Length of an Interval
The length of an interval is the distance between its two end points on the number plane. It comes from Pythagoras' theorem and is written as the distance formula \(d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\). In NSW Year 10 Advanced maths answers are given as exact surds or rounded, and the formula is used to prove facts about shapes (NESA outcome MA5‑LIN‑P‑01).
You will find distances between points with negative, decimal and surd coordinates, and with pronumerals in the coordinates. You will find perimeters of triangles and quadrilaterals, use side and diagonal lengths to classify a quadrilateral, decide whether a point is inside a circle, and find unknown coordinates from a given distance. A fencing cost on a scaled grid and an equilateral triangle on the axes put the formula to work. It is part of Year 10 Advanced Straight Line Graphs.
Theory
The length of an interval joining \((x_1,\ y_1)\) and \((x_2,\ y_2)\) is \(d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\), which is Pythagoras' theorem on the number plane. This Year 10 Advanced maths page covers exact surd answers, perimeters, classifying shapes, and working backwards to an unknown coordinate.
An interval is the part of a line between two points, its endpoints. Its length is the distance between the endpoints, always a positive number of units.
Join the two points with a horizontal side and a vertical side and you get a right-angled triangle. The horizontal side is the change in \(x\), \(x_2 - x_1\), and the vertical side is the change in \(y\), \(y_2 - y_1\). The interval is the hypotenuse, so Pythagoras' theorem gives its length.
When the square root is not a whole number, give an exact answer in simplest surd form, such as \(2\sqrt{13}\), unless the question asks for a decimal.
Lengths let you classify shapes: equal sides show an isosceles triangle, a rhombus or a kite, and the converse of Pythagoras' theorem shows a right angle.
The distance between \((x_1,\ y_1)\) and \((x_2,\ y_2)\):
Squared form, useful for comparing lengths and for equations:
How to find the length of an interval
- Label the points \((x_1,\ y_1)\) and \((x_2,\ y_2)\). Either point can go first.
- Subtract to get the change in \(x\) and the change in \(y\). Use brackets around negative numbers.
- Square, add, take the square root. Simplify the surd, or round if a decimal is asked for.
- Add units: units, or metres if the grid has a scale (multiply grid lengths by the scale).
Using lengths to solve problems
- Perimeter: find each side, simplify each surd, then collect like surds: \(2\sqrt{5} + 2\sqrt{5} = 4\sqrt{5}\).
- Classify a shape: compare side lengths (and diagonals). Use \(a^2 + b^2 = c^2\) to test for a right angle.
- Inside or outside a circle: compare the distance from the centre with the radius.
- Unknown coordinate: write the unknown point with a pronumeral, e.g. \((0,\ y)\) on the \(y\)-axis, set the distance equal to the given value, square both sides and solve. Expect two answers (\(\pm\)), then use any condition such as \(h \gt 0\).
- Same distance from two points: set \(PA^2 = PB^2\). The squared terms cancel and a linear equation is left (Example 4).
- General proofs: with coordinates such as \((a,\ b)\) and \((a - 6,\ b + 8)\), the pronumerals cancel in the differences, so the distance is the same for every point.
(i) exactly, in simplest surd form
(ii) correct to \(1\) decimal place.
Label the points, then substitute. Brackets keep the negative signs safe.
| \(x_1 = -4,\ y_1 = 3\) | \(\) | \(x_2 = 2,\ y_2 = -1\) |
(i)
| \(PQ\) | \(=\) | \(\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\) |
| \(\) | \(=\) | \(\sqrt{(2 - (-4))^2 + (-1 - 3)^2}\) |
| \(\) | \(=\) | \(\sqrt{6^2 + (-4)^2}\) |
| \(\) | \(=\) | \(\sqrt{36 + 16} = \sqrt{52}\) |
| \(\) | \(=\) | \(\sqrt{4 \times 13} = \boldsymbol{2\sqrt{13}}\text{ units}\) |
(ii)
| \(2\sqrt{13}\) | \(=\) | \(7.2111\ldots\) |
| \(\) | \(\approx\) | \(\boldsymbol{7.2}\text{ units}\) |
To simplify a surd, take out the largest square factor: \(52 = 4 \times 13\).
