Length of an Interval
The length of an interval on the number plane comes from Pythagoras' theorem, which gives the distance formula \(d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\). This NSW Year 9 Advanced maths subtopic finds distances between points (NESA outcome MA5‑LIN‑C‑01).
You will find distances as whole numbers, surds in simplest form and rounded decimals, the sides and perimeters of triangles, and real distances on a map grid. You will show that a triangle is isosceles or right-angled, find an unknown coordinate from a given distance, find a point on the \(x\)-axis that is equally far from two points, and show that three points lie on a line. It is part of Year 9 Advanced Coordinate Geometry and Graphs.
Every question with a fully worked solution.
- Length of an Interval - Video - Distance between two points Watch
Theory
The length of an interval is the distance between its end points. Draw the run and the rise, and the interval becomes the hypotenuse of a right-angled triangle, so Pythagoras' theorem gives its length. This Year 9 Advanced maths page covers the distance formula, exact surd answers, scale grids, classifying triangles, and working backwards from a distance to an unknown coordinate.
An interval is the part of a line between two end points. Its length is the distance between those points.
If the interval is horizontal or vertical, only one coordinate changes. Subtract that coordinate: \((-4,3)\) to \((-1,3)\) is \(-1-(-4)=3\) units, and \((2,0.5)\) to \((2,4)\) is \(4-0.5=3.5\) units.
Any other interval is the hypotenuse of a right-angled triangle. The horizontal side is the run (change in \(x\)) and the vertical side is the rise (change in \(y\)). Pythagoras' theorem gives the length: \(d^2 = \text{run}^2 + \text{rise}^2\).
Writing the run and rise with coordinates gives the distance formula. It is the same calculation, written once for every pair of points.
An answer such as \(\sqrt{52}\) is exact. Simplify it to simplest surd form, \(2\sqrt{13}\), or round it as the question asks.
For the points \((x_1,y_1)\) and \((x_2,y_2)\):
Squared form, best for comparing lengths and for unknown coordinates:
| Interval | Length |
|---|---|
| horizontal, \(y\) the same | \(|x_2-x_1|\) |
| vertical, \(x\) the same | \(|y_2-y_1|\) |
| sloping | \(\sqrt{\text{run}^2+\text{rise}^2}\) |
How to find the length of an interval
- Sketch the two points and the right-angled triangle under the interval.
- Find the run and the rise by subtracting coordinates. Their signs do not matter once squared.
- Apply Pythagoras' theorem: \(d^2 = \text{run}^2 + \text{rise}^2\).
- Answer exactly (simplest surd) or rounded, as asked, with units if there is a scale.
Problems built on lengths
- Perimeter: find every side, keep exact values, and add. Round once, at the end.
- Isosceles or right-angled? Compare the three values of \(d^2\). Two equal means isosceles. If the two smaller add to the largest, the triangle is right-angled, with the right angle opposite the longest side. Its area is then half the product of the two shorter sides.
- Unknown coordinate: write \(d^2\) with the unknown in it, set it equal to the given length squared and solve. A squared bracket can give two answers (Example 3). Expand if both brackets contain the unknown.
- Same distance from two points: set the two squared lengths equal (Example 4). A point on the \(x\)-axis is \((x,0)\); on the \(y\)-axis it is \((0,y)\).
- Three points in a line: \(A\), \(B\) and \(C\) are collinear when \(AB + BC = AC\). Simplify each surd so they can be added.
Take the run and the rise from \(A\) to \(B\). A negative rise is fine, because it is squared.
| \(\text{run}\) | \(=\) | \(2-(-4) = 6\) |
| \(\text{rise}\) | \(=\) | \(-3-1 = -4\) |
| \(AB^2\) | \(=\) | \(6^2+(-4)^2\) |
| \(\) | \(=\) | \(36+16 = 52\) |
| \(AB\) | \(=\) | \(\sqrt{52} = \sqrt{4\times 13}\) |
| \(\) | \(=\) | \(\mathbf{2\sqrt{13}}\) |
| \(\) | \(\approx\) | \(\mathbf{7.2}\text{ units}\) |
Simplify the surd by taking out the largest square factor, \(4\). Round only the decimal answer.
(i) Show that the triangle is isosceles.
(ii) Is it right-angled?
(iii) Find its perimeter, correct to one decimal place.
Work with the squared lengths. They are whole numbers, and they are what the comparisons need.
(i)
| \(PQ^2\) | \(=\) | \(4^2+3^2 = 25\) |
| \(QR^2\) | \(=\) | \((-4)^2+2^2 = 20\) |
| \(PR^2\) | \(=\) | \(0^2+5^2 = 25\) |
\(PQ = PR = 5\), so the triangle is isosceles. \(PR\) is vertical, so its length is just \(3-(-2)=5\).
(ii) The longest sides are \(PQ\) and \(PR\). Test every pair: \(20+25=45 \ne 25\), and \(25+25=50\ne 20\). No pair of squares adds to the third, so it is not right-angled.
(iii)
| \(\text{perimeter}\) | \(=\) | \(5+5+\sqrt{20}\) |
| \(\) | \(=\) | \(14.472\ldots\) |
| \(\) | \(\approx\) | \(\mathbf{14.5}\text{ units}\) |
Write the squared length with \(k\) in it and set it equal to \(5^2\).
| \((k-(-1))^2+(-1-2)^2\) | \(=\) | \(5^2\) |
| \((k+1)^2+9\) | \(=\) | \(25\) |
| \((k+1)^2\) | \(=\) | \(16\) |
| \(k+1\) | \(=\) | \(4\ \text{ or }\ -4\) |
| \(k\) | \(=\) | \(\mathbf{3}\ \text{ or }\ \mathbf{-5}\) |
A square root has two signs. Both points are \(5\) units from \(A\): one to the right, one to the left. Check: run \(4\), fall \(3\), \(\sqrt{16+9}=5\).
A point on the \(y\)-axis has \(x=0\), so let it be \(P(0,y)\). Set \(PA^2=PB^2\); no square roots are needed.
| \(2^2+(y-1)^2\) | \(=\) | \(4^2+(y-3)^2\) |
| \(4+y^2-2y+1\) | \(=\) | \(16+y^2-6y+9\) |
| \(5-2y\) | \(=\) | \(25-6y\) |
| \(4y\) | \(=\) | \(20\) |
| \(y\) | \(=\) | \(5\) |
So \(\boldsymbol{P(0,5)}\). Check: \(PA^2 = 4+16 = 20\) and \(PB^2 = 16+4 = 20\). The \(y^2\) terms cancel, which is why the equation is linear.
Common pitfalls
Frequently asked questions
How do you find the length of an interval?
Find the run (change in \(x\)) and the rise (change in \(y\)). The interval is the hypotenuse of a right-angled triangle, so its length is \(\sqrt{\text{run}^2+\text{rise}^2}\).
What is the distance formula?
The distance between \((x_1,y_1)\) and \((x_2,y_2)\) is \(\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\). It is Pythagoras' theorem written with coordinates.
Does it matter which point you start from?
No. Swapping the points changes the signs of the run and rise, but squaring removes the sign, so the length is the same.
How do you show a triangle is right-angled using coordinates?
Find \(d^2\) for each side. If the two smaller values add up to the largest one, the triangle is right-angled, and the right angle is opposite the longest side.
How do you write a distance in simplest surd form?
Find the largest square number that divides the number under the root and take its square root outside: \(\sqrt{52} = \sqrt{4\times 13} = 2\sqrt{13}\).