Why is "Confusing a 'line of invariant points' with an 'invariant line' and using the wrong method" a common mistake in Invariant Points and Lines?

Why it happens: The phrases sound almost identical, so students apply (M-I)p=0 when the question wants the y=mx+c substitution, or vice versa. How to avoid it: Line of invariant points = every point fixed (use (M-I)p=0). Invariant line = line maps to itself, points may slide (use the y=mx+c method). Match the method to the command word. Source: 9231 Paper 1 Examiner Reports — recurring comment

Why is "Forgetting that the origin is always an invariant point" a common mistake in Invariant Points and Lines?

Why it happens: Students hunt for 'interesting' fixed points and overlook the trivial solution p=0, especially when (M-I)≠0. How to avoid it: State that M0=0 so the origin is always fixed. When (M-I)≠0, the origin is the ONLY invariant point — say so explicitly. Source: 9231 Paper 1 mark schemes

Why is "Not treating $y'=mx'+c$ as an identity true for ALL $x$" a common mistake in Invariant Points and Lines?

Why it happens: Students substitute a single point or solve for one x instead of comparing coefficients, getting a spurious or incomplete answer. How to avoid it: Keep x as a free variable. Equate the coefficients of x on both sides AND equate the constant terms — two separate equations for m and c. Source: 9231 Paper 1 Examiner Reports

Why is "Solving only the gradient equation for $m$ and ignoring the constant equation for $c$" a common mistake in Invariant Points and Lines?

Why it happens: After finding m from the x-coefficients, students assume any c works and claim infinitely many parallel invariant lines. How to avoid it: The constant terms give a separate equation (e.g. c(1-m)=0). Solve it for each value of m — usually it forces c=0 unless m is special. Source: 9231 Paper 1 Examiner Reports

Why is "Trying to solve $(\mathbf{M}-\mathbf{I})\mathbf{p}=\mathbf{0}$ without checking the determinant first" a common mistake in Invariant Points and Lines?

Why it happens: Students grind through the two equations and either miss that they coincide (a line) or claim a line when only the origin exists. How to avoid it: Compute (M-I) first: non-zero origin only; zero the single equation IS the line of invariant points. Source: 9231 Paper 1 mark schemes

Why is "Quoting an invariant line without verifying it maps onto itself" a common mistake in Invariant Points and Lines?

Why it happens: Time pressure and over-confidence; an arithmetic slip in forming x' or y' goes unnoticed. How to avoid it: Take one concrete point on your line, transform it, and confirm the image satisfies the same line equation. A two-line check protects the accuracy marks. Source: 9231 Paper 1 Examiner Reports

Cambridge International A Levels Further Mathematics (9231)

Invariant Points and Lines

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Detailed Study Notes

Detailed notes on Matrices (paper 1) for Cambridge International A Levels Further Mathematics, covering key concepts, explanations, examples, and exam-focused revision points.

Invariant Points and Invariant Lines — Cambridge International AS & A Level Further Mathematics 9231 Paper 1 (2026-2027 syllabus)

A matrix transformation can leave some points exactly where they are (invariant points) and can map whole lines onto themselves (invariant lines). Learn the two-letter test $(\mathbf{M}-\mathbf{I})\mathbf{p}=\mathbf{0}$ for fixed points, and the substitution method that finds invariant lines $y=mx+c$ .

At a glance

An invariant (fixed) point satisfies $\mathbf{M}\mathbf{p}=\mathbf{p}$ , i.e. $(\mathbf{M}-\mathbf{I})\mathbf{p}=\mathbf{0}$ .
The origin is always invariant for a linear transformation, because $\mathbf{M}\mathbf{0}=\mathbf{0}$ .
Non-trivial fixed points exist only when $\det(\mathbf{M}-\mathbf{I})=0$ — and then there is a whole line of invariant points.
A line of invariant points (every point fixed) is NOT the same as an invariant line (the line maps onto itself, but points may slide along it).
Find invariant lines by putting $(x,\,mx+c)$ through $\mathbf{M}$ and forcing the image back onto $y=mx+c$ .

What you’ll learn

Mapped to the Cambridge International A Level 9231 syllabus (2026-2027).

1.4a — Understand the meaning of an invariant point and of an invariant line of a transformation represented by a $2\times2$ matrix.
1.4b — Find the invariant points and any line of invariant points of a given transformation using $(\mathbf{M}-\mathbf{I})\mathbf{p}=\mathbf{0}$ .
1.4c — Find the invariant line(s) $y=mx+c$ of a given transformation by mapping a general point and forcing the image back onto the line.

Invariant points — the equation (M − I)p = 0

A fixed point goes nowhere under the transformation; solve (M − I)p = 0 to find them all.

What "invariant point" means. A point $P$ with position vector $\mathbf{p}=\begin{pmatrix}x\\y\end{pmatrix}$ is invariant (or fixed) under the transformation with matrix $\mathbf{M}$ if the transformation leaves it exactly where it is:

$\mathbf{M}\mathbf{p}=\mathbf{p}$

Turn it into something you can solve. Subtract $\mathbf{p}$ from both sides and write $\mathbf{p}=\mathbf{I}\mathbf{p}$ (where $\mathbf{I}$ is the $2\times2$ identity):

$\mathbf{M}\mathbf{p}-\mathbf{I}\mathbf{p}=\mathbf{0}\quad\Longrightarrow\quad(\mathbf{M}-\mathbf{I})\mathbf{p}=\mathbf{0}$

This is a homogeneous pair of simultaneous equations in $x$ and $y$ .

