ICSE Class 10 Maths Reflection Solutions Selina Ch 12

Written by Greya LakshmiPublished on 19 May 2026Updated on 10 June 2026

ICSE Class 10 Maths Reflection Solutions: Chapter 12

ICSE Class 10 Maths Reflection in Concise Mathematics Selina Chapter 12 teaches how a point or figure changes when reflected in the x-axis, y-axis, origin, the lines $x=a$ , $y=a$ , and how to identify invariant points. The main idea is simple: a reflected image is the same distance from the mirror line as the original point, but on the opposite side.

This replacement page is written as a study-and-solution page for Selina Concise Mathematics Class 10 Chapter 12. It gives the rules first, then worked examples, then the Exercise 12 and Test Yourself answers in school-style steps. The article is text-only because images are disabled for this page.

For the official syllabus context, use the CISCE website and the prescribed Selina Concise Mathematics textbook as the reference points: CISCE official website and Selina Publishers.

Reflection Formula Table for Concise Mathematics Selina Chapter 12

Use this table before solving any question. In coordinate geometry, the abscissa is the $x$ -coordinate and the ordinate is the $y$ -coordinate.

Reflection in	Rule for point $(x,y)$	What changes?	Invariant points
x-axis, or $y=0$	$(x,y)\to(x,-y)$	Only $y$ changes sign	All points $(x,0)$
y-axis, or $x=0$	$(x,y)\to(-x,y)$	Only $x$ changes sign	All points $(0,y)$
origin	$(x,y)\to(-x,-y)$	Both coordinates change sign	Only $(0,0)$
vertical line $x=a$	$(x,y)\to(2a-x,y)$	$x$ changes using the mirror line	All points on $x=a$
horizontal line $y=a$	$(x,y)\to(x,2a-y)$	$y$ changes using the mirror line	All points on $y=a$
line $y=x$	$(x,y)\to(y,x)$	Coordinates interchange	All points where $x=y$

Concept Snapshot: The Mirror-Distance Test

Think of the mirror line as a fold in graph paper. After folding, the original point and its image must overlap. That means the mirror line is exactly halfway between the point and its image, and the joining segment is perpendicular to the mirror line. For example, when $P(4,1)$ is reflected in $y=3$ , the $x$ -coordinate stays $4$ . The old $y$ -value is $1$ , two units below $3$ , so the image has $y=5$ , two units above $3$ . Therefore $P'(4,5)$ .

Worked Examples Before Exercise 12

Worked Example 1: Reflection in the line $x=4$

Find the image of $P(2,3)$ under reflection in the line $x=4$ .

Step 1: The mirror line is vertical, so use $(x,y)\to(2a-x,y)$ .

Step 2: Here $a=4$ , $x=2$ , and $y=3$ .

$x' = 2a-x = 2(4)-2 = 8-2 = 6$

$y' = y = 3$

Final answer: The image is $P'(6,3)$ .

Worked Example 2: Two reflections as one transformation

A point $A(-3,2)$ is reflected in the x-axis and then in the origin. Name the single transformation from $A$ to the final image.

Step 1: Reflect $A(-3,2)$ in the x-axis.

$(-3,2)\to(-3,-2)$

Step 2: Reflect $(-3,-2)$ in the origin.

$(-3,-2)\to(3,2)$

Step 3: Compare the starting point and final point.

$(-3,2)\to(3,2)$

Final answer: The single transformation is reflection in the y-axis.

Worked Example 3: Area after reflection in axes

Let $P(2,-4)$ be reflected in $x=0$ to $Q$ , and then $Q$ be reflected in $y=0$ to $R$ . Find the area of $\triangle PQR$ .

Step 1: Reflection in $x=0$ means reflection in the y-axis.

$P(2,-4)\to Q(-2,-4)$

Step 2: Reflection in $y=0$ means reflection in the x-axis.

$Q(-2,-4)\to R(-2,4)$

Step 3: $PQ$ is horizontal and $QR$ is vertical.

$PQ=4,\qquad QR=8$

Step 4: Use the area of a right-angled triangle.

$\text{Area}=\frac{1}{2}\times PQ\times QR=\frac{1}{2}\times4\times8=16$

Final answer: Area of $\triangle PQR=16$ square units.

