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Year 10 Maths Core Straight Line Graphs

Length of an Interval

20 practice questions 2 video lessons Theory + worked examples

Length of an Interval in NSW Year 10 Core maths finds the distance between two points on the number plane (NESA outcome MA5‑LIN‑C‑01). The interval forms the longest side of a right-angled triangle whose other sides are the horizontal and vertical differences, so Pythagoras' theorem gives its length.

You will find the distance between two points, give lengths exactly as surds or rounded to one decimal place, and find perimeters of triangles and parallelograms drawn on a grid. You will also use a map scale to find real distances, check whether a triangle is isosceles or right-angled, find the radius of a circle from its centre, and find a missing coordinate from a given length. It is part of Year 10 Core Straight Line Graphs.

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Practice questions

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  • Length of an Interval - Video - Distance Between Points Watch
  • Length of an Interval - Video - How to Find the Distance Between Two Points - How to Use the Distance Formula Watch
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Theory

The length of an interval is the distance between its two endpoints. This Year 10 maths page shows how to find it for horizontal and vertical intervals, how the distance formula comes from Pythagoras' theorem, and how to use it for exact surd answers, triangle types and missing coordinates.

An interval is the part of a line between two points, its endpoints. Its length is the distance between them, always positive.

A horizontal interval has endpoints with the same \(y\)-value; its length is the difference of the \(x\)-values. A vertical interval has the same \(x\)-value; its length is the difference of the \(y\)-values.

Any other interval is the hypotenuse of a right-angled triangle with a horizontal side (the run) and a vertical side (the rise). Pythagoras' theorem then gives the length.

An exact length is left as a surd, such as \(\sqrt{13}\) or \(2\sqrt{5}\); otherwise round the decimal as asked.

An interval as the hypotenuse of a right-angled trianglePoints A and B on a number plane joined by an interval of length d. Dashed red lines go across from A then up to B, forming a right angle. The run is x2 minus x1 and the rise is y2 minus y1, so d squared equals run squared plus rise squared. x y −1 1 2 3 4 5 6 7 8 −1 1 2 3 4 5 6 O A(x₁, y₁) B(x₂, y₂) run = x₂ − x₁ rise = y₂ − y₁ d
Any interval is the hypotenuse of a right-angled triangle. Pythagoras' theorem gives its length from the run and the rise.
Horizontal and vertical intervalsA horizontal interval from (minus 3, 2) to (4, 2) has length 4 minus minus 3, which is 7. A vertical interval from (5, minus 2) to (5, 4) has length 4 minus minus 2, which is 6. x y −4 −3 −2 −1 1 2 3 4 5 6 −3 −2 −1 1 2 3 4 5 O (−3, 2) (4, 2) 4 − (−3) = 7 (5, −2) (5, 4) 6
Horizontal: same \(y\), subtract the \(x\)-values. Vertical: same \(x\), subtract the \(y\)-values. Take the bigger minus the smaller.

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

\[d = \sqrt{(x_2 - x_1)^{2} + (y_2 - y_1)^{2}}\]
d=(x2−x1)2+(y2−y1)2

It is Pythagoras' theorem with run \(= x_2 - x_1\) and rise \(= y_2 - y_1\):

\[d^{2} = \text{run}^{2} + \text{rise}^{2}\]
d2=run2+rise2
Order does not matter. \((x_2 - x_1)^{2}\) and \((x_1 - x_2)^{2}\) are equal, so either point can be first. Use brackets around negative numbers.

How to find the length of an interval

  1. Check for a horizontal or vertical interval: if so, subtract the coordinates that differ.
  2. Label the points \((x_1, y_1)\) and \((x_2, y_2)\), or find the run and rise from a sketch.
  3. Substitute into the distance formula, square, add, then take the square root.
  4. Simplify a surd for an exact answer, or round as asked.

Other questions you will meet

  1. Perimeter of a shape: find each side, using the quick method for horizontal and vertical sides, then add. A parallelogram needs only two sides.
  2. Map or grid scale: find the distance in grid units, then multiply by the real length of one unit.
  3. Which is closer? Compare the distances, or simply compare their squares.
  4. Triangle type: equal sides mean isosceles. If the two shorter squares add to the longest square, it is right-angled, with the right angle opposite the longest side (Example 3).
  5. Circles: the radius is the distance from the centre to any point on the circle; the diameter is twice the radius. Points all the same distance from a centre lie on one circle.
  6. Missing coordinate: substitute the length, square both sides and solve. There are usually two answers (Example 4).
Example 1 — Using the run and the rise
\(G\) is \((-4, -1)\) and \(H\) is \((4, 5)\). Find
(i) the length of \(GH\)
(ii) the distance from \(G\) to the point \((4, -1)\).
Solution

(i) Sketch the interval and make a right-angled triangle under it.

Interval GH on a number planePoints G(minus 4, minus 1) and H(4, 5). A right-angled triangle is drawn with a horizontal side of 8 and a vertical side of 6. x y −5 −4 −3 −2 −1 1 2 3 4 5 −2 −1 1 2 3 4 5 6 O G H 8 6
\(\text{run}\)\(=\)\(4 - (-4) = 8\)
\(\text{rise}\)\(=\)\(5 - (-1) = 6\)
\(GH\)\(=\)\(\sqrt{8^{2} + 6^{2}}\)
\(\)\(=\)\(\sqrt{100} = \mathbf{10} \text{ units}\)
GH=82+62=10

(ii) \((-4, -1)\) and \((4, -1)\) have the same \(y\)-value, so the interval is horizontal: \(4 - (-4) = \mathbf{8}\) units. No formula is needed.

