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Selina Concise Mathematics - Part II Solution for Class 10 Mathematics Chapter 12 - Reflection (In x-axis, y-axis, x=a, y=a and the origin ; Invariant Points)

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Selina Textbook Solutions Chapter 12 - Reflection (In x-axis, y-axis, x=a, y=a and the origin ; Invariant Points)

Selina Textbook Solutions are a perfect way to ace your examination with high marks. These Textbook Solutions are extremely helpful for solving difficult questions in the ICSE Class 10 Mathematics exam. Our Selina Textbook Solutions are written by our subject experts. Find all the answers to the Selina textbook questions of Chapter 12 - Reflection (In x-axis, y-axis, x=a, y=a and the origin ; Invariant Points).

All Selina textbook questions of Chapter 12 - Reflection (In x-axis, y-axis, x=a, y=a and the origin ; Invariant Points) solutions are created in accordance with the latest ICSE syllabus. These free Textbook Solutions for ICSE Class 10 Selina Concise Mathematics will give you a deeper insight on the fundamentals in this chapter and will help you to score more marks in the final examination. ICSE Class 10 students can refer to these solutions while doing their homework and while studying and revising for the Mathematics exam. 

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Selina Concise Mathematics - Part II Solution for Class 10 Mathematics Chapter 12 - Reflection (In x-axis, y-axis, x=a, y=a and the origin ; Invariant Points) Page/Excercise 12(A)

Solution 1

Point

Transformation

Image

(5, -7)

Reflection in origin

(-5, 7)

(4, 2)

Reflection in x-axis

(4, -2)

(0, 6)

Reflection in y-axis

(0, 6)

(6, -6)

Reflection in origin

(-6, 6)

(4, -8)

Reflection in y-axis

(-4, -8)

Solution 2

Since, the point P is its own image under the reflection in the line l. So, point P is an invariant point.

Hence, the position of point P remains unaltered.

Solution 3

(i) (3, 2)

The co-ordinate of the given point under reflection in the x-axis is (3, -2).

(ii) (-5, 4)

The co-ordinate of the given point under reflection in the x-axis is (-5, -4).

(iii) (0, 0)

The co-ordinate of the given point under reflection in the x-axis is (0, 0).

Solution 4

(i) (6, -3)

The co-ordinate of the given point under reflection in the y-axis is (-6, -3).

(ii) (-1, 0)

The co-ordinate of the given point under reflection in the y-axis is (1, 0).

(iii) (-8, -2)

The co-ordinate of the given point under reflection in the y-axis is (8, -2).

Solution 5

(i) (-2, -4)

The co-ordinate of the given point under reflection in origin is (2, 4).

(ii) (-2, 7)

The co-ordinate of the given point under reflection in origin is (2, -7).

(iii) (0, 0)

The co-ordinate of the given point under reflection in origin is (0, 0).

Solution 6

(i) (-6, 4)

The co-ordinate of the given point under reflection in the line x = 0 is (6, 4).

(ii) (0, 5)

The co-ordinate of the given point under reflection in the line x = 0 is (0, 5).

(iii) (3, -4)

The co-ordinate of the given point under reflection in the line x = 0 is (-3, -4).

Solution 7

(i) (-3, 0)

The co-ordinate of the given point under reflection in the line y = 0 is (-3, 0).

(ii) (8, -5)

The co-ordinate of the given point under reflection in the line y = 0 is (8, 5).

(iii) (-1, -3)

The co-ordinate of the given point under reflection in the line y = 0 is (-1, 3).

Solution 8

(i) Since, Mx (-4, -5) = (-4, 5)

So, the co-ordinates of P are (-4, -5).

(ii) Co-ordinates of the image of P under reflection in the y-axis (4, -5).

Solution 9

(i) Since, MO (2, -7) = (-2, 7)

So, the co-ordinates of P are (2, -7).

(ii) Co-ordinates of the image of P under reflection in the x-axis (2, 7).

Solution 10

MO (a, b) = (-a, -b)

My (-a, -b) = (a, -b)

Thus, we get the co-ordinates of the point P' as (a, -b). It is given that the co-ordinates of P' are (4, 6).

On comparing the two points, we get,

a = 4 and b = -6

Solution 11

Mx (x, y) = (x, -y)

MO (x, -y) = (-x, y)

Thus, we get the co-ordinates of the point P' as (-x, y). It is given that the co-ordinates of P' are (-8, 5).

On comparing the two points, we get,

x = 8 and y = 5

Solution 12

(i) The reflection in x-axis is given by Mx (x, y) = (x, -y).

A' = reflection of A (-3, 2) in the x- axis = (-3, -2).

The reflection in origin is given by MO (x, y) = (-x, -y).

A'' = reflection of A' (-3, -2) in the origin = (3, 2)

(ii) The reflection in y-axis is given by My (x, y) = (-x, y).

The reflection of A (-3, 2) in y-axis is (3, 2).

Thus, the required single transformation is the reflection of A in the y-axis to the point A''.