Exercise 12 Solutions

Question 1(a): A point $P$ is its own image in a line $l$

Step 1: A point is unchanged after reflection only if it lies on the mirror line.

Step 2: Such a point is called an invariant point.

Final answer: Option $1$ : $P$ is on the line $l$ .

Question 1(b): $P(-7,8)$ reflected in origin and then in x-axis

Step 1: Reflect in the origin: $(x,y)\to(-x,-y)$ .

$(-7,8)\to(7,-8)$

Step 2: Reflect the result in the x-axis: $(x,y)\to(x,-y)$ .

$(7,-8)\to(7,8)$

Final answer: Option $2$ : $Q(7,8)$ .

Question 1(c): Find $P$ when it is reflected in the y-axis and then in the origin to get $(6,6)$

Step 1: Let $P=(x,y)$ .

Step 2: Reflection in the y-axis gives $(-x,y)$ .

Step 3: Reflection of $(-x,y)$ in the origin gives $(x,-y)$ .

$(x,-y)=(6,6)$

$x=6,\qquad -y=6\Rightarrow y=-6$

Final answer: Option $1$ : $P(6,-6)$ .

Question 1(d): Reflect $P(4,-8)$ in $x=0$

Step 1: The line $x=0$ is the y-axis.

Step 2: Reflection in the y-axis changes the sign of the $x$ -coordinate.

$(4,-8)\to(-4,-8)$

Final answer: Option $4$ : $R(-4,-8)$ .

Question 1(e): $P(5,5)$ reflected in $y=0$ and then in x-axis

Step 1: The line $y=0$ is the x-axis.

Step 2: First reflection in the x-axis gives $(5,5)\to(5,-5)$ .

Step 3: Reflect again in the x-axis.

$(5,-5)\to(5,5)$

Final answer: Option $1$ : $Q(5,5)$ .

Question 2: Reflect points in the line $x=0$

Step 1: Since $x=0$ is the y-axis, use $(x,y)\to(-x,y)$ .

$(-6,4)\to(6,4)$

$(0,5)\to(0,5)$

$(3,-4)\to(-3,-4)$

Final answer: (i) $(6,4)$ , (ii) $(0,5)$ , (iii) $(-3,-4)$ .

Question 3: Reflect points in the line $y=0$

Step 1: Since $y=0$ is the x-axis, use $(x,y)\to(x,-y)$ .

$(-3,0)\to(-3,0)$

$(8,-5)\to(8,5)$

$(-1,-3)\to(-1,3)$

Final answer: (i) $(-3,0)$ , (ii) $(8,5)$ , (iii) $(-1,3)$ .

Question 4: Image of $P$ in x-axis is $(-4,5)$

Step 1: Let $P=(x,y)$ . Reflection in the x-axis gives $(x,-y)$ .

$(x,-y)=(-4,5)$

$x=-4,\qquad -y=5\Rightarrow y=-5$

Step 2: So $P=(-4,-5)$ .

Step 3: Reflect $P$ in the y-axis.

$(-4,-5)\to(4,-5)$

Final answer: (i) $P(-4,-5)$ ; (ii) image in the y-axis is $(4,-5)$ .

Question 5: Image of $P$ in origin is $(-2,7)$

Step 1: Let $P=(x,y)$ . Reflection in the origin gives $(-x,-y)$ .

$(-x,-y)=(-2,7)$

$-x=-2\Rightarrow x=2,\qquad -y=7\Rightarrow y=-7$

Step 2: So $P=(2,-7)$ .

Step 3: Reflect $P$ in the x-axis.

$(2,-7)\to(2,7)$

Final answer: (i) $P(2,-7)$ ; (ii) image in the x-axis is $(2,7)$ .

Question 6: $P(a,b)$ reflected in origin and then in y-axis gives $P'(4,6)$

Step 1: Reflect $P(a,b)$ in the origin.

$(a,b)\to(-a,-b)$

Step 2: Reflect $(-a,-b)$ in the y-axis.

$(-a,-b)\to(a,-b)$

Step 3: Compare with the given final point.

$(a,-b)=(4,6)$

$a=4,\qquad -b=6\Rightarrow b=-6$

Final answer: $a=4,\ b=-6$ .

Question 7: $A(-3,2)$ reflected in x-axis and then in origin

Step 1: Reflect $A(-3,2)$ in the x-axis.

