often does not need the square root at all. If you only want to know whether two lengths are equal, comparing $d^2$ values is enough and avoids surds entirely.

Tips and Tricks

Square the differences, do not subtract the squares. $\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$ is not $\sqrt{x_2^2 - x_1^2 + \ldots}$ . Work out each difference first, then square it.

Brackets around negative coordinates. $y_2 - y_1$ where $y_1 = -4$ becomes $y_2 + 4$ . Writing the subtraction out in full prevents the sign error.

A distance cannot be negative. If you get one, you have made a sign or arithmetic error — go back rather than dropping the minus.

To show two lengths are equal, you can compare $d^2$ instead of $d$ and skip the surds altogether. State that you are doing so, since it is a genuine simplification worth communicating.

Sketch the points. A rough plot tells you whether a distance of 10 is plausible between points that look 3 units apart, and it catches transcription errors before they propagate.

Worked ExampleUsing distances to classify a triangle

The points are $A(1, 2)$ , $B(5, 6)$ and $C(9, 2)$ . Show that triangle $ABC$ is isosceles, and determine whether it contains a right angle.

Step 1 — Sketch the points

$A$ and $C$ both have $y = 2$ , so they lie on a horizontal line, 8 units apart. $B$ sits above and between them. The triangle looks symmetric — which is worth confirming.

Step 2 — Find $AB$

$AB = \sqrt{(5-1)^2 + (6-2)^2}$

$= \sqrt{4^2 + 4^2} = \sqrt{16 + 16} = \sqrt{32}$

Simplify: $\sqrt{32} = \sqrt{16 \times 2} = 4\sqrt{2}$

Step 3 — Find $BC$

$BC = \sqrt{(9-5)^2 + (2-6)^2}$

$= \sqrt{4^2 + (-4)^2} = \sqrt{16 + 16} = \sqrt{32} = 4\sqrt{2}$

Step 4 — Find $AC$

$AC = \sqrt{(9-1)^2 + (2-2)^2} = \sqrt{64 + 0} = 8$

Step 5 — Conclude that the triangle is isosceles

$AB = BC = 4\sqrt{2}$

Two sides are equal in length, so triangle $ABC$ is isosceles, with the equal sides meeting at $B$ .

Step 6 — Test for a right angle

The longest side is $AC = 8$ , since $4\sqrt{2} \approx 5.66$ . If there is a right angle, it must be opposite the longest side — that is, at $B$ .

Apply the converse of Pythagoras. Compare $AB^2 + BC^2$ with $AC^2$ :

$AB^2 + BC^2 = 32 + 32 = 64$

$AC^2 = 8^2 = 64$

Step 7 — Conclude

$AB^2 + BC^2 = AC^2$

Since the sum of the squares of the two shorter sides equals the square of the longest side, the converse of Pythagoras' theorem applies and the angle at $B$ is a right angle.

Note how working with $d^2$ made Step 6 trivial — no surds appeared at all, because squaring $4\sqrt{2}$ returns the 32 we already had.

Distance between two points

The formula

Where it comes from

Why the order does not matter

Working through an example

Leaving the answer exact

What distance is used for

Step 1 — Sketch the points

Step 2 — Find ABABAB

Step 3 — Find BCBCBC

Step 4 — Find ACACAC

Step 5 — Conclude that the triangle is isosceles

Step 6 — Test for a right angle

Step 7 — Conclude

Distance between two points

The formula

Where it comes from

Why the order does not matter

Working through an example

Leaving the answer exact

What distance is used for

Step 1 — Sketch the points

Step 2 — Find ABABAB

Step 3 — Find BCBCBC

Step 4 — Find ACACAC

Step 5 — Conclude that the triangle is isosceles

Step 6 — Test for a right angle

Step 7 — Conclude

Step 2 — Find $AB$

Step 3 — Find $BC$

Step 4 — Find $AC$

Step 2 — Find $AB$

Step 3 — Find $BC$

Step 4 — Find $AC$