MathsICSE Class 10 · Reflection

Invariant Points under Reflection: Meaning, Rules and Solved Questions (ICSE Class 10)

Updated 15 October 2026 · Fact-checked

A point is invariant under a reflection if its image is the point itself. For reflection in a line, the invariant points are exactly the points on the mirror line: y = 0 for the x-axis, x = 0 for the y-axis. Under reflection in the origin, only (0, 0) is invariant.

A point off the mirror jumps to the other side, but a point on the mirror is at distance 0, so its image is itself.

Illustration drawn for QuestionBuddy.

Understand Invariant Points under Reflection

Meaning. 'Invariant' means unchanged. A point $P$ is invariant under a transformation if its image is $P$ itself.

Why points on the mirror stay put. The image of $P$ is as far beyond the mirror as $P$ is in front of it. If $P$ is on the mirror, that distance is 0, so the image is $P$ . Conversely, a point off the mirror always moves to the other side. So:

Under $M_{x}$ : $(x, y) = (x, - y)$ forces $y = 0$ . Invariant points: the x-axis, points $(k, 0)$ .
Under $M_{y}$ : $- x = x$ forces $x = 0$ . Invariant points: the y-axis, points $(0, k)$ .
Under reflection in $x = a$ : $2 a - x = x$ forces $x = a$ . Invariant points: $(a, k)$ .
Under reflection in $y = a$ : invariant points $(k, a)$ .
Under $M_{O}$ : $(- x, - y) = (x, y)$ forces $x = 0$ and $y = 0$ . Only the origin is invariant.

Typical questions.

1. Test a point. Is $(0, - 3)$ invariant under $M_{x}$ ? Its image is $(0, 3)$ , so no; it is invariant under $M_{y}$ instead.
2. Find an unknown. $(k, - 4)$ is invariant under reflection in $x = 5$ , so $k = 5$ .
3. Invariant point on a segment or figure. It is where the segment crosses the mirror line. For $A (- 2, 3)$ and $B (4, - 3)$ and the x-axis, the segment crosses $y = 0$ halfway (the $y$ -values 3 and $- 3$ are equal and opposite), at $(1, 0)$ .
4. Invariant under two reflections. The point must lie on both mirror lines: their intersection. Invariant under both $M_{x}$ and reflection in $x = - 4$ : $(- 4, 0)$ .
5. Mirror from invariant points. If two points are both invariant, the mirror line passes through both.

In graph questions invariant points save work: vertices on the mirror need no reflecting, and they often become vertices shared by a figure and its image (as in a rhombus formed by reflecting a triangle in its base).

Invariant Points under Reflection: formulas to remember

Invariant under reflection in the x-axis: $(k, 0)$; Points on the x-axis.
Invariant under reflection in the y-axis: $(0, k)$; Points on the y-axis.
Invariant under reflection in x = a or y = a: $(a, k) or (k, a)$; Points on that line.
Invariant under reflection in the origin: $(0, 0) only$; No other point.

How to solve Invariant Points under Reflection questions

Invariant means 'image equals point'; for a line mirror, that means 'on the line'.

1Write the mirror line's equation ( $y = 0$ , $x = 0$ , $x = a$ , $y = a$ ) or note it is the origin.
2To test a point, check whether it satisfies the mirror's equation (or find its image and compare).
3To find unknowns, substitute the point into the mirror's equation.
4For a segment, find where it meets the mirror line (often by symmetry or by a ratio).
5For two reflections, find the intersection of the two mirror lines.
6State the answer as 'the invariant point is …' or 'there is no invariant point other than …'.

Invariant Points under Reflection in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Points on the mirror line are invariant

Suppose a toy ruler lies across a flat mirror placed along the y-axis of a grid, touching the mirror at the point $(0, 2)$ , with its other end at $(3, 4)$ .

Which point of the ruler coincides with its own image, and where is the image of the far end?

Try it first, then show the solutionSolution

Points on the mirror ( $x = 0$ ) are invariant, so $(0, 2)$ is its own image.
The far end reflects to $(- 3, 4)$ .

Answer: $(0, 2)$ stays fixed; the far end's image is $(- 3, 4)$ .

Memory hook: Touch the mirror and you meet yourself.

Invariant points in paper folding

Imagine folding a sheet of graph paper along the line $x = 3$ while wet ink dots at $(3, 1)$ , $(3, - 2)$ and $(5, 1)$ print onto the other half.

Which dots print exactly on top of themselves, and where does the third print?

Try it first, then show the solutionSolution

Dots on the fold $x = 3$ are invariant: $(3, 1)$ and $(3, - 2)$ .
$(5, 1)$ prints at $(2 \times 3 - 5, 1) = (1, 1)$ .

Answer: $(3, 1)$ and $(3, - 2)$ print on themselves; $(5, 1)$ prints at $(1, 1)$ .

Memory hook: The fold line is the set of invariant points.

Tricks to remember Invariant Points under Reflection

Short memory aids. Say each one aloud once, then use it on a question the same day.