$A(-3,2)\to A'(-3,-2)$

Step 2: Reflect $A'(-3,-2)$ in the origin.

$A'(-3,-2)\to A''(3,2)$

Step 3: Directly, $A(-3,2)\to A''(3,2)$ , so only the $x$ -coordinate changes sign.

Final answer: (i) $A''(3,2)$ ; (ii) single transformation: reflection in the y-axis.

Question 8: Triangle reflected in y-axis and then in origin

Step 1: The combined transformation is $(x,y)\to(-x,y)\to(x,-y)$ .

Step 2: Apply it to each vertex.

$A(2,6)\to A''(2,-6)$

$B(-3,5)\to B''(-3,-5)$

$C(4,7)\to C''(4,-7)$

Step 3: The direct change is $(x,y)\to(x,-y)$ .

Final answer: (i) $A''(2,-6)$ , $B''(-3,-5)$ , $C''(4,-7)$ ; (ii) single transformation: reflection in the x-axis.

Question 9: Graph of $A(3,2)$ , $B(5,4)$ and their images in the x-axis

Step 1: Reflection in the x-axis changes $y$ to $-y$ .

$A(3,2)\to A'(3,-2),\qquad B(5,4)\to B'(5,-4)$

Step 2: The opposite non-parallel sides are equal in the plotted quadrilateral $ABB'A'$ , so the figure is an isosceles trapezium.

Step 3: Segment $BB'$ is vertical and $BA$ has equal horizontal and vertical changes, so the acute angle at $B$ with $BB'$ is $45^\circ$ .

Step 4: Reflect $A(3,2)$ in the origin.

$A(3,2)\to A''(-3,-2)$

Step 5: $A'(3,-2)\to A''(-3,-2)$ changes only the $x$ -coordinate sign.

Final answer: (i) isosceles trapezium; (ii) $\angle ABB'=45^\circ$ ; (iii) $A''(-3,-2)$ ; (iv) reflection in the y-axis.

Question 10: Invariant points under $L_1$ and $L_2$

Step 1: Points $(3,0)$ and $(-1,0)$ lie on the x-axis, so $L_1$ is the x-axis.

$L_1:\ y=0$

Step 2: Points $(0,-3)$ and $(0,1)$ lie on the y-axis, so $L_2$ is the y-axis.

$L_2:\ x=0$

Step 3: Reflect $P(3,4)$ and $Q(-5,-2)$ in $L_1$ .

$P(3,4)\to P'(3,-4),\qquad Q(-5,-2)\to Q'(-5,2)$

Step 4: Reflect $P(3,4)$ and $Q(-5,-2)$ in $L_2$ .

$P(3,4)\to P''(-3,4),\qquad Q(-5,-2)\to Q''(5,-2)$

Step 5: $P'(3,-4)\to P''(-3,4)$ changes both signs, so it is reflection in the origin.

Final answer: $L_1:y=0$ , $L_2:x=0$ ; $P'(3,-4)$ , $Q'(-5,2)$ ; $P''(-3,4)$ , $Q''(5,-2)$ ; single transformation: reflection in the origin.

Question 11: Find the mirror line and image of $(-8,-5)$

Step 1: The mappings $(-2,0)\to(2,0)$ and $(5,-6)\to(-5,-6)$ change only the sign of the $x$ -coordinate.

Step 2: Therefore the mirror line is the y-axis.

$x=0$

Step 3: Reflect $(-8,-5)$ in the y-axis.

$(-8,-5)\to(8,-5)$

Final answer: Mirror line: y-axis, $x=0$ ; image is $(8,-5)$ .

Question 12: Reflect $P(4,1)$ and $Q(-2,4)$ in $y=3$

Step 1: For reflection in $y=a$ , use $(x,y)\to(x,2a-y)$ .

Step 2: Here $a=3$ .

$P(4,1)\to P'(4,2(3)-1)=(4,5)$

$Q(-2,4)\to Q'(-2,2(3)-4)=(-2,2)$

Final answer: $P'(4,5)$ and $Q'(-2,2)$ .

Question 13: Reflect $P(-2,3)$ in $x=2$

Step 1: For reflection in $x=a$ , use $(x,y)\to(2a-x,y)$ .

Step 2: Here $a=2$ , $x=-2$ , $y=3$ .