Solution 13

(i) The reflection in origin is given by MO (x, y) = (-x, -y).

A' = reflection of A (4, 6) in the origin = (-4, -6)

The reflection in y-axis is given by My (x, y) = (-x, y).

A'' = reflection of A' (-4, -6) in the y-axis = (4, -6)

(ii) The reflection in x-axis is given by Mx (x, y) = (x, -y).

The reflection of A (4, 6) in x-axis is (4, -6).

Thus, the required single transformation is the reflection of A in the x-axis to the point A''.

Solution 14

(i) Reflection in y-axis is given by My (x, y) = (-x, y)

A' = Reflection of A (2, 6) in y-axis = (-2, 6)

Similarly, B' = (3, 5) and C' = (-4, 7)

Reflection in origin is given by MO (x, y) = (-x, -y)

A'' = Reflection of A' (-2, 6) in origin = (2, -6)

Similarly, B'' = (-3, -5) and C'' = (4, -7)

(ii) A single transformation which maps triangle ABC to triangle A''B''C'' is reflection in x-axis.

Solution 15

Reflection in x-axis is given by Mx (x, y) = (x, -y)

P' = Reflection of P(-2, 3) in x-axis = (-2, -3)

Reflection in y-axis is given by My (x, y) = (-x, y)

Q' = Reflection of Q(5, 4) in y-axis = (-5, 4)

Thus, the co-ordinates of points P' and Q' are (-2, -3) and (-5, 4) respectively.

Solution 16

The graph shows triangle ABC and triangle A'B'C' which is obtained when ABC is reflected in the origin.

Solution 17

(i) Mx . My (4, -6) = Mx (-4, -6) = (-4, 6)

Single transformation equivalent to Mx . My is MO.

(ii) My . Mx (4, -6) = My (4, 6) = (-4, 6)

Single transformation equivalent to My . Mx is MO.

(iii) MO . Mx (4, -6) = MO (4, 6) = (-4, -6)

Single transformation equivalent to MO . Mx is My.

(iv) Mx . MO (4, -6) = Mx (-4, 6) = (-4, -6)

Single transformation equivalent to Mx . MO is My.

(v) MO . My (4, -6) = MO (-4, -6) = (4, 6)

Single transformation equivalent to MO . My is Mx.

(vi) My . MO (4, -6) = My (-4, 6) = (4, 6)

Single transformation equivalent to Mx . MO is Mx.

From (iii) and (iv), it is clear that MO . Mx = Mx . MO.

From (v) and (vi), it is clear that My . MO = MO . My.

Solution 18

Reflection in y-axis is given by My (x, y) = (-x, y)

A' = Reflection of A(4, -1) in y-axis = (-4, -1)

Reflection in x-axis is given by Mx (x, y) = (x, -y)

B' = Reflection of B in x-axis = (-2, 5)

Thus, B = (-2, -5)

Solution 19

(a) We know that reflection in the line x = 0 is the reflection in the y-axis.

It is given that:

Point (-5, 0) on reflection in a line is mapped as (5, 0).

Point (-2, -6) on reflection in the same line is mapped as (2, -6).

Hence, the line of reflection is x = 0.

(b) It is known that My (x, y) = (-x, y)

Co-ordinates of the image of (5, -8) in the line x = 0 are (-5, -8).

Selina Concise Mathematics - Part II Solution for Class 10 Mathematics Chapter 12 - Reflection (In x-axis, y-axis, x=a, y=a and the origin ; Invariant Points) Page/Excercise 12(B)

Solution 1

(c)

(i) From graph, it is clear that ABB'A' is an isosceles trapezium.

(ii) The measure of angle ABB' is 45°.

(iii) A'' = (-3, -2)

(iv) Single transformation that maps A' to A" is the reflection in y-axis.

Solution 2

(i) We know that every point in a line is invariant under the reflection in the same line.

Since points (3, 0) and (-1, 0) lie on the x-axis.

So, (3, 0) and (-1, 0) are invariant under reflection in x-axis.

Hence, the equation of line L1 is y = 0.

Similarly, (0, -3) and (0, 1) are invariant under reflection in y-axis.

Hence, the equation of line L2 is x = 0.

(ii) P' = Image of P (3, 4) in L1 = (3, -4)

Q' = Image of Q (-5, -2) in L1 = (-5, 2)

(iii) P'' = Image of P (3, 4) in L2 = (-3, 4)

Q'' = Image of Q (-5, -2) in L2 = (5, -2)

(iv) Single transformation that maps P' onto P" is reflection in origin.

Solution 3

(i) We know Mx (x, y) = (x, -y)

P' (5, -2) = reflection of P (a, b) in x-axis.

Thus, the co-ordinates of P are (5, 2).

Hence, a = 5 and b = 2.

(ii) P" = image of P (5, 2) reflected in y-axis = (-5, 2)

(iii) Single transformation that maps P' to P" is the reflection in origin.

Solution 4

(i) We know reflection of a point (x, y) in y-axis is (-x, y).