Find all three side lengths. Leaving them squared makes the Pythagoras check easy.
| \(AB^2\) | \(=\) | \((3 - (-1))^2 + (1 - (-2))^2 = 16 + 9 = 25\) |
| \(BC^2\) | \(=\) | \((0 - 3)^2 + (5 - 1)^2 = 9 + 16 = 25\) |
| \(AC^2\) | \(=\) | \((0 - (-1))^2 + (5 - (-2))^2 = 1 + 49 = 50\) |
\(AB = BC = 5\), 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.
| \(AC\) | \(=\) | \(\sqrt{50} = 5\sqrt{2}\) |
| \(\text{Perimeter}\) | \(=\) | \(5 + 5 + 5\sqrt{2} = \boldsymbol{10 + 5\sqrt{2}}\text{ units}\) |
Any point on the \(x\)-axis has \(y = 0\), so write it as \(P(x,\ 0)\). Set the distance equal to \(10\) and square both sides to remove the square root.
| \(\sqrt{(x - 3)^2 + (0 - 6)^2}\) | \(=\) | \(10\) |
| \((x - 3)^2 + 36\) | \(=\) | \(100\) |
| \((x - 3)^2\) | \(=\) | \(64\) |
| \(x - 3\) | \(=\) | \(8 \text{ or } -8\) |
| \(x\) | \(=\) | \(11 \text{ or } -5\) |
So \(\boldsymbol{P(11,\ 0)}\) or \(\boldsymbol{P(-5,\ 0)}\). There are two answers because a circle of radius \(10\) around \(Q\) cuts the \(x\)-axis twice. Check: \(\sqrt{8^2 + 6^2} = 10\).
Write \(P(0,\ y)\). Equal distances have equal squares, so set \(PA^2 = PB^2\) and avoid square roots altogether.
| \((0 - 1)^2 + (y - 4)^2\) | \(=\) | \((0 - 5)^2 + (y - 2)^2\) |
| \(1 + y^2 - 8y + 16\) | \(=\) | \(25 + y^2 - 4y + 4\) |
| \(-8y + 17\) | \(=\) | \(-4y + 29\) |
| \(-4y\) | \(=\) | \(12\) |
| \(y\) | \(=\) | \(-3\) |
The \(y^2\) terms cancel, leaving a linear equation. So \(\boldsymbol{P(0,\ -3)}\).
| \(PA\) | \(=\) | \(\sqrt{1^2 + 7^2} = \sqrt{50} = \boldsymbol{5\sqrt{2}}\) |
| \(PB\) | \(=\) | \(\sqrt{5^2 + 5^2} = \sqrt{50}\ \checkmark\) |
Common pitfalls
Frequently asked questions
What is the distance formula?
The distance between \((x_1,\ y_1)\) and \((x_2,\ y_2)\) is \(d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\). It is Pythagoras' theorem applied to the change in \(x\) and the change in \(y\).
How do you find the length of a line segment between two points?
Subtract the \(x\)-coordinates and the \(y\)-coordinates, square both differences, add them and take the square root. For \((1,\ 2)\) and \((4,\ 6)\) the length is \(\sqrt{9 + 16} = 5\).
Does it matter which point is first in the distance formula?
No. Swapping the points changes the sign of each difference, but squaring removes the sign, so the length is the same.
How do you give a distance in exact form?
Leave the answer as a surd and simplify it by taking out the largest square factor. For example, \(\sqrt{52} = \sqrt{4 \times 13} = 2\sqrt{13}\).
How do you prove a triangle is right-angled using coordinates?
Find the square of each side length. If the two smaller squares add to the largest, the triangle is right-angled, by the converse of Pythagoras' theorem.