$x'=2(2)-(-2)=4+2=6$

$y'=3$

Final answer: $P'(6,3)$ .

Question 14: $P(a,b)$ reflected in x-axis to $P'(2,-3)$

Step 1: Reflection in the x-axis gives $(a,b)\to(a,-b)$ .

$(a,-b)=(2,-3)$

$a=2,\qquad -b=-3\Rightarrow b=3$

Step 2: So $P=(2,3)$ .

Step 3: Reflect $P$ in the y-axis.

$(2,3)\to(-2,3)$

Step 4: Reflect $P$ in $x=4$ .

$x'=2(4)-2=6,\qquad y'=3$

Final answer: $a=2$ , $b=3$ , $P''(-2,3)$ , $P'''(6,3)$ .

Question 15: Images of $A(3,4)$ and $B(0,2)$

Step 1: Reflect $A(3,4)$ in the x-axis.

$A(3,4)\to A'(3,-4)$

Step 2: The line $AA'$ is vertical through $x=3$ . Reflect $B(0,2)$ in $x=3$ .

$B(0,2)\to B'(2(3)-0,2)=(6,2)$

Step 3: Reflect $A(3,4)$ in the y-axis.

$A(3,4)\to A''(-3,4)$

Step 4: The line $AA''$ is horizontal through $y=4$ . Reflect $B(0,2)$ in $y=4$ .

$B(0,2)\to B''(0,2(4)-2)=(0,6)$

Final answer: $A'(3,-4)$ , $B'(6,2)$ , $A''(-3,4)$ , $B''(0,6)$ .

Question 16: Plot $A(3,5)$ , $B(-2,-4)$ and find images

Step 1: Reflect $A(3,5)$ in the x-axis.

$A(3,5)\to A'(3,-5)$

Step 2: Reflect $B(-2,-4)$ in the y-axis.

$(-2,-4)\to(2,-4)$

Step 3: Reflect $(2,-4)$ in the origin.

$(2,-4)\to B'(-2,4)$

Step 4: The plotted figure $AA'BB'$ has one pair of parallel sides and equal non-parallel sides, so it is an isosceles trapezium.

Step 5: Points invariant under reflection in the x-axis have form $(x,0)$ .

Final answer: $A'(3,-5)$ , $B'(-2,4)$ ; figure $AA'BB'$ is an isosceles trapezium; examples of invariant points are $(5,0)$ and $(-17,0)$ .

Question 17: $P(5,3)$ reflected in the origin

Step 1: Reflect $P(5,3)$ in the origin.

$P(5,3)\to P'(-5,-3)$

Step 2: The foot from $P(5,3)$ to the x-axis has the same $x$ -coordinate and $y=0$ .

$M=(5,0)$

Step 3: The foot from $P'(-5,-3)$ to the x-axis is $N=(-5,0)$ .

Step 4: $PMP'N$ is a parallelogram.

Step 5: Use base $NM=10$ and perpendicular height $3$ .

$\text{Area}=10\times3=30$

Final answer: $P'(-5,-3)$ , $M(5,0)$ , $N(-5,0)$ , $PMP'N$ is a parallelogram, and its area is $30$ square units.

Test Yourself Solutions

Question 1(a): $(5,6)$ reflected to $(-5,6)$

Step 1: Only the $x$ -coordinate changes sign.

Step 2: This is reflection in the y-axis.

Final answer: Option $4$ : y-axis.

Question 1(b): $A\to B$ in y-axis, $B\to C$ in origin, $C\to P$ in $y=0$

Step 1: Let $A=(x,y)$ .

$A(x,y)\to B(-x,y)$

$B(-x,y)\to C(x,-y)$

$C(x,-y)\to P(x,y)$

Final answer: Option $1$ : $P$ coincides with point $A$ .

Question 1(c): $P(3,4)$ reflected in $y=x$

Step 1: Reflection in $y=x$ interchanges the coordinates.

$(3,4)\to(4,3)$

Step 2: Thus $a=4$ and $b=3$ .

$a-b=4-3=1,\qquad b-a=3-4=-1$

Final answer: Option $2$ : $a-b=1$ .

Question 1(d): $P(5,7)$ , $P'$ in x-axis, and $O'$ in line $PP'$

Step 1: Reflect $P(5,7)$ in the x-axis.

$P(5,7)\to P'(5,-7)$

Step 2: The line $PP'$ is the vertical line $x=5$ .