Hence, the point (-2, 0) when reflected in y-axis is mapped to (2, 0).

Thus, the mirror line is the y-axis and its equation is x = 0.

(ii) Co-ordinates of the image of (-8, -5) in the mirror line (i.e., y-axis) are (8, -5).

Solution 5

The line y = 3 is a line parallel to x-axis and at a distance of 3 units from it.

Mark points P (4, 1) and Q (-2, 4).

From P, draw a straight line perpendicular to line CD and produce. On this line mark a point P' which is at the same distance above CD as P is below it.

The co-ordinates of P' are (4, 5).

Similarly, from Q, draw a line perpendicular to CD and mark point Q' which is at the same distance below CD as Q is above it.

The co-ordinates of Q' are (-2, 2).

Solution 6

The line x = 2 is a line parallel to y-axis and at a distance of 2 units from it.

Mark point P (-2, 3).

From P, draw a straight line perpendicular to line CD and produce. On this line mark a point P' which is at the same distance to the right of CD as P is to the left of it.

The co-ordinates of P' are (6, 3).

Solution 7

A point P (a, b) is reflected in the x-axis to P' (2, -3).

We know Mx (x, y) = (x, -y)

Thus, co-ordinates of P are (2, 3). Hence, a = 2 and b = 3.

P" = Image of P reflected in the y-axis = (-2, 3)

P''' = Reflection of P in the line (x = 4) = (6, 3)

Solution 8

(a) A' = Image of A under reflection in the x-axis = (3, -4)

(b) B' = Image of B under reflection in the line AA' = (6, 2)

(c) A" = Image of A under reflection in the y-axis = (-3, 4)

(d) B" = Image of B under reflection in the line AA" = (0, 6)

Solution 9

(i) The points A (3, 5) and B (-2, -4) can be plotted on a graph as shown.

(ii) A' = Image of A when reflected in the x-axis = (3, -5)

(iii) C = Image of B when reflected in the y-axis = (2, -4)

B' = Image when C is reflected in the origin = (-2, 4)

(iv) Isosceles trapezium

(v) Any point that remains unaltered under a given transformation is called an invariant.

Thus, the required two points are (3, 0) and (-2, 0).

Solution 10

(a) Co-ordinates of P' = (-5, -3)

(b) Co-ordinates of M = (5, 0)

(c) Co-ordinates of N = (-5, 0)

(d) PMP'N is a parallelogram.

(e) Are of PMP'N = 2 (Area of D PMN)

Solution 11

(i) Co-ordinates of P' and O' are (3, -4) and (6, 0) respectively.

(ii) PP' = 8 units and OO' = 6 units.

(iii) From the graph it is clear that all sides of the quadrilateral POP'O' are equal.

In right PO'Q,

PO' =

So, perimeter of quadrilateral POP'O' = 4 PO' = 4 5 units = 20 units

(iv) Quadrilateral POP'O' is a rhombus.

Solution 12

Quadrilateral ABCD is an isosceles trapezium.

Co-ordinates of A', B', C' and D' are A'(-1, -1), B'(-5, -1), C'(-4, -2) and D'(-2, -2) respectively.

It is clear from the graph that D, A, A' and D' are collinear.

Solution 13

(a) Any point that remains unaltered under a given transformation is called an invariant.

It is given that P (0, 5) is invariant when reflected in an axis. Clearly, when P is reflected in the y-axis then it will remain invariant. Thus, the required axis is the y-axis.

(b) The co-ordinates of the image of Q (-2, 4) when reflected in y-axis is (2, 4).

(c) (0, k) on reflection in the origin is invariant. We know the reflection of origin in origin is invariant. Thus, k = 0.

(d) Co-ordinates of image of Q (-2, 4) when reflected in origin = (2, -4)

Co-ordinates of image of (2, -4) when reflected in x-axis = (2, 4)

Thus, the co-ordinates of the point are (2, 4).

Solution 14

(i) P (2, -4) is reflected in (x = 0) y-axis to get Q.

P(2, -4) Q (-2, -4)

(ii) Q (-2, -4) is reflected in (y = 0) x-axis to get R.

Q (-2, -4) R (-2, 4)

(iii) The figure PQR is right angled triangle.

(iv) Area of

Solution 15

(a)

 

Solution 16

  

 

 i. A' = (4, 4) AND B' = (3, 0)

 ii. The figure is an arrow head.

 iii. The y-axis i.e. x = 0 is the line of symmetry of figure OABCB'A'.

TopperLearning provides step-by-step solutions for each question in each chapter. Access Chapter 12 - Reflection (In x-axis, y-axis, x=a, y=a and the origin ; Invariant Points)  for ICSE Class 10 Mathematics free of cost. The solutions are provided by our subject matter experts. Refer to our solutions for the Selina Concise Mathematics textbook to revise the whole chapter and clear your fundamentals before the examination. By referring to the solutions for this chapter and the others, we hope that you are able to write your exams well. 

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