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Linear Toolkit

Three small formulas do most of the work in coordinate geometry: the midpoint, the distance between two points, and the gradient. Learn them once and you can describe any straight line.

MYP 4Extended MathsCoordinate geometryCriteria A · C · D~10 min read

Written by Fayek Jawad Haider · MYP student

Points and labels

Every tool here starts the same way: take two points and give them names. Call the first \((x_1, y_1)\) and the second \((x_2, y_2)\). The little numbers below are just labels, they are not powers, so \(x_1\) simply means "the \(x\) of the first point".

Once the points are labelled, each formula is a matter of careful substitution. Keeping the same point as "point 1" all the way through is the single habit that prevents most mistakes.

Ordered pair
A point written as \((x, y)\), where the \(x\)-coordinate always comes first. The order matters: \((3, 5)\) and \((5, 3)\) are different places on the grid.

The midpoint

The midpoint is the point exactly halfway between two others. In plain terms, you average the two \(x\)-values and average the two \(y\)-values:

\[ M = \left(\frac{x_1 + x_2}{2},\ \frac{y_1 + y_2}{2}\right) \]
Worked example

Find the midpoint of the segment joining \((2, 3)\) and \((8, 11)\).

1
Label the points: \((x_1, y_1) = (2, 3)\) and \((x_2, y_2) = (8, 11)\).
2
Average the \(x\)-values: \(\dfrac{2 + 8}{2} = \dfrac{10}{2} = 5\).
3
Average the \(y\)-values: \(\dfrac{3 + 11}{2} = \dfrac{14}{2} = 7\).
\(M = (5, 7)\)

Sanity check the midpoint

The midpoint should sit between the two points. If your answer is larger than both \(x\)-values or smaller than both, you have added where you should have divided.

The distance formula

The distance between two points is the straight-line length of the segment joining them. It comes straight from Pythagoras: the horizontal gap is \(x_2 - x_1\), the vertical gap is \(y_2 - y_1\), and the distance is the hypotenuse.

\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
Worked example

Find the distance between \((1, 2)\) and \((4, 6)\).

1
Horizontal gap: \(x_2 - x_1 = 4 - 1 = 3\).
2
Vertical gap: \(y_2 - y_1 = 6 - 2 = 4\).
3
Substitute and square: \(d = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25}\).
\(d = 5\) units

The squares hide the sign

Because each gap is squared, a negative gap does no harm: \((-4)^2 = 16\) just like \(4^2\). So you never need to worry which point you subtract first for distance, though you must square before you add.

The gradient

The gradient measures steepness: the change in \(y\) for each step in \(x\), remembered as "rise over run".

\[ m = \frac{y_2 - y_1}{x_2 - x_1} \]
Worked example

Find the gradient of the line through \((2, 1)\) and \((6, 9)\).

1
Label the points: \((x_1, y_1) = (2, 1)\) and \((x_2, y_2) = (6, 9)\).
2
Rise over run: \(m = \dfrac{9 - 1}{6 - 2} = \dfrac{8}{4}\).
\(m = 2\)

Where this is assessed

These three tools appear under Criterion A (Knowing and understanding) when you calculate, and Criterion C (Communicating) when you set out your substitution clearly. In modelling tasks they support Criterion D (Applying maths in real-life contexts), for example finding how far apart two locations are on a map grid.

Check yourself

1. Find the midpoint of \((0, 0)\) and \((6, 4)\). +

Average each coordinate: \(\left(\tfrac{0+6}{2}, \tfrac{0+4}{2}\right) = \left(\tfrac{6}{2}, \tfrac{4}{2}\right)\). So the midpoint is \(\mathbf{(3, 2)}\).

2. Find the distance between \((0, 0)\) and \((6, 8)\). +

The gaps are 6 and 8, so \(d = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100}\). The distance is \(\mathbf{10}\) units.

3. Find the gradient of the line through \((3, 7)\) and \((5, 3)\). +

Rise over run: \(m = \dfrac{3 - 7}{5 - 3} = \dfrac{-4}{2}\). The gradient is \(\mathbf{-2}\), a line that slopes downwards.

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