Coordinate Geometry

Perpendicular bisectors and coordinate problems

Equation of a Perpendicular Bisector AS Level 14 min read

On this page

What you need to be able to do
What you should already know
The perpendicular bisector
Finding unknowns: the standard conditions
Key facts at a glance
How this is examined
Mark-scheme focus
Worked examples
Summary
Can I do this?
Quiz

What you need to be able to do

Syllabus 1.3 Coordinate geometry

1.3Find the perpendicular bisector of a line segment, and use it (where it meets an axis or another line; finding the other end of a segment).
1.3Find an unknown coordinate from a condition: a right angle, equal lengths, a point on a line.
1.3Solve multi-step problems with triangles and quadrilaterals, using midpoints, gradients and distances.

Before you start: Straight lines. Next: Circles (the centre of a circle is on the perpendicular bisector of every chord).

Part 1Learn the topic

The perpendicular bisector and the standard conditions (right angle, equal lengths, a point on a line), from the basics, with our own examples. Past-paper questions come in Part 2.

What you should already know

Midpoint and gradient

Midpoint = average of the coordinates; gradient = \(\tfrac{\text{change in }y}{\text{change in }x}\) (Straight lines).

Quick check: midpoint and gradient of \((0,\,4)\) and \((6,\,0)\)?

\((3,\,2)\); \(-\tfrac23\)

Perpendicular gradients

\(m_1m_2=-1\): flip the fraction and change the sign. \(\tfrac25\to-\tfrac52\).

Quick check: perpendicular to gradient \(-4\)?

\(\tfrac14\)

Equation of a line

\(y-y_1=m(x-x_1)\) through a known point.

Quick check: line through \((3,\,5)\), gradient 2

\(y=2x-1\)

Simultaneous linear equations

Two lines meet where both equations hold: substitute one into the other.

Quick check: where do \(y=x+1\) and \(y=-x+5\) meet?

\((2,\,3)\)

The perpendicular bisector

\(A(2,5)\), \(B(8,1)\): midpoint \(M(5,3)\), \(m_{AB}=-\tfrac23\), so the bisector has gradient \(\tfrac32\) and meets the \(y\)-axis at \(C\).

Midpoint \(M\) of \(AB\).
Gradient of \(AB\), then the perpendicular gradient \(-\dfrac{1}{m_{AB}}\).
Line through the midpoint \(M\) (not \(A\), not \(B\)) with that gradient.

The perpendicular bisector of \(AB\) is the line through the midpoint of \(AB\) at right angles to \(AB\). Every point on it is the same distance from \(A\) as from \(B\).

That last fact is why it matters: the centre of any circle through \(A\) and \(B\) lies on it (a circle's centre is equidistant from every point on the circle). So "show that the centre lies on the line …" is a perpendicular-bisector question.

Why every point on it is equidistant. Take any point \(P\) on the bisector. Triangles \(PMA\) and \(PMB\) share the side \(PM\), have \(MA=MB\) (\(M\) is the midpoint) and both have a right angle at \(M\). So they are congruent, and \(PA=PB\). Conversely, the set of all points equidistant from \(A\) and \(B\) is exactly this line.

The two classic errors: drawing the line through \(A\) or \(B\) instead of the midpoint, and forgetting the negative reciprocal.

Try it yourself

Find the perpendicular bisector of \((1,\,3)\) and \((5,\,7)\).
Answer
\(M(3,5)\), \(m=1\), so gradient \(-1\): \(y=-x+8\)
Check that \((4,\,4)\) on that line is equidistant from both points.
Answer
\(\sqrt{9+1}=\sqrt{10}\) and \(\sqrt{1+9}=\sqrt{10}\) ✓

Finding unknowns: the standard conditions

A right angle at R

\(m_{PR}\times m_{RQ}=-1\), or Pythagoras \(PR^2+RQ^2=PQ^2\), or (circle questions) \(R\) is on the circle with diameter \(PQ\). Any of the three works.

Equal lengths

\(AC=BC\): write both squared distances and equate (no square roots needed). Or: \(C\) is on the perpendicular bisector of \(AB\).