The origin is always a solution. Putting $\mathbf{p}=\mathbf{0}$ gives $\mathbf{0}=\mathbf{0}$ , true for every $\mathbf{M}$ . So every linear (matrix) transformation fixes the origin — reflections in lines through $O$ , rotations about $O$ , enlargements centred at $O$ and shears all leave $O$ untouched. The origin is rarely the interesting answer; examiners want the other fixed points.

When are there more fixed points? A homogeneous system $(\mathbf{M}-\mathbf{I})\mathbf{p}=\mathbf{0}$ has solutions other than $\mathbf{0}$ only when the matrix $\mathbf{M}-\mathbf{I}$ is singular, i.e.

$\det(\mathbf{M}-\mathbf{I})=0.$

If $\det(\mathbf{M}-\mathbf{I})\neq0$ the only invariant point is the origin.
If $\det(\mathbf{M}-\mathbf{I})=0$ the two equations are multiples of one another, leaving a single line of solutions — a whole line of invariant points.

Cambridge tip. Always test $\det(\mathbf{M}-\mathbf{I})$ first. If it is non-zero, write "the only invariant point is $(0,0)$ " and stop — hunting for more wastes marks and time.

Fixed point $\Leftrightarrow(\mathbf{M}-\mathbf{I})\mathbf{p}=\mathbf{0}$ .
The origin is invariant for every matrix transformation.
Extra fixed points exist iff $\det(\mathbf{M}-\mathbf{I})=0$ .

See the full worked example for invariant points and lines →

Line of invariant points vs invariant line — the key distinction

Every point fixed (line of invariant points) versus the line as a whole fixed (invariant line).

These two phrases sound alike but mean different things, and examiners deliberately test whether you can tell them apart.

A line of invariant points. Every single point on the line stays exactly where it is. Nothing moves. A reflection has its mirror line as a line of invariant points: each point on the mirror reflects to itself.

An invariant line. The line as a whole maps onto itself — but the individual points are allowed to slide along the line. A point may move; it just lands somewhere else on the same line. For a reflection, any line perpendicular to the mirror is an invariant line: a point on it reflects to a different point, but still on that same perpendicular line.

The logical relationship. Every line of invariant points is an invariant line (if every point stays put, the line certainly maps to itself). But the converse fails: an invariant line need not be a line of invariant points.

Left: a line of invariant points — P maps to itself. Right: an invariant line — P moves to P′, but P′ is still on the same line.

Cambridge tip. Read the command word carefully. "Find the invariant point(s)" wants $(\mathbf{M}-\mathbf{I})\mathbf{p}=\mathbf{0}$ . "Find the invariant line(s)" wants the $y=mx+c$ substitution method below. Mixing the two is the single biggest mark-loser in this subtopic.

Line of invariant points: every point is fixed (e.g. a mirror line).
Invariant line: the line maps to itself but points may move along it.
Every line of invariant points is an invariant line; not vice versa.

See the full worked example for invariant points and lines →

Finding invariant lines y = mx + c

Map a general point on the line, then force the image back onto the same line; solve for m and c.

To find the invariant lines of a transformation $\mathbf{M}=\begin{pmatrix}a&b\\c&d\end{pmatrix}$ , take a general point on a candidate line $y=mx+c$ and demand that its image also lies on that same line.

The method, step by step.

Step 1 — a general point. Write the point as $(x,\;mx+c)$ — any point on $y=mx+c$ .

Step 2 — apply the matrix. Compute the image $\begin{pmatrix}x'\\y'\end{pmatrix}=\mathbf{M}\begin{pmatrix}x\\mx+c\end{pmatrix}$ . You get $x'$ and $y'$ each as an expression in $x$ (with $m$ and $c$ as parameters).

Step 3 — force the image back onto the line. The image must satisfy the same equation, so set

$y'=m\,x'+c.$

Step 4 — make it hold for every point on the line. Both sides are now linear in $x$ . For the equation to be true for all $x$ , the coefficients of $x$ must match and the constant terms must match. This gives:

a relation that determines $m$ (often a quadratic in $m$ ),
a relation that determines $c$ (for each valid $m$ ).

Lines through the origin. If the question asks only for invariant lines through $O$ , set $c=0$ from the start. Then $y=mx$ and you solve a single equation for $m$ — usually a quadratic, so up to two such lines.

Always verify. Pick the matrix, the value of $m$ (and $c$ ), take any concrete point on the line, transform it, and check the image is back on the line. A two-line check at the end catches arithmetic slips that the mark scheme punishes.

The image P′ of any point P on an invariant line lands back on the same line — that requirement (y′ = mx′ + c) is what fixes m and c.

Cambridge tip. Keep $x$ as a free variable throughout. The condition "true for all $x$ " — comparing coefficients of $x$ and the constants separately — is exactly what separates an invariant line from a line that just happens to contain one fixed point.

Put $(x,\,mx+c)$ through $\mathbf{M}$ , then set $y'=mx'+c$ .
Compare coefficients of $x$ and the constants separately (true for all $x$ ).
Through the origin: set $c=0$ and solve the quadratic in $m$ .

