Year 9 Maths Core Straight Line

Length of an Interval

Q: How do you find the length of an interval?

Find the run, the difference of the x-coordinates, and the rise, the difference of the y-coordinates. Square both, add them, and take the square root.

Q: What is the distance formula?

The distance d between two points is the square root of (x2 minus x1) squared plus (y2 minus y1) squared. It is Pythagoras' theorem with the run and the rise as the shorter sides.

20 practice questions 1 video lesson Theory + worked examples

The length of an interval on the number plane is found with Pythagoras' theorem: the interval is the longest side of a right-angled triangle whose two shorter sides are the run and the rise. This NSW Year 9 Core maths subtopic is Stage 5 Core content under NESA outcome MA5‑LIN‑C‑01.

You will find the run and rise from two points, including negative coordinates where subtracting a negative makes the run longer, and calculate lengths such as the distance from \((1,-2)\) to \((8,22)\), which is \(25\) units. You will handle horizontal and vertical intervals, round answers or leave them as surds, find perimeters of shapes on the plane, decide whether a triangle is isosceles or right-angled, use a map grid where \(1\) unit is \(100\) m, and find a missing coordinate. It is part of Year 9 Core Straight Line.

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Theory

The length of an interval is the distance between its two end points. Draw the run across and the rise up, and the interval becomes the hypotenuse of a right-angled triangle. This Year 9 maths page uses Pythagoras' theorem on the Cartesian plane to find distances, with negative coordinates, surds and rounding.

An interval is the part of a line between two points. Its length is the straight-line distance from one end to the other.

Go across from one point and up (or down) to the other. The distance across is the run, the difference of the \(x\)-coordinates. The distance up is the rise, the difference of the \(y\)-coordinates.

The run and the rise are the two shorter sides of a right-angled triangle. The interval is the hypotenuse, so Pythagoras' theorem gives its length.

If the two points have the same \(y\)-coordinate, the interval is horizontal and its length is just the run. If they have the same \(x\)-coordinate, it is vertical and its length is just the rise.

A length is never negative. It does not matter which point you start from.

Go across, then up. The run and the rise are the two shorter sides, and the interval is the hypotenuse: \(d^2 = 4^2 + 2^2 = 20\), so \(d = \sqrt{20} \approx 4.5\) units.

With negative coordinates, subtract a negative: run \(= 2 - (-3) = 5\), rise \(= 2 - (-2) = 4\). Counting squares on the grid checks it.

For the points \((x_1, y_1)\) and \((x_2, y_2)\):

\[\text{run} = x_2 - x_1 \qquad \text{rise} = y_2 - y_1\]

run = x_{2} - x_{1}

\[d^2 = \text{run}^2 + \text{rise}^2 \qquad d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\]

d = \sqrt{{(x_{2} - x_{1})}^{2} + {(y_{2} - y_{1})}^{2}}

Surd or decimal? Leave the answer as a surd such as \(\sqrt{41}\) unless the question asks for decimal places. Then round only at the very end.

Perimeter and comparing sides. For a shape drawn on the plane, find each side the same way. Add them for the perimeter. To compare sides, compare their squares.

Grids with a scale. If \(1\) unit stands for a real distance, such as \(50\) m, find the length in grid units first, then multiply by the scale at the end.

A missing coordinate. If you know the length, put it in for \(d\) and solve. For \(A(2,1)\) and \(B(k,5)\) that are \(5\) units apart with \(k > 2\): the rise is \(4\), so \(\text{run}^2 = 25 - 16 = 9\), the run is \(3\), and \(k = 2 + 3 = 5\).

How to find the length of an interval

Sketch the two points. If they share a coordinate, the interval is horizontal or vertical: just subtract.
Find the run (\(x_2 - x_1\)) and the rise (\(y_2 - y_1\)). Take care when subtracting a negative.
Square the run and the rise separately, then add them to get \(d^2\).
Take the square root to get \(d\). Leave it as a surd, or round if the question asks.
Check by counting squares on a grid, or by making sure \(d\) is longer than both the run and the rise.

Example 1 — Horizontal and vertical intervals

Find the length of
(i) the interval from \(A(-3,4)\) to \(B(5,4)\)
(ii) the interval from \(P(4,-4)\) to \(Q(4,3)\).