A point on a line

Substitute its coordinates into the line's equation. \(C(2p,\ p)\) is on \(y=\tfrac12x\) for every \(p\): a hint that one unknown is enough.

The other end of a segment

If \(M\) is the midpoint of \(AB\): \(B=2M-A\), i.e. \((2m_x-a_x,\ 2m_y-a_y)\). Same for reflecting a point through a point.

1 Find B from the perpendicular bisector

\(A(6,1)\), \(B(p,q)\); the perpendicular bisector of \(AB\) is \(y=-3x+9\). Find \(p\) and \(q\).

Askedp and qKnowthe bisector has gradient −3, so AB has gradient ⅓Routeline AB through A; meet the bisector at M; double the step from A to M

\(m_{AB}=\tfrac13\): \(y-1=\tfrac13(x-6)\)
Meet \(y=-3x+9\): \(-3x+9=\tfrac13x-1\Rightarrow\tfrac{10}3x=10\Rightarrow x=3\), \(y=0\): \(M(3,0)\)
\(B=2M-A=(6-6,\ 0-1)=(0,-1)\)

B is NOT on the perpendicular bisector, and the meeting point is the midpoint, not B. A sketch prevents both mistakes.

2 Find B from A and the bisector's equation

\(A(-1,7)\); the perpendicular bisector of \(AB\) is \(y=2x+4\). Find \(B\).

AskedAB, then BKnowbisector gradient 2RouteAB gradient −½ through A; meet; double

\(AB\): \(y-7=-\tfrac12(x+1)\), i.e. \(2y=-x+13\)
With \(y=2x+4\): \(2(2x+4)=-x+13\Rightarrow x=1,\ y=6\): \(M(1,6)\)
\(B=(2-(-1),\ 12-7)=(3,5)\)

3 C on the bisector and on a parallel line

\(A(1,2)\), \(B(7,6)\). \(C\) is on the perpendicular bisector of \(AB\) and on the line through \((5,9)\) parallel to \(AB\).

AskedCKnowtwo lines, both through known pointsRoutewrite both lines, solve simultaneously

\(M(4,4)\), \(m_{AB}=\tfrac23\): bisector \(y-4=-\tfrac32(x-4)\), i.e. \(y=-\tfrac32x+10\)
parallel line: \(y-9=\tfrac23(x-5)\)
solve: \(-\tfrac32x+10=\tfrac23x+\tfrac{17}3\Rightarrow\tfrac{13}6x=\tfrac{13}3\Rightarrow x=2\): \(C(2,7)\)

4 Right angle: an unknown coordinate

\(P(-3,1)\), \(Q(7,1)\), \(R(5,k)\); angle \(PRQ=90°\). Find \(k\).

Askedk (two values)KnowPR ⟂ RQRoutegradients in k, product −1, quadratic

\(m_{PR}=\dfrac{k-1}{8}\), \(m_{RQ}=\dfrac{1-k}{2}\)
\(\dfrac{(k-1)(1-k)}{16}=-1\Rightarrow(k-1)^2=16\)
\(k-1=\pm4\): \(k=5\) or \(-3\)

Same answer by the circle route: R is on the circle with diameter PQ, centre \((2,1)\), radius 5: \((5-2)^2+(k-1)^2=25\).

5 Equal lengths: an unknown coordinate

\(A(1,4)\), \(B(7,2)\), \(C(p,2p)\); \(AC=BC\). Find \(p\).

AskedpKnowtwo distances equalRouteequate the squared distances; the squares cancel

\((p-1)^2+(2p-4)^2=(p-7)^2+(2p-2)^2\)
\(5p^2-18p+17=5p^2-22p+53\Rightarrow4p=36\)
\(p=9\): \(C(9,18)\). Check: \(AC^2=64+196=260=BC^2\) ✓

Try it yourself

\(A(2,\,1)\), \(B(6,\,k)\). \(AB\) is perpendicular to \(y=2x\). Find \(k\).
Answer
\(\tfrac{k-1}4=-\tfrac12\Rightarrow k=-1\)
\(M(3,\,-2)\) is the midpoint of \(A(-1,\,4)\) and \(B\). Find \(B\).
Answer
\(B=2M-A=(7,\,-8)\)
\(A(0,\,0)\), \(B(4,\,0)\), \(C(2,\,c)\) with \(AC=BC=\sqrt{13}\). Find \(c\).
Answer
\(4+c^2=13\): \(c=\pm3\)