On the mirror, no move
Invariant points of a line reflection = the mirror line itself.
Use it when: Every 'which point is invariant' question.
Origin is lonely
Reflection in the origin fixes only one point, the origin.
Use it when: MCQs on $M_{O}$ .
Two mirrors meet once
Invariant under two reflections = where the two mirror lines cross.
Use it when: Combined-condition questions.

Invariant Points under Reflection: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Spotting an invariant point · 1 mark · MCQBoard level

Which point is invariant under reflection in the y-axis?
(a) $(6, 0)$
(b) $(0, - 6)$
(c) $(- 6, 6)$
(d) $(6, 6)$

Try it first, then show the solutionSolution

Invariant points lie on the y-axis, where $x = 0$ : only $(0, - 6)$ .

Answer: (b) $(0, - 6)$

Example 2: Finding unknowns · 3 marks · Short answerBoard level

(a) The point $(k, - 4)$ is invariant under reflection in the line $x = 5$ . Find $k$ .
(b) The point $(2, m)$ is invariant under reflection in the line $y = - 3$ . Find $m$ .
(c) Write the image of $(2, 1)$ in $y = - 3$ .

Try it first, then show the solutionSolution

(a) It must lie on $x = 5$ : $k = 5$ .
(b) It must lie on $y = - 3$ : $m = - 3$ .
(c) $(2, 2 (- 3) - 1) = (2, - 7)$ .

Answer: (a) $k = 5$ (b) $m = - 3$ (c) $(2, - 7)$

Example 3: Invariant point on a segment · 3 marks · Short answerHarder board level

The segment joining $A (- 2, 3)$ and $B (4, - 3)$ is reflected in the x-axis. Find the point on $A B$ that is invariant.

Try it first, then show the solutionSolution

The invariant point is where $A B$ meets the x-axis ( $y = 0$ ).
$y$ goes from 3 to $- 3$ , so $y = 0$ is exactly halfway along $A B$ .
Halfway in $x$ : $\frac{- 2 + 4}{2} = 1$ . The point is $(1, 0)$ .

Answer: $(1, 0)$

Example 4: Invariant vertices of a triangle · 4 marks · Short answerHarder board level

Triangle $A B C$ has $A (3, 0)$ , $B (0, 2)$ , $C (- 1, - 1)$ .
(a) Which vertex is invariant under reflection in the x-axis? Which under reflection in the y-axis?
(b) Reflect the triangle in the x-axis and write the coordinates of the image.

Try it first, then show the solutionSolution

(a) $A$ lies on the x-axis, so it is invariant under $M_{x}$ . $B$ lies on the y-axis, so it is invariant under $M_{y}$ . $C$ lies on neither.
(b) $A^{'} = A (3, 0)$ , $B^{'} (0, - 2)$ , $C^{'} (- 1, 1)$ .

Answer: (a) $A$ under $M_{x}$ ; $B$ under $M_{y}$ (b) $A^{'} (3, 0)$ , $B^{'} (0, - 2)$ , $C^{'} (- 1, 1)$

Example 5: Invariant under a combined reflection · 3 marks · Short answerHOTS / competency

A point $P$ is reflected in the x-axis to $P^{'}$ , and $P^{'}$ is reflected in the origin to $P^{''}$ . For which points $P$ is $P^{''} = P$ ?

Try it first, then show the solutionSolution

$P (x, y) \to P^{'} (x, - y) \to P^{''} (- x, y)$ .
So the combined effect is reflection in the y-axis.
$P^{''} = P$ when $- x = x$ , that is $x = 0$ : every point on the y-axis.

Answer: All points $(0, k)$ on the y-axis.

Example 6: Case study: a rhombus from a reflection · 5 marks · Case-basedHOTS / competency

Use graph paper (2 cm = 1 unit). Triangle $A B C$ has $A (- 3, 0)$ , $B (0, 4)$ , $C (3, 0)$ .
(a) Reflect it in the x-axis. Which vertices are invariant? Write the image of $B$ .
(b) Name the figure $A B C B^{'}$ and give a reason.
(c) Find its area and the length of a side.

Try it first, then show the solutionSolution

(a) $A$ and $C$ lie on the x-axis, so they are invariant. $B^{'} = (0, - 4)$ .
(b) Diagonals $A C$ (on the x-axis, length 6) and $B B^{'}$ (on the y-axis, length 8) are perpendicular and bisect each other at $O$ , so $A B C B^{'}$ is a rhombus.
(c) Area $= \frac{1}{2} \times 6 \times 8 = 24$ square units. Side $A B = 3^{2} + 4^{2} = 5$ units.

Answer: (a) $A$ and $C$ ; $B^{'} (0, - 4)$ (b) Rhombus (c) 24 square units; side 5 units

Where students lose marks in Invariant Points under Reflection

The slips examiners see most often, and what to write instead.

