MathsICSE Class 10 · Reflection

Reflection in the Lines x = a and y = a: Rules, Mirror Lines and Graph Questions (ICSE Class 10)

Updated 15 October 2026 · Fact-checked

Reflection in the vertical line x = a sends (x, y) to (2a − x, y); reflection in the horizontal line y = a sends (x, y) to (x, 2a − y). The mirror line passes through the midpoint of a point and its image, so a mirror line can be found by averaging the coordinates that change.

The mirror x = 2 sits exactly midway between P and P', so y stays 1 and x becomes 2(2) − 5 = −1.

Illustration drawn for QuestionBuddy.

Understand Reflection in the Lines x = a and y = a

The axes are special mirror lines ( $x = 0$ and $y = 0$ ). The ICSE syllabus also includes reflection in any line parallel to an axis: $x = a$ (vertical) and $y = a$ (horizontal).

**Reflection in $x = a$ .** The line is vertical, so a point moves horizontally; $y$ is unchanged. The mirror is the midpoint of $x$ and its image $x^{'}$ : $\frac{x + x ^{'}}{2} = a$ , so $x^{'} = 2 a - x$ .

$M_{x = a} : (x, y) \to (2 a - x, y) .$

**Reflection in $y = a$ .** The line is horizontal, so the point moves vertically; $x$ is unchanged:

$M_{y = a} : (x, y) \to (x, 2 a - y) .$

Check with $a = 0$ : these become $(- x, y)$ and $(x, - y)$ , the y-axis and x-axis rules.

Counting squares. On graph paper, count how far the point is from the mirror and go the same distance beyond it. $P (5, 1)$ is 3 units right of $x = 2$ , so its image is 3 units left of it, at $x = - 1$ ; indeed $2 (2) - 5 = - 1$ .

Finding the mirror line. If $P$ and its image $P^{'}$ share the same $y$ , the mirror is vertical: $x = \frac{x _{P} + x _{P^{'}}}{2}$ . If they share the same $x$ , it is horizontal: $y = \frac{y _{P} + y _{P^{'}}}{2}$ . Once the line is known, reflect other points in it.

Lines named by points. A question may say 'reflect $A$ in the line $B B^{'}$ '. Work out that line first: in 2024 Q3(iii), $B (2, 1)$ and $B^{'} (- 2, 1)$ lie on $y = 1$ , so $A (0, 3)$ reflects to $A^{'} (0, - 1)$ . The examiners reported candidates who reflected $A$ in the x-axis instead, getting $(0, - 3)$ .

Two parallel mirrors. Reflecting in $x = a$ and then in $x = b$ shifts every point sideways by $2 (b - a)$ . Two perpendicular mirrors $x = a$ and $y = b$ act like a 'reflection in the point' $(a, b)$ .

Reflection in the Lines x = a and y = a: formulas to remember

Reflection in x = a: $(x, y) \to (2 a - x, y)$; Vertical mirror; y unchanged.
Reflection in y = a: $(x, y) \to (x, 2 a - y)$; Horizontal mirror; x unchanged.
Vertical mirror from P and P′: $x = \frac{x _{P} + x _{P^{'}}}{2}$; When $y_{P} = y_{P^{'}}$ .
Horizontal mirror from P and P′: $y = \frac{y _{P} + y _{P^{'}}}{2}$; When $x_{P} = x_{P^{'}}$ .

How to solve Reflection in the Lines x = a and y = a questions

Decide whether the mirror is vertical or horizontal, change only the matching coordinate, and check by midpoint.

1Write the mirror line as $x = a$ or $y = a$ (work it out if it is named by points, like $B B^{'}$ ).
2For $x = a$ : new $x = 2 a - x$ , keep $y$ . For $y = a$ : new $y = 2 a - y$ , keep $x$ .
3Check: the midpoint of the point and its image lies on the mirror.
4To find a mirror line, average the coordinate that changed.
5On graph paper, draw the mirror line as a dotted line and label it with its equation.