See the full worked example for invariant points and lines →

Quick recap

Invariant (fixed) point: $\mathbf{M}\mathbf{p}=\mathbf{p}$ , i.e. $(\mathbf{M}-\mathbf{I})\mathbf{p}=\mathbf{0}$ .
The origin is invariant for every matrix transformation.
Non-trivial fixed points (a line of them) exist iff $\det(\mathbf{M}-\mathbf{I})=0$ .
Line of invariant points = every point fixed; invariant line = the line maps to itself (points may slide).
Find invariant lines: map $(x,mx+c)$ , set $y'=mx'+c$ , compare coefficients of $x$ and constants.
Always verify the resulting line maps onto itself with one concrete point.

Memorise this

Verbatim phrases and definitions Cambridge mark schemes credit.

Invariant point: $(\mathbf{M}-\mathbf{I})\mathbf{p}=\mathbf{0}$ (the origin is always a solution).
Condition for more than the origin: $\det(\mathbf{M}-\mathbf{I})=0$ gives a line of invariant points.
Invariant-line method: apply $\mathbf{M}$ to $(x,\,mx+c)$ , then force the image onto $y'=mx'+c$ .
"True for all $x$ ": equate coefficients of $x$ and the constant terms separately to solve for $m$ and $c$ .
Line of invariant points $\neq$ invariant line — read the command word.

How it’s examined

Topic 1.4 appears in Paper 1 (typically 5-8 marks), often tagged onto a question that first asks for a transformation matrix. Examiners deliberately probe the distinction between a 'line of invariant points' and an 'invariant line' — many candidates conflate the two and apply the wrong method. A part asking for invariant points wants $(\mathbf{M}-\mathbf{I})\mathbf{p}=\mathbf{0}$ (and recognition that the origin is always fixed); a part asking for invariant lines wants the substitution of $(x, mx+c)$ followed by 'true for all $x$ ' coefficient-matching. Marks are lost through (i) forgetting the origin, (ii) using one method when the other is required, and (iii) failing to treat the line equation as an identity in $x$ . Showing the verification step protects the accuracy marks.

Sources: Cambridge International AS & A Level Further Mathematics 9231 syllabus (2026-2027); 9231 Examiner Reports 2022-2024 (Paper 1); 9231/11, 9231/12, 9231/13 Further Pure Mathematics 1 — past papers and mark schemes. Last reviewed 2026-06-23.

Take this whole topic with you

Step-by-step worked examples — Invariant Points and Lines

Step-by-step solutions to past-paper-style questions on invariant points and lines, written exactly the way a tutor would explain them at the board.

Question type:

