Extended. Two coordinates are all you need. Pythagoras turns the steps across and up into the length of the segment, and averaging the coordinates gives the point exactly halfway.
For \(A(x_1, y_1)\) and \(B(x_2, y_2)\): length \(AB = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\), and the midpoint is \(\left(\dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2}\right)\). Length uses differences; midpoint uses sums.
Fig. 1: from A(-2, 1) to B(4, 9) is 6 across and 8 up, so AB = 10; the midpoint M(1, 5) averages the coordinates
Section 01
Pythagoras on a grid.
In the lab, drag the end points and watch the right-angled triangle under the segment. Read off the steps across and up, square them, add, and square-root. Then check your answer against the length the lab shows. Try a segment that runs downhill: do the negative steps matter?
The segment is a hypotenuse. The run (change in \(x\)) and rise (change in \(y\)) are the two shorter sides.
Square, add, root: \(AB = \sqrt{(\text{change in } x)^2 + (\text{change in } y)^2}\).
Signs vanish when you square: \((-6)^2 = 36\), so the order of the points does not matter.
Give an exact surd or 3 significant figures, as the question asks.
Worked example A whole-number length
Find the length of the line segment joining \(A(-2, 1)\) and \(B(4, 9)\).
Adding before squaring. \(\sqrt{6^2 + 8^2}\) is 10, but \(6 + 8 = 14\) and \(\sqrt{6 + 8} = 3.74\) are both wrong. Square each step first, add, then take the square root of the total.
Section 02
Halfway between.
The midpoint sits halfway across and halfway up, so each of its coordinates is the mean of the two end coordinates: add and halve.
Worked example Midpoint from two points
Find the midpoint of the line segment joining \((-3, 8)\) and \((9, -2)\).
\(x = \dfrac{-5 + 1}{2} = -2\) and \(y = \dfrac{6 + (-4)}{2} = 1\), so the midpoint is \((-2, 1)\).
Examiner trap
Halving the difference instead of the sum. For \((-3, 8)\) and \((9, -2)\), \(\dfrac{9 - (-3)}{2} = 6\) is half the run, not the midpoint's \(x\)-coordinate. The midpoint is an average: ADD, then halve.
Stage 1 · Learn
Check your understanding
Four quick checks. Each correct answer earns XP and records this skill on your profile.
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What is the length of the line segment joining \((1, 1)\) and \((4, 5)\)?
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What is the midpoint of \((2, 6)\) and \((8, 0)\)?
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Which calculation gives the length of \(AB\), where \(A(-2, 3)\) and \(B(4, -5)\)?
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\(M(3, 7)\) is the midpoint of \(AB\), where \(A\) is \((1, 4)\). What are the coordinates of \(B\)?
Section 03
Putting them together.
Exam questions often combine both skills: the centre of a circle is the midpoint of a diameter, the radius is half its length, and a known distance can be turned into an equation for a missing coordinate.
Worked example Centre and radius of a circle
\(AB\) is a diameter of a circle, with \(A(1, 2)\) and \(B(7, 10)\). Find the centre and the radius.
Centre = midpoint of \(AB\) \(= \left(\dfrac{1 + 7}{2}, \dfrac{2 + 10}{2}\right) = (4, 6)\).
Forgetting the negative square root. \((k - 2)^2 = 25\) gives \(k - 2 = \pm 5\): a point 13 units away can be on either side, so there are usually two answers.
Stage 2 · Exam
Exam-style questions
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Find the length of \(PQ\), where \(P(-3, 4)\) and \(Q(5, -2)\).
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Find the midpoint of the line segment joining \((-5, 7)\) and \((9, -3)\).
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\(AB\) is a diameter of a circle, where \(A(-1, 3)\) and \(B(7, 9)\). Find the centre and radius of the circle.
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The point \(C(k, 1)\) is 13 units from \(D(2, -11)\), and \(k > 0\). Find \(k\).
Skill unlocked
Lengths and midpoints, found.
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