Reflection in the Lines x = a and y = a 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.

Reflection in a vertical line

Suppose a stage plan is drawn on a grid with a wall mirror along the line $x = 4$ , and a dancer stands at $(1, 3)$ .

Where does the dancer's image appear on the plan?

Try it first, then show the solutionSolution

Vertical mirror: $x^{'} = 2 \times 4 - 1 = 7$ ; $y$ stays 3.
Check: the midpoint of $(1, 3)$ and $(7, 3)$ is $(4, 3)$ , on the mirror.

Answer: At $(7, 3)$ .

Memory hook: As far behind the mirror as in front of it.

Reflection in a horizontal line

Imagine a lake whose straight shore lies along $y = - 2$ on a map grid, with a lamp post at $(5, 3)$ reflected in the still water.

Where does the lamp's reflection appear, and how far is it from the lamp?

Try it first, then show the solutionSolution

Horizontal mirror: $y^{'} = 2 (- 2) - 3 = - 7$ ; $x$ stays 5.
Distance $= 3 - (- 7) = 10$ units, twice the lamp's distance of 5 from the shore.

Answer: At $(5, - 7)$ , 10 units from the lamp.

Memory hook: Image distance = twice the distance to the mirror.

Tricks to remember Reflection in the Lines x = a and y = a

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

Twice the mirror minus you
New coordinate $= 2 \times (mirror) - (old)$ , for the coordinate the mirror's equation names.
Use it when: Both $x = a$ and $y = a$ .
x = a moves x
The letter in the mirror's equation is the letter that changes.
Use it when: Avoiding changing the wrong coordinate.
Mirror in the middle
The mirror line is the average of a point and its image.
Use it when: Finding a mirror line, and checking any answer.

Reflection in the Lines x = a and y = a: solved examples

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

Example 1: Image in a vertical line · 1 mark · MCQBoard level

The image of $(3, - 2)$ in the line $x = 1$ is:
(a) $(- 3, - 2)$
(b) $(- 1, - 2)$
(c) $(3, 4)$
(d) $(1, - 2)$

Try it first, then show the solutionSolution

$x^{'} = 2 (1) - 3 = - 1$ ; $y$ stays $- 2$ .

Answer: (b) $(- 1, - 2)$

Example 2: Images in two different lines · 3 marks · Short answerBoard level

Find the image of $P (- 2, 5)$ in (a) the line $y = 2$ (b) the line $x = - 3$ .

Try it first, then show the solutionSolution

(a) $y = 2$ is horizontal: $(- 2, 2 (2) - 5) = (- 2, - 1)$ .
(b) $x = - 3$ is vertical: $(2 (- 3) - (- 2), 5) = (- 4, 5)$ .

Answer: (a) $(- 2, - 1)$ (b) $(- 4, 5)$

Example 3: Unknowns from an image · 3 marks · Short answerHarder board level

The point $P (m, 7)$ is reflected in the line $y = b$ and its image is $(- 3, - 1)$ . Find $m$ and $b$ .

Try it first, then show the solutionSolution

In $y = b$ , $x$ does not change: $m = - 3$ .
$y^{'} = 2 b - 7 = - 1 \Rightarrow b = 3$ (equivalently, $b$ is the average of 7 and $- 1$ ).

Answer: $m = - 3$ , $b = 3$

Example 4: Find the mirror, then use it · 4 marks · Short answerHarder board level

$A (3, - 2)$ is reflected in a line $L$ to $A^{'} (3, 6)$ .
(a) Find the equation of $L$ .
(b) Find the image of $B (- 1, 0)$ in $L$ .

Try it first, then show the solutionSolution

(a) $x$ is unchanged, so $L$ is horizontal: $y = \frac{- 2 + 6}{2} = 2$ .
(b) $B^{'} = (- 1, 2 (2) - 0) = (- 1, 4)$ .