Step 3: Reflect the origin $O(0,0)$ in $x=5$ .

$O(0,0)\to O'(2(5)-0,0)=(10,0)$

Final answer: Option $1$ : $O'(10,0)$ .

Question 1(e): $P\to P'$ in $x=0$ , then $P'\to P''$ in $y=0$

Step 1: Let $P=(x,y)$ .

$P(x,y)\to P'(-x,y)\to P''(-x,-y)$

Step 2: For a general point, no two of $P$ , $P'$ , and $P''$ coincide.

Final answer: Option $4$ : no-one.

Question 1(f): Triangle reflected in y-axis and then in $y=0$

Step 1: A reflection preserves side lengths and angles.

Step 2: After two reflections, the final triangle is still congruent to the original triangle and to the intermediate triangle.

Step 3: Congruent triangles are also similar.

Final answer: Option $4$ : $\triangle ABC\ne \triangle A''B''C''$ is the statement that is not true.

Question 1(g): Assertion-reason on invariant transformation

Step 1: If the image of $M(x,y)$ in line $AB$ is $M$ itself, then $M$ is invariant.

Step 2: The reason correctly explains the assertion because an invariant point is its own image.

Final answer: Option $3$ : Both A and R are true, and R is the correct reason for A.

Question 1(h): Triangle reflected in the origin

Step 1: Reflection in the origin maps $(x,y)\to(-x,-y)$ .

Step 2: Reflection preserves lengths and angles, so the image triangle is congruent to the original triangle.

Step 3: Every pair of congruent triangles is also similar.

Final answer: Option $1$ : Both statements are true.

Question 1(i): Invariant points $(-5,1)$ and $(4,1)$

Step 1: Both given invariant points have the same $y$ -coordinate, $1$ .

Step 2: Therefore they lie on the horizontal line $y=1$ , not $x=1$ .

Step 3: The definition of invariant point is correct: a point is invariant if its image is itself.

Final answer: Option $4$ : Statement 1 is false, Statement 2 is true.

Question 2: $A(4,-1)$ reflected in y-axis; $B$ reflected in x-axis to $B'(-2,5)$

Step 1: Reflect $A(4,-1)$ in the y-axis.

$A(4,-1)\to A'(-4,-1)$

Step 2: Let $B=(a,b)$ . Reflection in x-axis gives $B'=(a,-b)$ .

$(a,-b)=(-2,5)$

$a=-2,\qquad b=-5$

Final answer: $A'(-4,-1)$ and $B(-2,-5)$ .

Question 3: Name the reflection line and image of $(5,-8)$

Step 1: The mappings $(-5,0)\to(5,0)$ and $(-2,-6)\to(2,-6)$ change only the $x$ -coordinate sign.

Step 2: Hence the reflection line is the y-axis, $x=0$ .

Step 3: Reflect $(5,-8)$ in the y-axis.

$(5,-8)\to(-5,-8)$

Final answer: (a) y-axis, $x=0$ ; (b) $(-5,-8)$ .

Question 4: $P(3,4)$ , $P'$ in x-axis and $O'$ in line $PP'$

Step 1: Reflect $P(3,4)$ in the x-axis.

$P(3,4)\to P'(3,-4)$

Step 2: The line $PP'$ is $x=3$ . Reflect $O(0,0)$ in $x=3$ .

$O(0,0)\to O'(6,0)$

Step 3: Find $PP'$ and $OO'$ .

$PP'=|4-(-4)|=8,\qquad OO'=|0-6|=6$

Step 4: Find side length $PO$ .

$PO=\sqrt{(3-0)^2+(4-0)^2}=\sqrt{9+16}=5$

Step 5: By symmetry, all sides of $POP'O'$ are $5$ units.

$\text{Perimeter}=4\times5=20$

Final answer: (i) $P'(3,-4)$ , $O'(6,0)$ ; (ii) $PP'=8$ units, $OO'=6$ units; (iii) perimeter $=20$ units; (iv) $POP'O'$ is a rhombus.

Question 5: Quadrilateral $ABCD$ reflected in origin

Step 1: The given points are $A(1,1)$ , $B(5,1)$ , $C(4,2)$ , and $D(2,2)$ . Since $AB\parallel DC$ and the non-parallel sides are equal, $ABCD$ is an isosceles trapezium.