1Confirming the origin is invariant
Getting startedDirect calculation• invariant-point, origin
Question
The transformation $T$ has matrix $\mathbf{M}=\begin{pmatrix}3&-2\\1&4\end{pmatrix}$ . Show that the origin $(0,0)$ is an invariant point of $T$ .
Step-by-step solution
1. Step 1
  An invariant point $\mathbf{p}$ satisfies $\mathbf{M}\mathbf{p}=\mathbf{p}$ . Test $\mathbf{p}=\begin{pmatrix}0\\0\end{pmatrix}$ .
2. Step 2
  Multiply the matrix by the zero vector.
  $\mathbf{M}\begin{pmatrix}0\\0\end{pmatrix}=\begin{pmatrix}3(0)-2(0)\\1(0)+4(0)\end{pmatrix}=\begin{pmatrix}0\\0\end{pmatrix}$
3. Step 3
  Compare with $\mathbf{p}$ : the image equals the original, so $(0,0)$ is fixed.
  $\mathbf{M}\begin{pmatrix}0\\0\end{pmatrix}=\begin{pmatrix}0\\0\end{pmatrix}=\mathbf{p}$
Answer
$\mathbf{M}\begin{pmatrix}0\\0\end{pmatrix}=\begin{pmatrix}0\\0\end{pmatrix}$ , so the origin is invariant. (This holds for every matrix transformation.)
Examiner tip
Any linear transformation maps $\mathbf{0}$ to $\mathbf{0}$ , so the origin is always invariant. Stating this fact alone earns the mark.
2Line of invariant points vs invariant line for a reflection
Getting startedIdentify & classify• reflection, distinction
Question
A reflection $T$ has mirror line $y=x$ . Describe (a) a line of invariant points of $T$ and (b) an invariant line of $T$ that is NOT a line of invariant points.
Step-by-step solution
1. Step 1
  Line of invariant points = every point on it is fixed. For a reflection, each point of the mirror line reflects to itself.
  $\text{Line of invariant points: } y=x$
2. Step 2
  Invariant line = the line maps onto itself, though points may move along it. Any line perpendicular to the mirror reflects onto itself.
  $\text{Invariant line: any } y=-x+c$
3. Step 3
  Check (b) is not a line of invariant points. Take $y=-x+2$ , i.e. point $(0,2)$ . Its reflection in $y=x$ is $(2,0)$ , which is still on $y=-x+2$ but is a different point — so points slide along the line.
Answer
(a) The mirror line $y=x$ is a line of invariant points. (b) Any perpendicular line $y=-x+c$ is an invariant line on which points move (e.g. $(0,2)\mapsto(2,0)$ on $y=-x+2$ ).
Examiner tip
The discriminating idea: on a line of invariant points nothing moves; on an invariant line the line is fixed but the points generally are not.
3When the origin is the only invariant point
Building confidenceMulti-step problem• invariant-point, determinant
Question
Find all invariant points of the transformation with matrix $\mathbf{M}=\begin{pmatrix}2&0\\0&3\end{pmatrix}$ .
Step-by-step solution
1. Step 1
  Form $\mathbf{M}-\mathbf{I}$ by subtracting $1$ from each diagonal entry.
  $\mathbf{M}-\mathbf{I}=\begin{pmatrix}2-1&0\\0&3-1\end{pmatrix}=\begin{pmatrix}1&0\\0&2\end{pmatrix}$
2. Step 2
  Test the determinant to decide whether non-trivial fixed points exist.
  $\det(\mathbf{M}-\mathbf{I})=(1)(2)-(0)(0)=2\neq0$
3. Step 3
  Non-zero determinant $\Rightarrow$ the only solution of $(\mathbf{M}-\mathbf{I})\mathbf{p}=\mathbf{0}$ is $\mathbf{p}=\mathbf{0}$ .
  $\begin{pmatrix}1&0\\0&2\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}0\\0\end{pmatrix}\Rightarrow x=0,\;y=0$
Answer
The only invariant point is the origin $(0,0)$ .
Examiner tip
Computing $\det(\mathbf{M}-\mathbf{I})$ first is the efficient route: a non-zero value immediately tells you the origin is the sole fixed point.
4Finding a line of invariant points
Building confidenceMulti-step problem• line-of-invariant-points, determinant
Question
The matrix $\mathbf{M}=\begin{pmatrix}0&1\\1&0\end{pmatrix}$ represents a reflection. Find the line of invariant points of this transformation.
Step-by-step solution
1. Step 1
  Form $\mathbf{M}-\mathbf{I}$ and check its determinant.
  $\mathbf{M}-\mathbf{I}=\begin{pmatrix}-1&1\\1&-1\end{pmatrix},\quad\det=(-1)(-1)-(1)(1)=0$
2. Step 2
  Determinant zero $\Rightarrow$ a whole line of invariant points exists. Write out $(\mathbf{M}-\mathbf{I})\mathbf{p}=\mathbf{0}$ .
  $\begin{pmatrix}-1&1\\1&-1\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}0\\0\end{pmatrix}$
3. Step 3
  Both rows give the same equation. Row 1: $-x+y=0$ . Row 2: $x-y=0$ . These are identical, so the solution is the line $y=x$ .
  $-x+y=0\;\Longrightarrow\;y=x$
4. Step 4
  Verify. Take $(3,3)$ on $y=x$ : $\mathbf{M}\begin{pmatrix}3\\3\end{pmatrix}=\begin{pmatrix}3\\3\end{pmatrix}$ — fixed. ✓
Answer
The line of invariant points is $y=x$ (the mirror line of the reflection).
Examiner tip
Because $\det(\mathbf{M}-\mathbf{I})=0$ , the two equations collapse to one — that single equation IS the line of invariant points. Quote it as $y=x$ .
5Finding invariant lines y = mx + c (full method)
StretchMulti-step problem• invariant-line, substitution-method
Question
Find all invariant lines $y=mx+c$ of the transformation with matrix $\mathbf{M}=\begin{pmatrix}2&1\\0&1\end{pmatrix}$ .
Step-by-step solution
1. Step 1
  Take a general point $(x,\;mx+c)$ on the line and apply $\mathbf{M}$ to find its image $(x',y')$ .
  $\begin{pmatrix}x'\\y'\end{pmatrix}=\begin{pmatrix}2&1\\0&1\end{pmatrix}\begin{pmatrix}x\\mx+c\end{pmatrix}=\begin{pmatrix}2x+(mx+c)\\mx+c\end{pmatrix}$
2. Step 2
  Write $x'$ and $y'$ explicitly in terms of $x$ (with $m,c$ as parameters).
  $x'=(2+m)x+c,\qquad y'=mx+c$
3. Step 3
  Force the image back onto the line: $y'=m\,x'+c$ .
  $mx+c=m\big[(2+m)x+c\big]+c=m(2+m)x+mc+c$
4. Step 4
  True for all $x$ $\Rightarrow$ match coefficients of $x$ and the constant terms separately.
  $x:\;\; m=m(2+m)\qquad\text{const:}\;\; c=mc+c$
5. Step 5
  Solve the $x$ -coefficient equation. $m=2m+m^2\Rightarrow m^2+m=0\Rightarrow m(m+1)=0$ , so $m=0$ or $m=-1$ .
  $m(m+1)=0\;\Rightarrow\;m=0\ \text{or}\ m=-1$
6. Step 6
  Solve the constant equation $c=mc+c$ , i.e. $mc=0$ , for each $m$ . If $m=0$ : $0=0$ , so $c$ is free — every line $y=c$ works. If $m=-1$ : $-c=0\Rightarrow c=0$ , giving $y=-x$ .
  $m=0\Rightarrow c\text{ free }(y=c);\qquad m=-1\Rightarrow c=0\;(y=-x)$
7. Step 7
  Verify $y=-x$ . Take $(1,-1)$ : image $x'=(2-1)(1)+0=1$ , $y'=-1$ , so $(1,-1)\mapsto(1,-1)$ — on $y=-x$ . ✓ Verify $y=c$ : take $(0,c)$ , image $x'=c$ , $y'=c$ , on $y=c$ . ✓
Answer
The invariant lines are every horizontal line $y=c$ (the family with $m=0$ ) and the line $y=-x$ .
Examiner tip
Two separate equations come from 'true for all $x$ ': the $x$ -coefficient gives a quadratic in $m$ ; the constant gives the allowed $c$ . Note $y=c$ is invariant for all $c$ but is NOT a line of invariant points — points slide horizontally.
6Invariant lines through the origin (c = 0)
StretchMulti-step problem• invariant-line, through-origin
Question
Find the invariant lines through the origin of the transformation with matrix $\mathbf{M}=\begin{pmatrix}4&1\\2&3\end{pmatrix}$ .
Step-by-step solution
1. Step 1
  Lines through $O$ have the form $y=mx$ (set $c=0$ ). Take a general point $(x,\,mx)$ and apply $\mathbf{M}$ .
  $\begin{pmatrix}x'\\y'\end{pmatrix}=\begin{pmatrix}4&1\\2&3\end{pmatrix}\begin{pmatrix}x\\mx\end{pmatrix}=\begin{pmatrix}(4+m)x\\(2+3m)x\end{pmatrix}$
2. Step 2
  Force the image onto $y=mx$ : require $y'=m\,x'$ .
  $(2+3m)x=m(4+m)x$
3. Step 3
  True for all $x$ $\Rightarrow$ cancel $x$ and rearrange into a quadratic in $m$ .
  $2+3m=4m+m^2\;\Rightarrow\;m^2+m-2=0$
4. Step 4
  Factorise and solve the quadratic.
  $(m+2)(m-1)=0\;\Rightarrow\;m=-2\ \text{or}\ m=1$
5. Step 5
  Verify $y=x$ (m=1). Take $(1,1)$ : image $x'=4+1=5$ , $y'=2+3=5$ , so $(1,1)\mapsto(5,5)$ — on $y=x$ . ✓ Verify $y=-2x$ (m=-2). Take $(1,-2)$ : $x'=4-2=2$ , $y'=2-6=-4$ , so $(1,-2)\mapsto(2,-4)$ — on $y=-2x$ . ✓
Answer
The invariant lines through the origin are $y=x$ and $y=-2x$ .
Examiner tip
Setting $c=0$ reduces the problem to one quadratic in $m$ , giving up to two lines. These are the eigendirections of $\mathbf{M}$ — points on them are scaled, hence stay on the line.

