Length of a Line Segment

Find the midpoint of a line segment, work backwards from a midpoint to a missing end, and find a segment's length with Pythagoras.

In the exam for Cambridge IGCSE Maths 0580, Edexcel Further Pure Maths 4PM1, Edexcel IGCSE Maths 4MA1, Cambridge O Level Maths 4024, IGCSE Additional Maths 0606 · 71 real past-paper questions below

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The Middle and the Length

A line segment is the straight piece between two points. Exam questions ask two things about it.

Question	Tool
Where is the middle?	the midpoint: average the coordinates
How long is it?	Pythagoras on the grid

Key idea

The midpoint question is by far the most common. It is usually worth 2 marks.

Finding the Midpoint

The 2-mark question: “Find the coordinates of the midpoint of AB.” Find the middle of the x values, then of the y values.

Formula

M = (\frac{x _{1} + x _{2}}{2}, \frac{y _{1} + y _{2}}{2})

$(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ are the two ends. Add the two x values and halve. Then add the two y values and halve.

Worked example

1
$A (- 3, 5), B (7, - 2)$ — The two ends. The first number in each bracket is x, the second is y.
2
$x : \frac{- 3 + 7}{2} = \frac{4}{2}$ — Add the two x values: −3 + 7 = 4.
3
$= 2$ — Halve it. Half way between −3 and 7 is 2.
4
$y : \frac{5 + ( - 2 )}{2} = \frac{3}{2}$ — Add the two y values. Adding −2 is the same as taking 2: 5 − 2 = 3.
5
$= 1.5$ — Halve it. Half way between 5 and −2 is 1.5.
6
$M = (2, 1.5)$ — Write x first, then y. A decimal is fine, so do not round it.

Common mistake

Subtracting the coordinates. For the midpoint you add, then halve.

Key idea

If PQ is a diameter of a circle, the centre of the circle is the midpoint of PQ.

Working Back From the Midpoint

The 3-mark question gives the midpoint and one end, with letters for unknowns. Write an equation for each coordinate.

Worked example

1
$A (2, 9), B (14, k), M (p, 4)$ — Find p and k. M is the midpoint, so each of its coordinates is the average of A's and B's.
2
$p = \frac{2 + 14}{2} = \frac{16}{2}$ — Both x values are known, so add them: 2 + 14 = 16.
3
$p = 8$ — Halve 16. That is the midpoint's x value.
4
$\frac{9 + k}{2} = 4$ — The y values: k is unknown, so write their average and set it equal to the midpoint's 4.
5
$9 + k = 8$ — Multiply both sides by 2 to undo the halving: 4 × 2 = 8.
6
$k = - 1$ — Take 9 from both sides: 8 − 9 = −1.
7
$\frac{9 + ( - 1 )}{2} = \frac{8}{2} = 4$ — Check: put k = −1 back in. The average is 4, so k is right.

Common mistake

Halving the 4 instead of doubling it. The midpoint is the average, so undo it: times 2, then take away the known end.

Key idea

Quick check: from A to M is the same step as from M to B.

The Length of a Segment

The 3-mark question: “Work out the distance between P and Q.” Draw a right-angled triangle under the segment and use Pythagoras.

Formula

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

$d$ is the length. $x_{2} - x_{1}$ is the distance across; $y_{2} - y_{1}$ is the distance up. The segment is the long side.

Worked example

1
$P (- 2, - 1), Q (6, 5)$ — The two ends. Take Q's numbers minus P's numbers each time.
2
$across = 6 - (- 2)$ — The change in x. Keep −2 in a bracket so its minus is not lost.
3
$= 6 + 2 = 8$ — Taking away −2 is the same as adding 2. This is the flat side.
4
$up = 5 - (- 1) = 5 + 1 = 6$ — The change in y, the same way. This is the upright side.
5
$d^{2} = 8^{2} + 6^{2}$ — Pythagoras: the segment is the longest side, so its square is the two other squares added.
6
$d^{2} = 64 + 36 = 100$ — Square each side first, then add.
7
$d = 100 = 10$ — Square root to undo the square. A length is never negative, so take the positive root.

Common mistake

Adding 8 + 6 = 14. Square each side first, add, then take the square root.

Key idea

Write the line $8^{2} + 6^{2}$ in full. Most of the marks are for that working, so it still scores if your calculator slips.

Exact Lengths and Surds

Most lengths are not whole numbers. Read the question: it asks for an exact answer, a surd, or a decimal to 3 significant figures.

Worked example

1
$P (1, - 2), Q (7, 1)$ — Find the length PQ in its simplest surd form, then to 3 s.f.
2
$across = 7 - 1 = 6$ — The change in x.
3
$up = 1 - (- 2) = 1 + 2 = 3$ — The change in y. Taking away −2 is the same as adding 2.
4
$d^{2} = 6^{2} + 3^{2} = 36 + 9 = 45$ — Pythagoras: square each side, then add.
5
$d = 45$ — 45 is not a square number, so the root is not whole. $45$ is already exact.
6
$45 = 9 \times 5$ — For simplest form, split off the biggest square number: 45 = 9 × 5.
7
$= 35$ — √9 = 3, so 3 comes out in front. Use this when the question says “simplest surd form”.
8
$\approx 6.71$ — Only if the question asks for a decimal: √45 = 6.708…, which is 6.71 to 3 s.f.

Key idea

“Show that AB = AC”: work out both lengths with the formula, write both answers in full, then say in words that they are equal.

A Point From a Given Length

Sometimes the length is given and a coordinate is missing. Use Pythagoras with a letter, and expect two answers.

Worked example

1
$A (6, 5), A P = 10$ — P is on the y-axis. Find the two places P could be.
2
$P (0, k)$ — On the y-axis, x is always 0. Call the unknown y value k.
3
$across = 6 - 0 = 6$ — The change in x.
4
$up = k - 5$ — The change in y. Keep it as a bracket, because k is unknown.
5
$6^{2} + (k - 5)^{2} = 1 0^{2}$ — Pythagoras: across² + up² = length².
6
$36 + (k - 5)^{2} = 100$ — Square the numbers you know: 6² = 36 and 10² = 100.
7
$(k - 5)^{2} = 64$ — Take 36 from both sides: 100 − 36 = 64.
8
$k - 5 = \pm 8$ — Square root both sides. Both 8 and −8 square to 64, so keep the ±.
9
$k = 5 + 8 = 13 or k = 5 - 8 = - 3$ — Add 5 to both sides, once for each sign.
10
$P (0, 13) or P (0, - 3)$ — One point above A, one below. Both are 10 from A.

Common mistake

Writing only k − 5 = 8. The −8 gives the second point, and a mark is for giving both.

Key idea

If the question says P is above the x-axis, reject (0, −3), because its y value is negative. Write why you reject it.

Which Method? Read the Question

The word in the question picks the tool.

You see	You do
“Midpoint of AB”	add and halve, x then y
Midpoint given, one end missing	double the midpoint, take the known end
“Distance” or “length”	across, up, Pythagoras
Same x, or same y	just subtract; no Pythagoras needed
Length given, a coordinate missing	Pythagoras with a letter; give both answers
Midpoint and a right angle	The Perpendicular Bisector
Plotting or reading a point	Coordinates and Plotting Points

Key idea

Sketch the two points. A rough picture shows if your midpoint is really between them.

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