Key facts at a glance

Perpendicular bisector of \(AB\): through the midpoint, gradient \(-\tfrac1{m_{AB}}\). Every point on it is equidistant from \(A\) and \(B\).
Right angle at \(R\): \(m_{PR}m_{RQ}=-1\), or Pythagoras, or \(R\) on the circle with diameter \(PQ\).
Equal lengths: equate squared distances. A point on a line: substitute it in.
Other end of a segment: \(B=2M-A\).
Sketch every problem first and label what you know.

Part 2Past-paper questions, step by step

How these problems are examined, what the mark scheme wants, then real Paper 1 questions worked through slowly.

How this is examined

A perpendicular bisector is worth about 5 marks: B1 midpoint, B1 *M1 gradient and its negative reciprocal, DM1 A1 the line through the midpoint. The November 2022 report on 9709/12/O/N/22 Q1(a) named the two errors: using \(A\) or \(B\) instead of the midpoint, and forgetting the negative reciprocal. When \(B\) has to be found from the bisector, the June 2021 report saw candidates assume \(B\) lies on the bisector, or take the meeting point as \(B\). Schemes accept all three right-angle methods 9709/13/M/J/25 Q9(a), and "show the centre lies on …" is a bisector question 9709/13/O/N/24 Q10(a).

Mark-scheme focus

Find the equation of the perpendicular bisector of AB

Credited

midpoint \(\left(\tfrac{15}2,\ \tfrac12\right)\) (accept unsimplified) B1
\(m_{AB}=-\tfrac35\Rightarrow\) perpendicular \(\tfrac53\) M1
\(y-\tfrac12=\tfrac53\left(x-\tfrac{15}{2}\right)\), "any correct version", e.g. \(5x-3y=36\) A1

Not credited

a line through \(A\) or \(B\) with gradient \(\tfrac53\)
gradient \(-\tfrac35\) (that is \(AB\) itself)
\(\tfrac{\text{change in }x}{\text{change in }y}\)

From: 9709/12/O/N/22 Q1(a) and the November 2022 report.

Perpendicular bisector: 3-5 marks, often (a) of a circle question.
Unknown coordinate from a right angle / equal lengths: 3-4 marks; the resulting quadratic needs a seen method.
Multi-step coordinate problems (pre-2020 style, still possible): 6-8 marks.

Worked examples

Example 1 · Perpendicular bisector meeting an axis (Paper 1, 5 marks) ★ Easy

Two points A and B have coordinates (1, 3) and (9, −1) respectively. The perpendicular bisector of AB intersects the y-axis at the point C. Find the coordinates of C. [5] — 9709/12/M/J/19 Q2Paper 1 · Pure Mathematics 15 marksFull size

1 Read it with me what is actually happening?

Build the perpendicular bisector (midpoint + perpendicular gradient), then put \(x=0\) to see where it crosses the \(y\)-axis.

perpendicular bisectorthrough the midpoint, at right angles
y-axisx = 0

2 What is it asking for? the target, part by part

coordinates of C5 marks

3 What do we know? everything the question gives us

\(A(1,3)\)
\(B(9,-1)\)

4 Plan the route work backwards from what we want

FindMaths ideaAlready have itWork out firstTrap

FindC
- LineM(5, 1), gradient 2
- Thenx = 0

Gradient of AB?

\(\tfrac{-1-3}{9-1}=-\tfrac12\).

Perpendicular gradient?

5 Now solve it have a go on paper first, then open the worked solution

\(M(5,1)\) B1
\(m_{AB}=-\tfrac12\) B1; perpendicular gradient 2 *M1
\(y-1=2(x-5)\Rightarrow y=2x-9\) DM1
\(C(0,-9)\) A1

6 Check it does the answer make sense?

\(CA^2=1+144=145\), \(CB^2=81+64=145\): C is equidistant from A and B ✓.

