The Perpendicular Bisector

Find the equation of the line that cuts a segment in half at right angles, and spot it hidden in kites, squares and isosceles triangles.

In the exam for Cambridge IGCSE Maths 0580, Edexcel IGCSE Maths 4MA1, Cambridge O Level Maths 4024, IGCSE Additional Maths 0606, Edexcel Further Pure Maths 4PM1 · 36 real past-paper questions below

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Half Way, at a Right Angle

The perpendicular bisector of PQ is the line that cuts PQ in half and crosses it at a right angle.

Every point on it is the same distance from P as from Q. That is why it hides inside symmetrical shapes.

Steps

1
Find the midpoint of PQ: the line goes through it.
2
Find the gradient of PQ, then flip it and change the sign.
3
Put the midpoint into $y = m x + c$ to find c.

Key idea

Puzzle: three villages want to dig one well, the same distance from each. Where? Answer: draw the perpendicular bisectors of two sides of the triangle they make. Each line is equally far from its two villages, so the crossing point is equally far from all three.

The Equation of the Bisector

The 5- or 6-mark question: “Find an equation of the perpendicular bisector of PQ.” Follow the three steps, one move per line.

Worked example

1
$P (- 2, 1), Q (4, 5)$ — The two points. The bisector goes through their middle, at a right angle to PQ.
2
$M = (\frac{- 2 + 4}{2}, \frac{1 + 5}{2})$ — Midpoint: add the two x's and halve, then the two y's. Watch the sign: −2 + 4 = 2.
3
$M = (1, 3)$ — 2 ÷ 2 = 1 and 6 ÷ 2 = 3. This is the one point we know is on the bisector.
4
$m_{P Q} = \frac{5 - 1}{4 - ( - 2 )} = \frac{4}{6} = \frac{2}{3}$ — Gradient of PQ: change in y over change in x. 4 − (−2) = 6, because taking away a negative adds.
5
$m = - \frac{3}{2}$ — Flip 2/3 to 3/2 and change the sign, because perpendicular gradients multiply to −1: $\frac{2}{3} \times (- \frac{3}{2}) = - 1$ .
6
$3 = - \frac{3}{2} (1) + c$ — Put M into $y = m x + c$ : x = 1 and y = 3. Use M, not P or Q.
7
$3 = - \frac{3}{2} + c$ — −3/2 × 1 = −3/2.
8
$c = 3 + \frac{3}{2} = \frac{9}{2}$ — Add 3/2 to both sides to get c on its own.
9
$y = - \frac{3}{2} x + \frac{9}{2}$ — Write m and c into $y = m x + c$ . Stop here if the question asks for this form.
10
$2 y = - 3 x + 9$ — Times every term by 2 to clear the halves, for questions that want whole numbers.
11
$3 x + 2 y - 9 = 0$ — Add 3x and take 9 from both sides. This is the form $a x + b y + c = 0$ .

Key idea

Show every line. Most of the marks are for the working: one each for the midpoint, the gradient of PQ, the flipped gradient and putting M in. A slip at the end still scores most of them.

Common mistake

Putting P or Q into the equation instead of M. The bisector does not pass through P or Q, only through the midpoint.

When the Midpoint Is a Fraction

If two coordinates add to an odd number, the midpoint has a half in it. Keep it as a fraction and clear the fractions at the end.

Worked example

1
$P (- 3, 2), Q (2, 5)$ — The two points. −3 + 2 = −1 is odd, so expect a half.
2
$M = (\frac{- 3 + 2}{2}, \frac{2 + 5}{2}) = (- \frac{1}{2}, \frac{7}{2})$ — Midpoint: add and halve each pair. Keep the halves as fractions, not decimals.
3
$m_{P Q} = \frac{5 - 2}{2 - ( - 3 )} = \frac{3}{5}$ — Gradient of PQ: change in y over change in x. 2 − (−3) = 5.
4
$m = - \frac{5}{3}$ — Flip 3/5 and change the sign. Check: $\frac{3}{5} \times (- \frac{5}{3}) = - 1$ .
5
$\frac{7}{2} = - \frac{5}{3} \times (- \frac{1}{2}) + c$ — Put M into $y = m x + c$ . Brackets keep the minus on x.
6
$\frac{7}{2} = \frac{5}{6} + c$ — Minus times minus is plus: $- \frac{5}{3} \times (- \frac{1}{2}) = + \frac{5}{6}$ .
7
$c = \frac{21}{6} - \frac{5}{6} = \frac{16}{6} = \frac{8}{3}$ — Take 5/6 from both sides. Write 7/2 as 21/6 so the bottoms match, then simplify.
8
$y = - \frac{5}{3} x + \frac{8}{3}$ — Write m and c into $y = m x + c$ .
9
$3 y = - 5 x + 8$ — Times every term by 3 to clear the thirds.
10
$5 x + 3 y = 8$ — Add 5x to both sides. Whole numbers only, as “px + qy = r” questions ask.

Key idea

Check with M: $5 \times (- \frac{1}{2}) + 3 \times \frac{7}{2} = - \frac{5}{2} + \frac{21}{2} = 8$ . It fits, so the line is right.

Common mistake

Rounding −5/3 to −1.67. The answer is no longer exact, and you cannot clear the fractions to get whole numbers.

The Hidden Bisector

The 5-mark question does not say “perpendicular bisector”. It hides it in a shape. Spot it, then use the same three steps.