Mistake: Choosing the wrong axis for invariance
Invariant points of $M_{x}$ lie on the x-axis ( $y = 0$ ), not the y-axis.
Wrong working
$(0, 5)$ is invariant under reflection in the x-axis.
Right working
$M_{x} (0, 5) = (0, - 5)$ , so it is not; $(5, 0)$ would be.
Do this: Find the image and compare, if unsure.
Usually costs: The 1-mark MCQ.
Mistake: Thinking every point on an axis is invariant under M_O
Only the origin is fixed by reflection in the origin.
Wrong working
$(3, 0)$ is invariant under $M_{O}$ .
Right working
$M_{O} (3, 0) = (- 3, 0)$ .
Do this: Remember: one invariant point for the origin.
Usually costs: The mark.
Mistake: Reflecting invariant vertices anyway, and getting them wrong
A vertex on the mirror is its own image.
Wrong working
$A (- 3, 0)$ in the x-axis: $(3, 0)$ .
Right working
$A^{'} = A (- 3, 0)$ .
Do this: Check each vertex against the mirror's equation first.
Usually costs: Plotting marks and the figure's name in graph questions.

Exam tips

Invariant points are often asked as a 1-mark MCQ or as a part of a 5-mark graph question; state them explicitly.
When a figure is reflected in its own side, the vertices on that side are shared; the result is often a rhombus, kite or square.
Write 'invariant' with a reason: 'since $A$ lies on the x-axis'.
For segments, find the crossing point with the mirror by symmetry or ratio, then check its coordinates satisfy the mirror's equation.

Practice questions on Invariant Points under Reflection

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
Which point is invariant under reflection in the line $y = - 2$ ?
(a) $(- 2, 4)$
(b) $(4, - 2)$
(c) $(2, 4)$
(d) $(- 4, 2)$
Hint
Invariant points lie on the mirror line, where $y = - 2$ .
Answer
(b) $(4, - 2)$
Q2 · 2 marks · Short answerBoard level
Which of $(0, 4)$ , $(4, 0)$ and $(- 2, - 2)$ are invariant under reflection in the y-axis? Write the images of the others.
Hint
Invariant means $x = 0$ .
Answer
Only $(0, 4)$ . The images are $(- 4, 0)$ and $(2, - 2)$ .
Q3 · 3 marks · Short answerHarder board level
Find the point on the segment joining $P (- 4, - 2)$ and $Q (2, 4)$ that is invariant under reflection in the y-axis.
Hint
Where does $P Q$ meet $x = 0$ ? $x$ goes $- 4 \to 2$ , so $x = 0$ is $\frac{4}{6}$ of the way.
Answer
$y = - 2 + \frac{2}{3} \times 6 = 2$ : the point is $(0, 2)$ .
Q4 · 1 mark · Assertion-reasonHarder board level
Assertion (A): $(0, - 3)$ is invariant under reflection in the x-axis.
Reason (R): Points on the y-axis are invariant under reflection in the y-axis.
(a) (A) is true and (R) is false.
(b) (A) is false and (R) is true.
(c) Both (A) and (R) are true, and (R) is the correct explanation for (A).
(d) Both (A) and (R) are true, but (R) is not the correct explanation for (A).
Hint
$M_{x} (0, - 3) = (0, 3)$ .
Answer
(b)
Q5 · 3 marks · Short answerHarder board level
The points $(4, 1)$ and $(4, - 5)$ are both invariant under reflection in a line $L$ . Find $L$ and the image of $(1, 1)$ in $L$ .
Hint
Both points lie on $L$ .
Answer
$L$ is $x = 4$ ; the image of $(1, 1)$ is $(7, 1)$ .
Q6 · 2 marks · Short answerHarder board level
Prove that no point other than the origin is invariant under reflection in the origin.
Hint
Set the image equal to the point.
Answer
If $(- x, - y) = (x, y)$ then $2 x = 0$ and $2 y = 0$ , so $x = y = 0$ .
Q7 · 4 marks · Short answerHOTS / competency
Triangle $A B C$ has $A (2, 2)$ , $B (5, 2)$ , $C (5, 5)$ . It is reflected in the line $y = 2$ . Which vertices are invariant? Find $C^{'}$ , and show that triangle $A C C^{'}$ is isosceles; find its area.
Hint
$A$ and $B$ lie on $y = 2$ .
Answer
$A$ and $B$ are invariant; $C^{'} (5, - 1)$ . $A C = A C^{'} = 18$ , so isosceles. Base $C C^{'} = 6$ , height 3: area 9 square units.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.What is an invariant point?Show answer
A point that is its own image.
2.Invariant points under reflection in $y = - 1$ ?Show answer
All points $(k, - 1)$ .
3.Is $(2, 0)$ invariant under $M_{x}$ ?Show answer
Yes.
4.Point invariant under reflection in both $x = 2$ and $y = 5$ ?Show answer
$(2, 5)$
5.How many points are invariant under reflection in the origin?Show answer
One: the origin.

Invariant Points under Reflection: frequently asked questions

What are invariant points in reflection?

Points whose image is the point itself. For reflection in a line, they are the points on that line; for reflection in the origin, only the origin.

Which points are invariant under reflection in the x-axis?

All points on the x-axis, that is points of the form (k, 0).

How do I find an invariant point on a line segment?

Find where the segment meets the mirror line. For example, the segment from (−2, 3) to (4, −3) meets the x-axis at (1, 0), which is invariant under reflection in the x-axis.