Model Answers — Invariant Points and Lines

High-scoring sample answers for invariant points and lines on the Cambridge IGCSE paper, with examiner-style notes mapping each response to the mark scheme and assessment objectives.

Question 1
4 marks
The transformation $T$ is represented by $\mathbf{M}=\begin{pmatrix}5&-1\\2&2\end{pmatrix}$ . Determine all invariant points of $T$ .
Model answer
Invariant points satisfy $(\mathbf{M}-\mathbf{I})\mathbf{p}=\mathbf{0}$ : $\mathbf{M}-\mathbf{I}=\begin{pmatrix}4&-1\\2&1\end{pmatrix}$

Check the determinant: $\det(\mathbf{M}-\mathbf{I})=(4)(1)-(-1)(2)=4+2=6\neq0$

Since $\det(\mathbf{M}-\mathbf{I})\neq0$ , the matrix $\mathbf{M}-\mathbf{I}$ is non-singular, so the only solution of $(\mathbf{M}-\mathbf{I})\mathbf{p}=\mathbf{0}$ is $\mathbf{p}=\mathbf{0}$ .

Conclusion: the only invariant point is the origin $(0,0)$ .
Why this scores
M1 for forming $\mathbf{M}-\mathbf{I}$ , M1 for $\det(\mathbf{M}-\mathbf{I})$ , A1 for value $6$ , A1 for concluding the origin is the unique invariant point. A bare answer of $(0,0)$ without the determinant argument loses the reasoning marks.
Question 2
4 marks
Find the line of invariant points of the transformation with matrix $\mathbf{M}=\begin{pmatrix}3&-2\\1&0\end{pmatrix}$ .
Model answer
Set up $(\mathbf{M}-\mathbf{I})\mathbf{p}=\mathbf{0}$ : $\mathbf{M}-\mathbf{I}=\begin{pmatrix}2&-2\\1&-1\end{pmatrix},\qquad\det=(2)(-1)-(-2)(1)=-2+2=0$

Because the determinant is $0$ , a line of invariant points exists. Writing out the system: $\begin{pmatrix}2&-2\\1&-1\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}0\\0\end{pmatrix}$

Row 1: $2x-2y=0\Rightarrow y=x$ . Row 2: $x-y=0\Rightarrow y=x$ . Both rows give the same equation.