The shape	The hidden bisector
Kite with AB = AD	diagonal AC bisects BD
Square PQRS	diagonal QS bisects PR
Isosceles, AB = AC	line of symmetry bisects BC

Worked example

1
$Kite: A B = A D, C B = C D, B (1, 6), D (7, 2)$ — Find the line AC. A is the same distance from B and D, and so is C. So AC is the perpendicular bisector of BD.
2
$M = (\frac{1 + 7}{2}, \frac{6 + 2}{2}) = (4, 4)$ — Midpoint of BD: add and halve. AC passes through it.
3
$m_{B D} = \frac{2 - 6}{7 - 1} = \frac{- 4}{6} = - \frac{2}{3}$ — Gradient of BD: change in y over change in x. y goes down 4, so the gradient is negative.
4
$m = \frac{3}{2}$ — Flip 2/3 to 3/2 and change the sign: minus becomes plus. Check: $- \frac{2}{3} \times \frac{3}{2} = - 1$ .
5
$4 = \frac{3}{2} (4) + c$ — Put M(4, 4) into $y = m x + c$ .
6
$4 = 6 + c$ — 3/2 × 4 = 6.
7
$c = - 2$ — Take 6 from both sides: 4 − 6 = −2.
8
$y = \frac{3}{2} x - 2$ — Write m and c into $y = m x + c$ .
9
$2 y = 3 x - 4$ — Times every term by 2 to clear the half. This is line AC.

Common mistake

Using the midpoint of AC. You do not know A or C yet: it is BD that gets cut in half.

A Point Already on the Bisector

Sometimes you do not know both ends of the segment. Any point that is the same distance from both ends lies on the bisector, so use it instead of the midpoint.

Worked example

1
$A B = A C, A (5, 9), B and C on 2 y = x + 4$ — Find the line of symmetry of triangle ABC. A is the same distance from B and C, so A is on the perpendicular bisector of BC.
2
$y = \frac{1}{2} x + 2$ — Divide the line BC by 2 so y is on its own. Its gradient is the number in front of x: 1/2.
3
$m = - 2$ — The line of symmetry is perpendicular to BC: flip 1/2 to 2 and change the sign.
4
$9 = - 2 (5) + c$ — Put A(5, 9) into $y = m x + c$ . A takes the place of the midpoint, which we do not know.
5
$9 = - 10 + c$ — −2 × 5 = −10.
6
$c = 19$ — Add 10 to both sides.
7
$y = - 2 x + 19$ — Write m and c into $y = m x + c$ .
8
$2 x + y = 19$ — Add 2x to both sides, for the form px + qy = r.

Key idea

A circle works the same way. Its centre is the same distance from both ends of any chord, so the perpendicular bisector of a chord passes through the centre. For the chord from (−1, 7) to (7, 1) of a circle with centre (0, 0), the chord's gradient is $- \frac{3}{4}$ , so the bisector is $y = \frac{4}{3} x$ : c = 0 because it goes through (0, 0).

Common mistake

Treating A as an end of BC. A is the top of the triangle. It sits on the bisector, not on BC.

Working Backwards From the Bisector

The 6-mark question gives the bisector's equation and asks about A and B. Use both facts: AB is perpendicular to it, and the midpoint lies on it.

Worked example

1
$bisector: y = 2 x - 1$ — The bisector's gradient is the number in front of x: 2.
2
$m_{A B} = - \frac{1}{2}$ — AB crosses the bisector at a right angle, so flip 2 to 1/2 and change the sign.
3
$A (- 1, 7), B (7, k)$ — The question: find k.
4
$\frac{k - 7}{7 - ( - 1 )} = - \frac{1}{2}$ — Write the gradient of AB from the two points, and set it equal to −1/2.
5
$\frac{k - 7}{8} = - \frac{1}{2}$ — 7 − (−1) = 8, because taking away a negative adds.
6
$k - 7 = - 4$ — Times both sides by 8: 8 × (−1/2) = −4.
7
$k = 3$ — Add 7 to both sides.
8
$M = (\frac{- 1 + 7}{2}, \frac{7 + 3}{2}) = (3, 5)$ — Now that k is known, find the midpoint of AB: add and halve.
9
$2 (3) - 1 = 5 ✓$ — Put x = 3 into the bisector: y = 5, which matches M. So M is on the bisector, as it must be.

Key idea

The check is part of the answer. If the midpoint does not fit the bisector, something went wrong.

Drawing It With Compasses

Some questions say “use a straight edge and compasses only”, or ask for the points “equidistant from A and B” (the same distance from both). Both mean: construct the perpendicular bisector of AB.

Steps

1
Open the compasses to more than half of AB. Less than half, and the arcs will not meet.
2
Put the point on A. Draw one arc above AB and one below it.
3
Keep the same width. Put the point on B and draw two arcs that cross the first two.
4
With a ruler, join the two crossing points. That line is the perpendicular bisector. Where it crosses AB is the midpoint.

Common mistake

Rubbing out the arcs. They are your working: without them the line loses marks, even if it is in the right place.

Key idea

All the points closer to A than to B are on A's side of this line. Use that when a question asks you to shade a region.

Which Method? Read the Question

Look for “the same distance”, “bisect” or a symmetrical shape.

You see	You do
“Perpendicular bisector of PQ”	midpoint, flip gradient, find c
Kite, square or isosceles triangle	find the hidden bisector
The bisector's equation is given	perpendicular gradient, midpoint on it
The top of an isosceles triangle, or a circle's centre	use that point instead of the midpoint
“Compasses only” or “equidistant from A and B”	arcs from A and B, join the crossings
Answer as px + qy = r	keep fractions, clear them at the end
Just the midpoint	Length of a Line Segment
Perpendicular through any point	Perpendicular Lines

Key idea

Two things go into every bisector: the midpoint and the flipped gradient. Miss either and the line is wrong.

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