Step 2: Reflection in the origin changes both coordinate signs.

$A(1,1)\to A'(-1,-1)$

$B(5,1)\to B'(-5,-1)$

$C(4,2)\to C'(-4,-2)$

$D(2,2)\to D'(-2,-2)$

Step 3: Points $D(2,2)$ , $A(1,1)$ , $A'(-1,-1)$ , and $D'(-2,-2)$ all lie on $y=x$ .

Final answer: $ABCD$ is an isosceles trapezium; $A'(-1,-1)$ , $B'(-5,-1)$ , $C'(-4,-2)$ , $D'(-2,-2)$ ; yes, $D,A,A',D'$ are collinear.

Question 6: $P(0,5)$ invariant in an axis and images of $Q(-2,4)$

Step 1: $P(0,5)$ lies on the y-axis, so it is invariant under reflection in the y-axis.

Step 2: Reflect $Q(-2,4)$ in the y-axis.

$Q(-2,4)\to(2,4)$

Step 3: For $(0,k)$ to be invariant under reflection in the origin, it must equal its image.

$(0,k)\to(0,-k),\qquad (0,k)=(0,-k)\Rightarrow k=0$

Step 4: Reflect $Q(-2,4)$ in the origin, then in the x-axis.

$(-2,4)\to(2,-4)\to(2,4)$

Final answer: (a) y-axis; (b) $(2,4)$ ; (c) $k=0$ ; (d) $(2,4)$ .

Question 7: $P(2,-4)$ , $Q$ , $R$ , figure and area

Step 1: Reflect $P(2,-4)$ in $x=0$ .

$P(2,-4)\to Q(-2,-4)$

Step 2: Reflect $Q(-2,-4)$ in $y=0$ .

$Q(-2,-4)\to R(-2,4)$

Step 3: $PQ$ is horizontal and $QR$ is vertical, so $PQR$ is a right-angled triangle.

$PQ=4,\qquad QR=8$

$\text{Area}=\frac{1}{2}\times4\times8=16$

Final answer: $Q(-2,-4)$ , $R(-2,4)$ , $PQR$ is a right-angled triangle, and its area is $16$ square units.

Question 8: $A(6,4)$ , $B(0,4)$ , reflection in origin and perimeter

Step 1: Reflect $A$ and $B$ in the origin.

$A(6,4)\to A'(-6,-4),\qquad B(0,4)\to B'(0,-4)$

Step 2: The figure $ABA'B'$ is a parallelogram.

Step 3: From coordinates, $AB=6$ and $A'B'=6$ .

Step 4: Find $BA'$ .

$BA'=\sqrt{(-6-0)^2+(-4-4)^2}=\sqrt{36+64}=10$

Step 5: Opposite sides of a parallelogram are equal.

$\text{Perimeter}=6+10+6+10=32$

Final answer: $A'(-6,-4)$ , $B'(0,-4)$ , $ABA'B'$ is a parallelogram, and its perimeter is $32$ units.

Question 9: Reflect $A(-4,4)$ and $B(-3,0)$ in y-axis

Step 1: Reflection in the y-axis changes $x$ to $-x$ .

$A(-4,4)\to A'(4,4),\qquad B(-3,0)\to B'(3,0)$

Step 2: The closed figure $OABCB'A'$ , from the plotted points, is an arrowhead-shaped figure.

Step 3: The y-axis divides it into two matching halves.

Final answer: $A'(4,4)$ , $B'(3,0)$ ; figure $OABCB'A'$ is an arrowhead; line of symmetry is the y-axis, $x=0$ .

Question 10: Reflect $B(2,3)$ , $C(1,1)$ , $D(2,0)$ in y-axis

Step 1: Reflect each point in the y-axis using $(x,y)\to(-x,y)$ .

$B(2,3)\to B'(-2,3)$

$C(1,1)\to C'(-1,1)$

$D(2,0)\to D'(-2,0)$

Step 2: The closed figure $A-B-C-D-D'-C'-B'-A$ has the y-axis as its fold line.

Final answer: $B'(-2,3)$ , $C'(-1,1)$ , $D'(-2,0)$ ; the required line is $x=0$ .

Question 11: Reflect $B,C,D,E$ on the y-axis and name the closed figure

Step 1: Use reflection in the y-axis: $(x,y)\to(-x,y)$ .