Verify: take $(2,2)$ : $\mathbf{M}\begin{pmatrix}2\\2\end{pmatrix}=\begin{pmatrix}6-4\\2\end{pmatrix}=\begin{pmatrix}2\\2\end{pmatrix}$ — fixed. ✓

Line of invariant points: $y=x$ .
Why this scores
M1 for $\mathbf{M}-\mathbf{I}$ and showing $\det=0$ , M1 for reducing the system to a single equation, A1 for $y=x$ , A1 for a correct verification (or fully convincing reasoning). The fact that both rows coincide is the hallmark of a line of invariant points.
Question 3
6 marks
The transformation $T$ has matrix $\mathbf{M}=\begin{pmatrix}1&2\\3&2\end{pmatrix}$ . Find the equations of the two invariant lines of $T$ that pass through the origin.
Model answer
Lines through the origin have the form $y=mx$ . Take a general point $(x,\,mx)$ and apply $\mathbf{M}$ : $\begin{pmatrix}x'\\y'\end{pmatrix}=\begin{pmatrix}1&2\\3&2\end{pmatrix}\begin{pmatrix}x\\mx\end{pmatrix}=\begin{pmatrix}(1+2m)x\\(3+2m)x\end{pmatrix}$

For the image to lie on $y=mx$ we need $y'=m\,x'$ for all $x$ : $(3+2m)x=m(1+2m)x\;\Rightarrow\;3+2m=m+2m^2$

Rearranging: $2m^2-m-3=0\;\Rightarrow\;(2m-3)(m+1)=0\;\Rightarrow\;m=\tfrac{3}{2}\ \text{or}\ m=-1$

Verify $y=-x$ : $(1,-1)\mapsto\big(1-2,\,3-2\big)=(-1,1)$ , on $y=-x$ . ✓ Verify $y=\tfrac32 x$ : $(2,3)\mapsto\big(2+6,\,6+6\big)=(8,12)$ , and $12=\tfrac32(8)$ . ✓

Invariant lines: $y=\dfrac{3}{2}x$ and $y=-x$ .
Why this scores
M1 for image of $(x,mx)$ , M1 for imposing $y'=mx'$ , M1 A1 for the quadratic $2m^2-m-3=0$ correctly formed and solved, A1 A1 for the two lines. Treating the equation as 'true for all $x$ ' (cancelling $x$ ) is essential.
Question 4
5 marks
A transformation $T$ is a reflection in the line $y=2x$ . (a) State the line of invariant points of $T$ . (b) Explain why the line $y=-\tfrac12 x$ is an invariant line of $T$ but is not a line of invariant points.
Model answer
(a) Under a reflection, every point on the mirror line maps to itself. Hence the line of invariant points is the mirror line $y=2x$ .

(b) The line $y=-\tfrac12 x$ is perpendicular to the mirror $y=2x$ (gradients $2$ and $-\tfrac12$ multiply to $-1$ ). Reflecting any point of this perpendicular line gives another point of the same perpendicular line, so the line as a whole maps onto itself — it is an invariant line.

However, the individual points are not fixed. For example $(2,-1)$ lies on $y=-\tfrac12 x$ ; its reflection in $y=2x$ is $(-2,1)$ , a different point (still on $y=-\tfrac12 x$ ). Because points move along the line, it is not a line of invariant points.
Why this scores
(a) B1 for $y=2x$ . (b) B1 for stating perpendicularity, M1 for explaining the line maps to itself, A1 for showing a specific point moves, A1 for the explicit distinction. Examiners reward a concrete moving point as evidence the line is not pointwise fixed.
Question 5
8 marks
The transformation $T$ has matrix $\mathbf{M}=\begin{pmatrix}3&1\\0&2\end{pmatrix}$ . Find ALL invariant lines of $T$ (including any not through the origin).
Model answer
Take a general point $(x,\,mx+c)$ on the line $y=mx+c$ and apply $\mathbf{M}$ : $\begin{pmatrix}x'\\y'\end{pmatrix}=\begin{pmatrix}3&1\\0&2\end{pmatrix}\begin{pmatrix}x\\mx+c\end{pmatrix}=\begin{pmatrix}3x+(mx+c)\\2(mx+c)\end{pmatrix}$

So $x'=(3+m)x+c$ and $y'=2mx+2c$ .

Force the image onto the line: $y'=m\,x'+c$ for all $x$ : $2mx+2c=m\big[(3+m)x+c\big]+c=m(3+m)x+mc+c$

Match coefficients of $x$ : $2m=m(3+m)\;\Rightarrow\;m^2+m=0\;\Rightarrow\;m(m+1)=0\;\Rightarrow\;m=0\ \text{or}\ m=-1$

Match constants: $2c=mc+c\Rightarrow c=mc\Rightarrow c(1-m)=0$ .

If $m=0$ : $c(1-0)=0\Rightarrow c=0$ , giving the line $y=0$ (the $x$ -axis).

If $m=-1$ : $c(1-(-1))=2c=0\Rightarrow c=0$ , giving the line $y=-x$ .