Example 2 — Exact length and decimal
Find the length of the interval joining \(J(-3, 2)\) and \(K(3, -1)\),
(i) exactly, as a simplified surd
(ii) correct to one decimal place.
Solution

Label the points, then substitute into the distance formula. Brackets keep the negative signs safe.

\((x_1, y_1)\)\(=\)\((-3, 2), \quad (x_2, y_2) = (3, -1)\)
\(JK\)\(=\)\(\sqrt{(3 - (-3))^{2} + (-1 - 2)^{2}}\)
\(\)\(=\)\(\sqrt{6^{2} + (-3)^{2}}\)
\(\)\(=\)\(\sqrt{36 + 9} = \sqrt{45}\)

(i) \(45 = 9 \times 5\) and \(9\) is a perfect square.

\(\sqrt{45}\)\(=\)\(\sqrt{9} \times \sqrt{5} = \mathbf{3\sqrt{5}} \text{ units}\)

(ii)

\(\sqrt{45}\)\(=\)\(6.708\ldots\)
\(\)\(\approx\)\(\mathbf{6.7} \text{ units}\)
JK=45=35≈6.7

\((-3)^{2} = 9\), not \(-9\): a square is never negative.

Example 3 — What type of triangle?
Triangle \(ABC\) has vertices \(A(1, 1)\), \(B(5, 3)\) and \(C(3, 7)\). Show that it is an isosceles right-angled triangle, and say where the right angle is.
Solution

Work with the squares of the lengths: they are whole numbers, and Pythagoras' theorem uses squares anyway.

\(AB^{2}\)\(=\)\((5 - 1)^{2} + (3 - 1)^{2} = 16 + 4 = 20\)
\(BC^{2}\)\(=\)\((3 - 5)^{2} + (7 - 3)^{2} = 4 + 16 = 20\)
\(AC^{2}\)\(=\)\((3 - 1)^{2} + (7 - 1)^{2} = 4 + 36 = 40\)

\(AB = BC = \sqrt{20}\), so the triangle is isosceles. Test the longest side with Pythagoras' theorem:

\(AB^{2} + BC^{2}\)\(=\)\(20 + 20 = 40 = AC^{2}\)
AB2+BC2=AC2
Triangle ABCTriangle with vertices A(1, 1), B(5, 3) and C(3, 7). Sides AB and BC are marked equal and the angle at B is marked as a right angle. x y 2 4 6 2 4 6 8 O A(1, 1) B(5, 3) C(3, 7)

So the triangle is right-angled. The right angle is opposite the longest side \(AC\), at \(B\).

Example 4 — A missing coordinate
The points \(P(-2, 1)\) and \(Q(4, y)\) are \(10\) units apart. Find the two possible values of \(y\).
Solution

Put the known length into the distance formula, \(\sqrt{(4 - (-2))^{2} + (y - 1)^{2}} = 10\), and square both sides to remove the square root.

\((4 - (-2))^{2} + (y - 1)^{2}\)\(=\)\(10^{2}\)
\(36 + (y - 1)^{2}\)\(=\)\(100\)
\((y - 1)^{2}\)\(=\)\(64\)
\(y - 1\)\(=\)\(8 \text{ or } -8\)
\(y\)\(=\)\(\mathbf{9} \text{ or } \mathbf{-7}\)
(y−1)2=64

Two answers make sense: \(Q\) can be \(8\) units above or \(8\) units below the level of \(P\). Check: from \((-2, 1)\) to \((4, 9)\) the run is \(6\) and the rise is \(8\), and \(\sqrt{36 + 64} = 10\).

Common pitfalls

Square negatives in brackets. \((-4)^{2} = 16\). Typing \(-4^{2}\) on a calculator gives \(-16\).
Watch subtracting a negative. From \(x = -3\) to \(x = 4\) is \(4 - (-3) = 7\), not \(1\).
Square, add, then square root. \(\sqrt{9^{2} + 12^{2}} = \sqrt{225} = 15\), not \(9 + 12 = 21\). The square root of a sum is not the sum of the square roots.
Keep x with x and y with y. Subtract the \(x\)-values from each other and the \(y\)-values from each other. Mixing them up gives a wrong run and rise.

Frequently asked questions

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 using the run and the rise.

How do you find the length of a horizontal or vertical line?

For a horizontal interval the \(y\)-values are the same, so subtract the \(x\)-values. For a vertical interval the \(x\)-values are the same, so subtract the \(y\)-values. Take the bigger minus the smaller.

How is the distance formula related to Pythagoras' theorem?

Draw a right-angled triangle with the interval as its hypotenuse. The horizontal side is the run and the vertical side is the rise, so \(d^{2} = \text{run}^{2} + \text{rise}^{2}\).

What does exact length mean?

An exact length is left as a whole number or a surd, such as \(\sqrt{13}\) or \(2\sqrt{5}\), instead of a rounded decimal. Simplify the surd if it has a square factor.

Does it matter which point is first in the distance formula?

No. Swapping the points changes the sign of the run and the rise, but squaring makes them positive, so the length is the same.

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