Example 2 · The other end of a segment (Paper 1, 4 marks) ★★ Medium

Points A and B have coordinates (8, 3) and (p, q) respectively. The equation of the perpendicular bisector of AB is y = −2x + 4. Find the values of p and q. [4] — 9709/12/M/J/21 Q6Paper 1 · Pure Mathematics 14 marksFull size

1 Read it with me what is actually happening?

We know one end, \(A\), and the bisector. The bisector tells us the gradient of \(AB\) (perpendicular to −2: \(\tfrac12\)) and contains the midpoint. So draw line \(AB\), find where it hits the bisector: that's \(M\). Then \(B\) is as far beyond \(M\) as \(A\) is before it.

perpendicular bisector y = −2x + 4gradient −2; contains the midpoint of AB

2 What is it asking for? the target, part by part

p and q4 marks

3 What do we know? everything the question gives us

\(A(8,3)\)
\(y=-2x+4\)bisector

B is NOT on the bisector; M is

4 Plan the route work backwards from what we want

FindMaths ideaAlready have itWork out firstTrap

FindB
- ABgradient ½ through A
- MAB ∩ bisector
- ThenB = 2M − A

Solve −2x + 4 = ½x − 1.

\(5=\tfrac52x\Rightarrow x=2\), \(y=0\).

From A(8, 3) to M(2, 0) is a step of?

\((-6,-3)\); repeat it: \(B(-4,-3)\).

5 Now solve it have a go on paper first, then open the worked solution

\(m_{AB}=\tfrac12\) B1
\(-2x+4=\tfrac12(x-8)+3\Rightarrow x=2\) *M1
\(\tfrac{8+p}2=2\), \(\tfrac{3+q}2=0\) DM1
\(p=-4,\ q=-3\) A1

6 Check it does the answer make sense?

Midpoint of \((8,3)\) and \((-4,-3)\) is \((2,0)\), on \(y=-2x+4\) ✓; \(m_{AB}=\tfrac{6}{12}=\tfrac12\) ✓.

Example 3 · Equal lengths, then a right angle (Paper 1, 7 marks) ★★★ Hard

The coordinates of points A, B and C are A(5, −2), B(10, 3) and C(2p, p). (a) Given that AC and BC are equal in length, find the value of the fraction p. [3] (b) It is now given instead that AC is perpendicular to BC and that p is an integer. (i) Find the value of p. [4] (ii) Find the equation of the circle through A, B, C. [4] — 9709/13/O/N/22 Q11Paper 1 · Pure Mathematics 1Parts (a)–(b)(i)7 marksFull size

Why it is hard: two different conditions on the same point, a quadratic that must be solved with method shown, and a root to reject.

1 Read it with me what is actually happening?

Two separate conditions on the same unknown point \(C(2p,p)\). (a) equal lengths: equate squared distances. (b) a right angle at \(C\): gradients multiply to −1. "Instead" means (a)'s answer doesn't carry over.

equal in lengthAC² = BC²
perpendicular\(m_{AC}m_{BC}=-1\)
p is an integerreject the fraction root
insteadstart again

2 What is it asking for? the target, part by part

(a)p3
(b)(i)integer p4

3 What do we know? everything the question gives us

\(A(5,-2)\), \(B(10,3)\)
\(C(2p,p)\)

4 Plan the route work backwards from what we want

FindMaths ideaAlready have itWork out firstTrap

(a)p
- Use\((5-2p)^2+(p+2)^2=(10-2p)^2+(3-p)^2\)
(b)p
- Use\(\tfrac{p+2}{2p-5}\cdot\tfrac{p-3}{2p-10}=-1\)
- Trapequating the gradients instead of multiplying to −1

In (a), what happens to the p² terms?

\(5p^2\) on both sides: they cancel, leaving a linear equation.

Factorise 5p² − 31p + 44.

\((p-4)(5p-11)\): \(p=4\) (integer) or \(\tfrac{11}5\).