Solution

(i) The \(y\)-coordinates are both \(4\), so \(AB\) is horizontal. Its length is the run.

\(AB\)	\(=\)	\(5 - (-3)\)
\(\)	\(=\)	\(\mathbf{8} \text{ units}\)

(ii) The \(x\)-coordinates are both \(4\), so \(PQ\) is vertical. Its length is the rise.

\(PQ\)	\(=\)	\(3 - (-4)\)
\(\)	\(=\)	\(\mathbf{7} \text{ units}\)

A B = 8, P Q = 7

Subtracting a negative adds. Counting squares on the grid gives the same lengths.

Example 2 — Using the formula

Find the distance between \(A(1,-2)\) and \(B(8,22)\).

Solution

The points are too far apart to count squares, so use the run, the rise and Pythagoras' theorem.

\(\text{run}\)	\(=\)	\(8 - 1 = 7\)
\(\text{rise}\)	\(=\)	\(22 - (-2) = 24\)
\(d^2\)	\(=\)	\(7^2 + 24^2\)
\(\)	\(=\)	\(49 + 576 = 625\)
\(d\)	\(=\)	\(\sqrt{625} = \mathbf{25} \text{ units}\)

d = \sqrt{7^{2} + 24^{2}} = 25

Example 3 — A negative rise, rounded

Find the distance between \(C(-4,2)\) and \(D(1,-3)\), correct to \(1\) decimal place.

Solution

\(\text{run}\)	\(=\)	\(1 - (-4) = 5\)
\(\text{rise}\)	\(=\)	\(-3 - 2 = -5\)
\(d^2\)	\(=\)	\(5^2 + (-5)^2\)
\(\)	\(=\)	\(25 + 25 = 50\)
\(d\)	\(=\)	\(\sqrt{50} = 7.071\ldots\)
\(\)	\(\approx\)	\(\mathbf{7.1} \text{ units}\)

d = \sqrt{50} \approx 7.1

The negative rise does no harm, because squaring makes it positive. Leave \(\sqrt{50}\) as it is if the question does not ask for decimals.

Example 4 — Is the triangle isosceles?

A triangle has vertices \(A(0,0)\), \(B(6,2)\) and \(C(2,6)\). Show that it is isosceles.

Solution

Work out the square of each side. There is no need to take square roots to compare them.

\(AB^2\)	\(=\)	\(6^2 + 2^2 = 40\)
\(AC^2\)	\(=\)	\(2^2 + 6^2 = 40\)
\(BC^2\)	\(=\)	\((2 - 6)^2 + (6 - 2)^2\)
\(\)	\(=\)	\(16 + 16 = 32\)

{AB}^{2} = {AC}^{2} = 40

\(AB^2 = AC^2\), so \(AB = AC\) and the triangle is isosceles. Since \(32 + 40 \ne 40\), it is not right-angled.

Common pitfalls

Square first, then add. \((4 + 3)^2\) is not \(4^2 + 3^2\). Square the run and the rise separately.

Take the square root at the end. Stopping at \(d^2\) leaves the answer squared. The length is \(\sqrt{d^2}\).

Watch the double negative. A run from \(x = -4\) to \(x = 1\) is \(1 - (-4) = 5\), not \(-3\).

Keep x with x and y with y. The run uses only the \(x\)-coordinates and the rise uses only the \(y\)-coordinates. Mixing them gives a wrong triangle.

Frequently asked questions

How do you find the length of an interval?

Find the run, the difference of the \(x\)-coordinates, and the rise, the difference of the \(y\)-coordinates. Square both, add them, and take the square root.

What is the distance formula?

\(d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\). It is Pythagoras' theorem with the run and the rise as the shorter sides.

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 removes the sign, so the distance is the same.

How do you find the distance between two points with the same y-coordinate?

The interval is horizontal, so there is no rise. Subtract the \(x\)-coordinates. From \((-3, 4)\) to \((5, 4)\) the distance is \(5 - (-3) = 8\) units.

Can a distance be negative?

No. A distance is a length, so it is always positive or zero. A negative run or rise becomes positive when it is squared.

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Length of an Interval

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How to find the length of an interval

Common pitfalls

Frequently asked questions

Theory