$B(2,5)\to B'(-2,5)$

$C(5,2)\to C'(-5,2)$

$D(5,-2)\to D'(-5,-2)$

$E(2,-5)\to E'(-2,-5)$

Step 2: The plotted closed figure has eight sides.

Final answer: $B'(-2,5)$ , $C'(-5,2)$ , $D'(-5,-2)$ , $E'(-2,-5)$ ; the closed figure is an octagon with opposite sides equal.

Quick Answer Index

Section	Question	Answer / status
Exercise 12	1(a)	Option $1$ : on the line $l$
Exercise 12	1(b)	$(7,8)$
Exercise 12	1(c)	$(6,-6)$
Exercise 12	1(d)	$(-4,-8)$
Exercise 12	1(e)	$(5,5)$
Exercise 12	2	$(6,4)$ , $(0,5)$ , $(-3,-4)$
Exercise 12	3	$(-3,0)$ , $(8,5)$ , $(-1,3)$
Exercise 12	4	$P(-4,-5)$ , image $(4,-5)$
Exercise 12	5	$P(2,-7)$ , image $(2,7)$
Exercise 12	6	$a=4,\ b=-6$
Exercise 12	7	$A''(3,2)$ , reflection in y-axis
Exercise 12	8	$A''(2,-6)$ , $B''(-3,-5)$ , $C''(4,-7)$ ; reflection in x-axis
Exercise 12	9	Isosceles trapezium; $45^\circ$ ; $A''(-3,-2)$ ; y-axis
Exercise 12	10	$L_1:y=0$ , $L_2:x=0$ ; images as solved above
Exercise 12	11	Mirror line $x=0$ ; image $(8,-5)$
Exercise 12	12	$P'(4,5)$ , $Q'(-2,2)$
Exercise 12	13	$P'(6,3)$
Exercise 12	14	$a=2$ , $b=3$ , $P''(-2,3)$ , $P'''(6,3)$
Exercise 12	15	$A'(3,-4)$ , $B'(6,2)$ , $A''(-3,4)$ , $B''(0,6)$
Exercise 12	16	$A'(3,-5)$ , $B'(-2,4)$ , isosceles trapezium, invariant examples $(5,0)$ , $(-17,0)$
Exercise 12	17	$P'(-5,-3)$ , $M(5,0)$ , $N(-5,0)$ , area $30$ square units
Test Yourself	1(a)-1(i)	Options $4,1,2,1,4,4,3,1,4$
Test Yourself	2	$A'(-4,-1)$ , $B(-2,-5)$
Test Yourself	3	y-axis; $(-5,-8)$
Test Yourself	4	$P'(3,-4)$ , $O'(6,0)$ , $PP'=8$ , $OO'=6$ , perimeter $20$ , rhombus
Test Yourself	5	Isosceles trapezium; $A'(-1,-1)$ , $B'(-5,-1)$ , $C'(-4,-2)$ , $D'(-2,-2)$ ; collinear: yes
Test Yourself	6	y-axis; $(2,4)$ ; $k=0$ ; $(2,4)$
Test Yourself	7	$Q(-2,-4)$ , $R(-2,4)$ , right-angled triangle, area $16$
Test Yourself	8	$A'(-6,-4)$ , $B'(0,-4)$ , parallelogram, perimeter $32$
Test Yourself	9	$A'(4,4)$ , $B'(3,0)$ , arrowhead, y-axis
Test Yourself	10	$B'(-2,3)$ , $C'(-1,1)$ , $D'(-2,0)$ , line $x=0$
Test Yourself	11	$B'(-2,5)$ , $C'(-5,2)$ , $D'(-5,-2)$ , $E'(-2,-5)$ , octagon

Examiner’s Mindset: How to Present Reflection Answers

In coordinate-geometry reflection questions, marks are usually earned through correct transformation rules, correct substitution, and a clear final coordinate. For a graph-based question, the graph helps, but the written answer should still state the rule or the coordinate reason. For example, in reflection in $x=4$ , writing only $P'''(6,3)$ is weaker than showing $x'=2(4)-2=6$ . The short calculation proves that the image is equally far from the mirror line.

For geometry names such as trapezium, parallelogram or rhombus, state the coordinate evidence: parallel sides, equal sides, perpendicular height or equal distances. This is especially useful when the figure is not drawn to scale.