5 Now solve it have a go on paper first, then open the worked solution

(a) \(25-20p+5p^2+4p+4=100-40p+5p^2-6p+9\) M1 A1 \(\Rightarrow30p=80\Rightarrow p=\tfrac83\) A1
(b)(i) \(m_{AC}=\dfrac{p+2}{2p-5}\), \(m_{BC}=\dfrac{p-3}{2p-10}\) M1; product \(=-1\) M1
\(p^2-p-6=-(4p^2-30p+50)\Rightarrow5p^2-31p+44=0\) A1
\((p-4)(5p-11)=0\Rightarrow p=4\) A1

The November 2022 report: common errors were "to merely let the expressions for the gradients of AC and BC equal each other or to multiply one of the gradient expressions by –1 and then equate"; Pythagoras was a reliable alternative.

6 Check it does the answer make sense?

\(C(8,4)\): \(m_{AC}=\tfrac63=2\), \(m_{BC}=\tfrac{1}{-2}\); product \(-1\) ✓.

Summary

Perpendicular bisector = midpoint + negative reciprocal gradient.
Right angle: m₁m₂ = −1 (or Pythagoras). Equal lengths: equate squared distances.
Midpoint M of AB: B = 2M − A.
Draw a sketch: it stops you confusing B with M.

If you see…	do this
"perpendicular bisector"	midpoint, −1/m, line through the midpoint
"equidistant from A and B"	on the perpendicular bisector of AB
"angle ABC = 90°"	\(m_{AB}m_{BC}=-1\)
"AC = BC"	squared distances equal
"M is the mid-point", find B	B = 2M − A

Can I do this?

I can find the equation of a perpendicular bisector and use it.
I can find an unknown coordinate from a right angle or from equal lengths.
I can find the other end of a segment from a midpoint or a bisector.

Anything unticked: revise that section first. All ticked: take the quiz.

Quiz

The multiple-choice and guided questions need JavaScript. The concept checks and exam practice below work without it.

Concept checks

Answer each one in your head first, then open it.

1Why does the centre of a circle through \(A\) and \(B\) lie on the perpendicular bisector of \(AB\)?Show answer

The centre is the same distance (the radius) from A and from B, and the perpendicular bisector is exactly the set of such points.

2Can you find the circumcentre of a triangle with perpendicular bisectors?Show answer

Yes: it is where any two of the three perpendicular bisectors meet.

3Why is equating squared distances safer than equating distances?Show answer

No square roots to handle, and the equations are equivalent because distances are positive.

Exam-style practice

New questions, not used above. Do them on paper, then mark your work against the official mark scheme.

The point A has coordinates (−2, 6). The equation of the perpendicular bisector of the line AB is 2y = 3x + 5. (i) Find the equation of AB. [3] (ii) Find the coordinates of B. [3] — Q19709/12/M/J/17 Q2Paper 1 · Pure Mathematics 16 marksFull size

Show the mark scheme

Part	Answer / working	Marks	Guidance
(i)	bisector gradient \(\tfrac32\), so \(m_{AB}=-\tfrac23\)	B1
	\(y-6=-\tfrac23(x+2)\)	M1	line through A with their perpendicular gradient
	\(3y+2x=14\) (or any correct form)	A1
(ii)	solve with \(2y=3x+5\): midpoint \((1,4)\)	M1
	\(B=2M-A\)	M1	or vector steps from A to M, twice
	\(B(4,2)\)	A1
Total: 6 marks

The point C lies on the perpendicular bisector of the line joining A(4, 6) and B(10, 2). C also lies on the line parallel to AB through (3, 11). (i) Find the equation of the perpendicular bisector of AB. [4] (ii) Calculate the coordinates of C. [3] — Q29709/12/M/J/15 Q7Paper 1 · Pure Mathematics 17 marksFull size

Show the mark scheme

Part	Answer / working	Marks	Guidance
(i)	midpoint \(M(7,4)\)	B1
	\(m_{AB}=-\tfrac23\)	B1
	perpendicular gradient \(\tfrac32\)	M1	\(m_1m_2=-1\)
	\(y-4=\tfrac32(x-7)\)	A1	through the midpoint
(ii)	\(y-11=-\tfrac23(x-3)\)	M1	the parallel line
	solve the two equations together	DM1
	\(C(9,7)\)	A1
Total: 7 marks

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