Concise Mathematics Selina Solutions Class 10 ICSE Chapter 12 Reflection (in x-axis, y-axis, x = a, y = a and the origin; Invariant Points)

Our detailed ICSE Class 10 Maths Reflection Solutions help you master the key concepts of transforming points in the coordinate plane. This chapter from the Concise Mathematics Selina textbook for Class 10 introduces you to the geometric transformation of reflection. We’ll explore how to find the image of a point when it is reflected across the x-axis, the y-axis, and the origin. You will also learn the rules for reflection in lines parallel to the axes, such as x = a and y = a. A very important concept covered here is that of invariant points—points that do not change their position after a reflection. Understanding these rules is crucial for building a strong foundation in coordinate geometry.

If you are stuck on a specific question about finding the coordinates of a reflected point or identifying invariant points, you’ve come to the right place. This page provides clear, step-by-step solutions for all 40 questions in Exercise 12 and the Test Yourself section of the Selina textbook. Each solution is worked out using the exact same methods and formulas prescribed by the ICSE board, ensuring you learn the correct way to present your answers in exams. Here, you will find reliable and easy-to-understand solutions for every problem in the chapter.

Exercise 12

Question 1(a)

A point P is its own image in a line l, then point P is :

• (a) on the line l
• (b) below the line l
• (c) above the line l
• (d) none of the above

When a point P serves as its own reflection across a line l, it means P does not move during the reflection. This characteristic makes P an invariant point, indicating that it must lie directly on the line l itself.

Hence, Option 1 is the correct option.

Question 1(b)

A point P(-7, 8) is first reflected in the origin and then in x-axis to get the point Q. The co-ordinates of point Q are :

• (a) (-7, 8)
• (b) (7, 8)
• (c) (7, -8)
• (d) (-7, -8)

Reflecting a point in the origin involves changing the sign of both the x and y coordinates. Thus, when point ( P(-7, 8) ) is reflected in the origin, it becomes ( P'(7, -8) ).

Next, reflecting a point in the x-axis requires changing only the sign of the y-coordinate. Therefore, when ( P'(7, -8) ) is reflected in the x-axis, it becomes ( Q(7, 8) ).

Hence, Option 2 is the correct option.

Question 1(c)

A point P is reflected first in the y-axis and then in the origin to get the point (6, 6). The co-ordinates of the point P are :

• (a) (6, -6)
• (b) (-6, -6)
• (c) (-6, 6)
• (d) (6, 6)

Assume the coordinates of point P are (x, y) and after two reflections, the coordinates become P” = (6, 6).

First, consider the reflection of P in the y-axis. This reflection changes the sign of the x-coordinate, so P becomes P’ = (-x, y).

Next, this point P’ is reflected in the origin. This reflection changes the sign of both coordinates, resulting in P” = (x, -y).

Given P” = (6, 6), equate the coordinates:

⇒ (x, -y) = (6, 6)

From this, we get:

⇒ x = 6 and -y = 6

Solving for y gives:

⇒ y = -6

Thus, the coordinates of point P are (x, y) = (6, -6).

Hence, Option 1 is the correct option.

Question 1(d)

The point P(4, -8) is reflected in the line x = 0 to get the point R. The co-ordinates of point R are :

• (a) (4, -8)
• (b) (-4, 8)
• (c) (4, 8)
• (d) (-4, -8)

The line x = 0 represents the y-axis.

When a point is reflected across the y-axis, the x-coordinate changes its sign, while the y-coordinate remains unchanged.

Given that point P(4, -8) is reflected over the line x = 0 (which is the y-axis), the new coordinates of point R will be:

∴ P(4, -8) becomes R(-4, -8).

Hence, Option 4 is the correct option.

Question 1(e)

The point P(5, 5) is first reflected in line y = 0 and then in x-axis to get the point Q. The co-ordinates of point Q are :

• (a) (5, 5)
• (b) (-5, 5)
• (c) (5, -5)
• (d) (-5, -5)

The initial point given is P(5, 5), which undergoes two reflections.

First, consider the reflection across the line y = 0, which is indeed the x-axis. When reflecting a point across the x-axis, the y-coordinate changes its sign while the x-coordinate remains unchanged. Therefore, after the first reflection, point P becomes P'(5, -5).

Next, reflect point P'(5, -5) again across the x-axis. Applying the same rule, the y-coordinate changes its sign once more, resulting in point Q(5, 5).

Notice that after two successive reflections over the x-axis, the point returns to its original position.

∴ Option 1 is the correct option.

Question 2

State the co-ordinates of the following points under reflection in the line x = 0 :

(i) (-6, 4)

(ii) (0, 5)

(iii) (3, -4)

The equation x = 0 corresponds to the y-axis.

Thus, reflecting a point across the line x = 0 is the same as reflecting it across the y-axis.

For a reflection over the y-axis, the transformation is defined by:

M_y(x, y) = (-x, y) ……….(1)

(i) Applying this transformation to the point (-6, 4):

M_y(-6, 4) = (6, 4).