Common Mistakes Students Make in Reflection

Confusing x-axis and y-axis: Reflection in the x-axis changes $y$ , not $x$ . Reflection in the y-axis changes $x$ , not $y$ .
Treating $x=a$ like the y-axis: For $x=a$ , use $x'=2a-x$ . Do not simply change the sign of $x$ unless $a=0$ .
Forgetting that invariant points lie on the mirror line: Under reflection in $y=3$ , every point with $y=3$ is invariant; not every point on an axis is invariant.
Skipping the second reflection: In combined transformations, write the image after each step. This prevents sign errors.
Using a graph without calculation: For coordinates, calculation is safer than judging the diagram by sight.

After reflection, revise nearby coordinate geometry and formula-heavy chapters. The Section and Mid-Point Formula solutions strengthen coordinate calculations, while Heights and Distances solutions help with diagram-based reasoning. For formula proof practice, use the Trigonometrical Identities solutions.

Frequently Asked Questions

What is the fastest way to reflect a point in the x-axis or y-axis?

For the x-axis, keep the x-coordinate and change the sign of the y-coordinate: $(x,y) \to (x,-y)$ . For the y-axis, change the sign of the x-coordinate and keep the y-coordinate: $(x,y) \to (-x,y)$ .

How do I reflect a point in the line x = a in ICSE Class 10 Maths?

Use the rule $(x,y) \to (2a-x,y)$ . The y-coordinate stays the same because the mirror line is vertical, and the reflected point must be the same horizontal distance from $x=a$ .

Which points are invariant under reflection in a line?

The invariant points are exactly the points that lie on the mirror line. For example, under reflection in the x-axis, every point of the form $(x,0)$ is invariant.

Does reflection change the size of a triangle?

No. Reflection preserves lengths and angles, so the image triangle is congruent to the original triangle. Its orientation may be reversed, but the shape and size remain the same.

What should I write in graph-based reflection answers?

Write the plotted image coordinates, name the mirror line or transformation used, and show any required length, angle, perimeter or area calculation. Do not rely only on a diagram when a coordinate rule gives the answer.

ICSE Class 10 Maths Reflection Solutions: Chapter 12

Reflection Formula Table for Concise Mathematics Selina Chapter 12

Concept Snapshot: The Mirror-Distance Test

Worked Examples Before Exercise 12

Worked Example 1: Reflection in the line x=4

Worked Example 2: Two reflections as one transformation

Worked Example 3: Area after reflection in axes

Exercise 12 Solutions

Question 1(a): A point P is its own image in a line l

Question 1(b): P(-7,8) reflected in origin and then in x-axis

Question 1(c): Find P when it is reflected in the y-axis and then in the origin to get (6,6)

Question 1(d): Reflect P(4,-8) in x=0

Question 1(e): P(5,5) reflected in y=0 and then in x-axis

Question 2: Reflect points in the line x=0

Question 3: Reflect points in the line y=0

Question 4: Image of P in x-axis is (-4,5)

Question 5: Image of P in origin is (-2,7)

Question 6: P(a,b) reflected in origin and then in y-axis gives P'(4,6)

Question 7: A(-3,2) reflected in x-axis and then in origin

Question 8: Triangle reflected in y-axis and then in origin

Question 9: Graph of A(3,2), B(5,4) and their images in the x-axis

Question 10: Invariant points under L_1 and L_2

Question 11: Find the mirror line and image of (-8,-5)

Question 12: Reflect P(4,1) and Q(-2,4) in y=3

Question 13: Reflect P(-2,3) in x=2

Question 14: P(a,b) reflected in x-axis to P'(2,-3)

Question 15: Images of A(3,4) and B(0,2)

Question 16: Plot A(3,5), B(-2,-4) and find images

Question 17: P(5,3) reflected in the origin

Test Yourself Solutions

Question 1(a): (5,6) reflected to (-5,6)

Question 1(b): A\to B in y-axis, B\to C in origin, C\to P in y=0

Question 1(c): P(3,4) reflected in y=x

Question 1(d): P(5,7), P' in x-axis, and O' in line PP'

Question 1(e): P\to P' in x=0, then P'\to P'' in y=0

Question 1(f): Triangle reflected in y-axis and then in y=0