Answer: (a) $y = 2$ (b) $(- 1, 4)$

Example 5: Graph question: reflecting in a line through two points · 5 marks · Long answerHOTS / competency

Use graph paper (2 cm = 1 unit).
(a) Plot $A (1, 4)$ and $B (- 2, 1)$ .
(b) Reflect $B$ in the line $x = 1$ to $B^{'}$ .
(c) Reflect $A$ in the line $B B^{'}$ to $A^{'}$ .
(d) Join $A$ , $B$ , $A^{'}$ , $B^{'}$ in order; name the figure and find its area.

Try it first, then show the solutionSolution

(b) $B^{'} = (2 (1) - (- 2), 1) = (4, 1)$ .
(c) $B$ and $B^{'}$ both have $y = 1$ , so $B B^{'}$ is the line $y = 1$ . $A^{'} = (1, 2 (1) - 4) = (1, - 2)$ .
(d) Diagonals: $A A^{'}$ is vertical with length 6; $B B^{'}$ is horizontal with length 6. They bisect each other at $(1, 1)$ , are perpendicular and equal, so $A B A^{'} B^{'}$ is a square.
Area $= \frac{1}{2} \times 6 \times 6 = 18$ square units (side $18 = 32$ ).

Answer: $B^{'} (4, 1)$ , $A^{'} (1, - 2)$ ; a square of area 18 square units.

Example 6: Case study: a garden path as a mirror · 4 marks · Case-basedHOTS / competency

A garden plan uses the path along $x = - 1$ as a line of symmetry. A fountain is at $F (3, 2)$ and a bench at $B (- 4, - 2)$ .
(a) Where should the matching fountain go?
(b) Where should the matching bench go?
(c) How far apart are the two fountains?

Try it first, then show the solutionSolution

(a) $F^{'} = (2 (- 1) - 3, 2) = (- 5, 2)$ .
(b) $B^{'} = (2 (- 1) - (- 4), - 2) = (2, - 2)$ .
(c) $F F^{'} = 3 - (- 5) = 8$ units (twice the fountain's distance of 4 from the path).

Answer: (a) $(- 5, 2)$ (b) $(2, - 2)$ (c) 8 units

Where students lose marks in Reflection in the Lines x = a and y = a

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

Mistake: Reflecting in an axis instead of the stated line
The mirror is the line named, here $B B^{'}$ ( $y = 1$ ), not the x-axis.
Wrong working
$A (0, 3)$ reflected in $B B^{'}$ : $A^{'} (0, - 3)$ .
Right working
$B B^{'}$ is $y = 1$ : $A^{'} = (0, 2 - 3) = (0, - 1)$ .
Do this: Write the equation of the mirror line before reflecting.
Usually costs: Reported in 2024 Q3(iii); the image and the figure's name both go wrong.
Mistake: Changing the wrong coordinate
A vertical mirror $x = a$ changes $x$ only.
Wrong working
$(3, - 2)$ in $x = 1$ : $(3, 4)$ .
Right working
$(2 - 3, - 2) = (- 1, - 2)$ .
Do this: The letter in the mirror's equation is the one that changes.
Usually costs: The whole answer.
Mistake: Using a − x instead of 2a − x
The image is as far beyond the mirror as the point is before it.
Wrong working
$(5, 1)$ in $x = 2$ : $x^{'} = 2 - 5 = - 3$ .
Right working
$x^{'} = 4 - 5 = - 1$ (check: midpoint of 5 and $- 1$ is 2).
Do this: Check with the midpoint every time.
Usually costs: Accuracy marks.

Exam tips

2024 Q3(iii) (5 marks) reflected $B$ and $C$ in the y-axis, then $A$ in the line $B B^{'}$ ; the figure $A B A^{'} B^{'}$ was a square, and calling it a rhombus or kite lost the mark.
Draw and label the mirror line ( $x = 1$ , $y = 1$ , …) on the graph; it shows the examiner your method.
To prove a square, show the diagonals are equal, perpendicular and bisect each other, or that all sides are equal and an angle is $90°$ .
Read every 'reflect in' carefully: axis, origin, or a named line.