Verify $y=0$ : $(1,0)\mapsto(3,0)$ , on $y=0$ . ✓ Verify $y=-x$ : $(1,-1)\mapsto(3-1,\,-2)=(2,-2)$ , on $y=-x$ . ✓

Invariant lines: $y=0$ and $y=-x$ .
Why this scores
M1 image of $(x,mx+c)$ ; M1 imposing $y'=mx'+c$ ; M1 A1 for the $x$ -coefficient quadratic $m(m+1)=0$ ; M1 for the constant equation $c(1-m)=0$ ; A1 A1 for the two lines $y=0$ and $y=-x$ . Both equations (coefficients AND constants) must be used — many candidates solve only for $m$ and wrongly claim infinitely many parallel lines.
Question 6
8 marks
The transformation $T$ has matrix $\mathbf{M}=\begin{pmatrix}2&-1\\-1&2\end{pmatrix}$ . (a) Find the line of invariant points of $T$ . (b) Find the other invariant line of $T$ through the origin.
Model answer
(a) Invariant points: $(\mathbf{M}-\mathbf{I})\mathbf{p}=\mathbf{0}$ . $\mathbf{M}-\mathbf{I}=\begin{pmatrix}1&-1\\-1&1\end{pmatrix},\qquad\det=(1)(1)-(-1)(-1)=1-1=0$

Determinant $0$ , so a line of invariant points exists. Row 1: $x-y=0\Rightarrow y=x$ (row 2 gives the same). Verify: $(1,1)\mapsto(2-1,\,-1+2)=(1,1)$ , fixed. ✓ So the line of invariant points is $\boxed{y=x}$ .

(b) Other invariant line through $O$ : put $(x,mx)$ through $\mathbf{M}$ with $y'=mx'$ . $x'=2x-mx=(2-m)x,\qquad y'=-x+2mx=(-1+2m)x$ $y'=mx'\;\Rightarrow\;(-1+2m)x=m(2-m)x\;\Rightarrow\;-1+2m=2m-m^2$ $\Rightarrow\;m^2=1\;\Rightarrow\;m=\pm1$

$m=1$ gives $y=x$ (already found, the line of invariant points). The other line is $m=-1$ , i.e. $y=-x$ . Verify $y=-x$ : $(1,-1)\mapsto(2+1,\,-1-2)=(3,-3)$ , on $y=-x$ . ✓
Why this scores
(a) M1 $\det(\mathbf{M}-\mathbf{I})=0$ , A1 $y=x$ , B1 verification. (b) M1 image of $(x,mx)$ , M1 imposing $y'=mx'$ , A1 $m^2=1$ , A1 selecting the new line $y=-x$ . Recognising that $m=1$ recovers the line of invariant points (so the 'other' line is $y=-x$ ) is the discriminating step.

Key Formulae — Invariant Points and Lines

The formulae you need to memorise for invariant points and lines on the Cambridge IGCSE paper, with every variable defined in plain English and a note on when to use it.

Invariant (fixed) point condition
$\mathbf{M}\mathbf{p}=\mathbf{p}\quad\Longleftrightarrow\quad(\mathbf{M}-\mathbf{I})\mathbf{p}=\mathbf{0}$
$\mathbf{M}$
the $2\times2$ transformation matrix
$\mathbf{p}=\begin{pmatrix}x\\y\end{pmatrix}$
position vector of the point
$\mathbf{I}$
the $2\times2$ identity matrix
When to use
To locate every point fixed by the transformation. The origin $\mathbf{p}=\mathbf{0}$ is always a solution.
Condition for non-trivial invariant points
$\det(\mathbf{M}-\mathbf{I})=0$
$\mathbf{M}-\mathbf{I}$
the matrix with $1$ subtracted from each diagonal entry
When to use
If $\det(\mathbf{M}-\mathbf{I})\neq0$ the only invariant point is the origin; if $=0$ there is a whole line of invariant points.
Invariant-line method (line y = mx + c)
$\text{Map }(x,\,mx+c)\to(x',y');\quad\text{require } y'=m\,x'+c \text{ for all } x$
$m$
gradient of the candidate line
$c$
intercept of the candidate line
$(x',y')$
image of the general point under $\mathbf{M}$
When to use
To find invariant lines. 'True for all $x$ ' means equating coefficients of $x$ and the constant terms separately.
Invariant lines through the origin (c = 0)
$y=mx:\quad y'=m\,x'\ \text{ gives a quadratic in } m$
$m$
gradient of an invariant line through $O$
When to use
When only lines through the origin are required. Setting $c=0$ leaves one quadratic in $m$ , yielding up to two lines.
Image of a point under M
$\begin{pmatrix}x'\\y'\end{pmatrix}=\begin{pmatrix}a&b\\c&d\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}ax+by\\cx+dy\end{pmatrix}$
$a,b,c,d$
entries of the transformation matrix
$(x,y)$
the original point
When to use
The basic multiplication used to transform any point — the first step of every invariant-line calculation.

Key Definitions and Keywords — Invariant Points and Lines

Definitions to memorise and the exact keywords mark schemes credit for invariant points and lines answers — sharpened from recent examiner reports for the 2026 Cambridge IGCSE sitting.

Invariant point (fixed point)
Examiner keyword
A point $P$ that the transformation leaves exactly where it is: $\mathbf{M}\mathbf{p}=\mathbf{p}$ , equivalently $(\mathbf{M}-\mathbf{I})\mathbf{p}=\mathbf{0}$ .
Line of invariant points
Examiner keyword
A line on which EVERY point is invariant — each point maps to itself. Arises when $\det(\mathbf{M}-\mathbf{I})=0$ ; the equation $(\mathbf{M}-\mathbf{I})\mathbf{p}=\mathbf{0}$ then reduces to a single line equation.
Invariant line
Examiner keyword
A line that the transformation maps onto itself as a whole. Individual points may move ALONG the line; the set of points on the line is unchanged.
Invariance of the origin
Examiner keyword
Every linear (matrix) transformation fixes the origin, since $\mathbf{M}\mathbf{0}=\mathbf{0}$ . The origin is therefore an invariant point of every such transformation.
Singular matrix
A matrix with zero determinant, hence no inverse. $(\mathbf{M}-\mathbf{I})$ being singular is exactly the condition for a line of invariant points to exist.
True for all $x$ (identity condition)
Examiner keyword
When two linear expressions in $x$ must be equal for every value of $x$ , their $x$ -coefficients must be equal AND their constant terms must be equal. This is how the invariant-line method splits into equations for $m$ and $c$ .