Practice questions on Reflection in the Lines x = a and y = a

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
The image of $(5, 3)$ in the line $y = 3$ is:
(a) $(5, - 3)$
(b) $(1, 3)$
(c) $(5, 3)$
(d) $(5, 9)$
Hint
Where is the point compared with the mirror?
Answer
(c) $(5, 3)$ : the point lies on the mirror, so it does not move.
Q2 · 2 marks · Short answerBoard level
Find the image of $(- 4, - 1)$ in the line $x = 2$ .
Hint
$x^{'} = 4 - (- 4)$ .
Answer
$(8, - 1)$
Q3 · 2 marks · Short answerHarder board level
The point $(6, - 2)$ is reflected in a line to $(6, 4)$ . Find the equation of the line.
Hint
Average the $y$ -coordinates.
Answer
$y = 1$
Q4 · 3 marks · Short answerHarder board level
$P (a, b)$ is reflected in the line $x = 3$ to $(- 1, 5)$ . Find $a$ and $b$ .
Hint
$2 (3) - a = - 1$ .
Answer
$a = 7$ , $b = 5$
Q5 · 1 mark · Assertion-reasonHarder board level
Assertion (A): The image of $(2, 5)$ in the line $y = 1$ is $(2, - 3)$ .
Reason (R): Reflection in $y = a$ maps $(x, y)$ to $(x, 2 a - y)$ .
(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
$2 (1) - 5 = - 3$ .
Answer
(c)
Q6 · 3 marks · Short answerHarder board level
$A (1, 2)$ is reflected in the line $x = 3$ to $A^{'}$ , and $A^{'}$ is reflected in the line $x = 6$ to $A^{''}$ . Find $A^{'}$ and $A^{''}$ , and describe the single movement from $A$ to $A^{''}$ .
Hint
Apply the rule twice.
Answer
$A^{'} (5, 2)$ , $A^{''} (7, 2)$ : a shift of 6 units to the right, which is $2 \times (6 - 3)$ .
Q7 · 4 marks · Short answerHOTS / competency
Triangle $A B C$ has $A (2, 1)$ , $B (5, 1)$ , $C (2, 4)$ . Reflect it in the line $y = - 1$ . Write the images, and name and find the area of the quadrilateral $A B B^{'} A^{'}$ .
Hint
Only the $y$ -coordinates change.
Answer
$A^{'} (2, - 3)$ , $B^{'} (5, - 3)$ , $C^{'} (2, - 6)$ . $A B B^{'} A^{'}$ is a rectangle 3 by 4, area 12 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.Image of $(x, y)$ in $x = a$ ?Show answer
$(2 a - x, y)$
2.Image of $(x, y)$ in $y = a$ ?Show answer
$(x, 2 a - y)$
3.Image of $(1, 4)$ in $y = 3$ ?Show answer
$(1, 2)$
4. $(2, 7)$ is reflected to $(8, 7)$ . Mirror line?Show answer
$x = 5$
5.Which points do not move under reflection in $x = 4$ ?Show answer
The points on the line $x = 4$ .

Reflection in the Lines x = a and y = a: frequently asked questions

What is the image of a point in the line x = a?

(x, y) maps to (2a − x, y). For example, (5, 1) in x = 2 maps to (−1, 1), because 2 is midway between 5 and −1.

How do I find the mirror line from a point and its image?

If the y-coordinates are equal, the mirror is x = average of the x-coordinates; if the x-coordinates are equal, it is y = average of the y-coordinates.

Does the ICSE syllabus include reflection in lines like y = x?

No. The CISCE syllabus lists reflection in x = 0, y = 0, x = a, y = a and the origin only.