Common Mistakes and Misconceptions — Invariant Points and Lines

The traps other students keep falling into on invariant points and lines questions — taken from recent Cambridge IGCSE examiner reports and mark schemes — and how to avoid them.

✕Confusing a 'line of invariant points' with an 'invariant line' and using the wrong method
9231 Paper 1 Examiner Reports — recurring comment
Why it happens
The phrases sound almost identical, so students apply $(\mathbf{M}-\mathbf{I})\mathbf{p}=\mathbf{0}$ when the question wants the $y=mx+c$ substitution, or vice versa.
How to avoid it
Line of invariant points = every point fixed (use $(\mathbf{M}-\mathbf{I})\mathbf{p}=\mathbf{0}$ ). Invariant line = line maps to itself, points may slide (use the $y=mx+c$ method). Match the method to the command word.
✕Forgetting that the origin is always an invariant point
9231 Paper 1 mark schemes
Why it happens
Students hunt for 'interesting' fixed points and overlook the trivial solution $\mathbf{p}=\mathbf{0}$ , especially when $\det(\mathbf{M}-\mathbf{I})\neq0$ .
How to avoid it
State that $\mathbf{M}\mathbf{0}=\mathbf{0}$ so the origin is always fixed. When $\det(\mathbf{M}-\mathbf{I})\neq0$ , the origin is the ONLY invariant point — say so explicitly.
✕Not treating $y'=mx'+c$ as an identity true for ALL $x$
9231 Paper 1 Examiner Reports
Why it happens
Students substitute a single point or solve for one $x$ instead of comparing coefficients, getting a spurious or incomplete answer.
How to avoid it
Keep $x$ as a free variable. Equate the coefficients of $x$ on both sides AND equate the constant terms — two separate equations for $m$ and $c$ .
✕Solving only the gradient equation for $m$ and ignoring the constant equation for $c$
9231 Paper 1 Examiner Reports
Why it happens
After finding $m$ from the $x$ -coefficients, students assume any $c$ works and claim infinitely many parallel invariant lines.
How to avoid it
The constant terms give a separate equation (e.g. $c(1-m)=0$ ). Solve it for each value of $m$ — usually it forces $c=0$ unless $m$ is special.
✕Trying to solve $(\mathbf{M}-\mathbf{I})\mathbf{p}=\mathbf{0}$ without checking the determinant first
9231 Paper 1 mark schemes
Why it happens
Students grind through the two equations and either miss that they coincide (a line) or claim a line when only the origin exists.
How to avoid it
Compute $\det(\mathbf{M}-\mathbf{I})$ first: non-zero $\Rightarrow$ origin only; zero $\Rightarrow$ the single equation IS the line of invariant points.
✕Quoting an invariant line without verifying it maps onto itself
9231 Paper 1 Examiner Reports
Why it happens
Time pressure and over-confidence; an arithmetic slip in forming $x'$ or $y'$ goes unnoticed.
How to avoid it
Take one concrete point on your line, transform it, and confirm the image satisfies the same line equation. A two-line check protects the accuracy marks.

Invariant Points and Lines

Detailed Study Notes

At a glance

What you’ll learn

Quick recap

Memorise this

How it’s examined

Take this whole topic with you

1Confirming the origin is invariant

2Line of invariant points vs invariant line for a reflection

3When the origin is the only invariant point

4Finding a line of invariant points

5Finding invariant lines y = mx + c (full method)

6Invariant lines through the origin (c = 0)

Invariant (fixed) point condition

Condition for non-trivial invariant points

Invariant-line method (line y = mx + c)

Invariant lines through the origin (c = 0)

Image of a point under M

Invariant point (fixed point)

Line of invariant points

Invariant line

Invariance of the origin

Singular matrix

True for all xxx (identity condition)

✕Confusing a 'line of invariant points' with an 'invariant line' and using the wrong method

✕Forgetting that the origin is always an invariant point

✕Not treating y′=mx′+cy'=mx'+cy′=mx′+c as an identity true for ALL xxx

✕Solving only the gradient equation for mmm and ignoring the constant equation for ccc

✕Trying to solve (M−I)p=0(\mathbf{M}-\mathbf{I})\mathbf{p}=\mathbf{0}(M−I)p=0 without checking the determinant first

✕Quoting an invariant line without verifying it maps onto itself

True for all $x$ (identity condition)

✕Not treating $y'=mx'+c$ as an identity true for ALL $x$

✕Solving only the gradient equation for $m$ and ignoring the constant equation for $c$

✕Trying to solve $(\mathbf{M}-\mathbf{I})\mathbf{p}=\mathbf{0}$ without checking the determinant first