Therefore, the point (-6, 4) becomes (6, 4) when reflected over the line x = 0.

(ii) Applying the same to (0, 5):

M_y(0, 5) = (0, 5).

Therefore, the point (0, 5) remains unchanged as (0, 5) when reflected over the line x = 0.

(iii) Applying the transformation to (3, -4):

M_y(3, -4) = (-3, -4).

Therefore, the point (3, -4) becomes (-3, -4) when reflected over the line x = 0.

Question 3

State the co-ordinates of the following points under reflection in the line y = 0 :

(i) (-3, 0)

(ii) (8, -5)

(iii) (-1, -3)

The x-axis is represented by the line y = 0. Thus, reflecting a point across this line results in a reflection over the x-axis.

When reflecting a point (x, y) over the x-axis, the new coordinates become (x, -y). This transformation can be represented as:

M~x(x, y) = (x, -y) ……….(1)

(i) For the point (-3, 0), applying equation 1:

M~x(-3, 0) = (-3, 0).

Thus, the coordinates of (-3, 0) after reflection in the line y = 0 remain (-3, 0).

(ii) For the point (8, -5), using equation 1:

M~x(8, -5) = (8, 5).

Thus, the coordinates of (8, -5) after reflection in the line y = 0 are (8, 5).

(iii) For the point (-1, -3), applying equation 1:

M~x(-1, -3) = (-1, 3).

Thus, the coordinates of (-1, -3) after reflection in the line y = 0 are (-1, 3).

Question 4

A point P is reflected in the x-axis. Co-ordinates of its image are (-4, 5).

(i) Find the co-ordinates of P.

(ii) Find the co-ordinates of the image of P under reflection in the y-axis.

(i) When reflecting a point across the x-axis, the transformation is described by the rule:

M_x(x, y) = (x, -y) ………(1)

We know the image of point P(x, y) after reflection over the x-axis is (-4, 5).

By comparing with equation (1), we have:

⇒ x = -4 and -y = 5

⇒ x = -4 and y = -5.

Thus, the coordinates of P are (-4, -5).

Hence, co-ordinates of P = (-4, -5).

(ii) For reflection across the y-axis, the transformation is given by:

Substituting (-4, -5) into this equation, we get:

M_y(-4, -5) = (4, -5).

Hence, co-ordinates of the image of P under reflection in the y-axis = (4, -5).

Question 5

A point P is reflected in the origin. Co-ordinates of its image are (-2, 7).

(i) Find the co-ordinates of P.

(ii) Find the co-ordinates of the image of P under reflection in the x-axis.

(i) To find the reflection of a point in the origin, we use the formula:

M_o(x, y) = (-x, -y) ……….(1)

We’re given that the image of point P(x, y) after reflection in the origin is (-2, 7).

By comparing with equation (1), we have:

This results in two equations:

-x = -2 and -y = 7

Solving these, we find:

x = 2 and y = -7.

Thus, the coordinates of P are (x, y) = (2, -7).

Hence, coordinates of P = (2, -7).

(ii) For reflection in the x-axis, the formula is:

Substituting the coordinates (2, -7) into this formula gives:

M_x(2, -7) = (2, 7).

Hence, coordinates of the image of P under reflection in the x-axis = (2, 7).

Question 6

The point P(a, b) is first reflected in the origin and then reflected in the y-axis to P’. If P’ has co-ordinates (4, 6); evaluate a and b.

Consider the point P(a, b). When P is reflected across the origin, the new coordinates become (-a, -b). Next, reflect this new point across the y-axis. This reflection results in the coordinates (a, -b).

We know that after these two reflections, the coordinates are given as (4, 6). By matching these with the reflection result, we have:

⇒ (a, -b) = (4, 6)

From this, we deduce:

⇒ a = 4 and -b = 6

⇒ a = 4 and b = -6.

Hence, a = 4 and b = -6.

Question 7

The point A(-3, 2) is reflected in the x-axis to the point A’. Point A’ is then reflected in the origin to point A”.

(i) Write down the co-ordinates of A”.

(ii) Write down a single transformation that maps A onto A”.

(i) When reflecting a point across the x-axis, the transformation rule is:

∴ Reflecting point A(-3, 2) over the x-axis results in A'(-3, -2).

Next, reflecting a point through the origin involves the rule:

∴ Reflecting A'(-3, -2) through the origin gives us A''(3, 2).

Hence, co-ordinates of A'' = (3, 2).

(ii) To map A directly to A'', we observe:

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

The transformation rule for reflection across the y-axis is:

∴ Reflecting A(-3, 2) across the y-axis results in A''(3, 2).

Hence, the single transformation that maps A onto A'' is reflection in the y-axis.

Question 8

The triangle ABC, where A is (2, 6), B is (-3, 5) and C is (4, 7), is reflected in the y-axis to triangle A’B’C’. Triangle A’B’C’ is then reflected in the origin to triangle A”B”C”.

(i) Write down the co-ordinates of A”, B” and C”.

(ii) Write down a single transformation that maps triangle ABC onto triangle A”B”C”.

(i) Let’s determine the coordinates after each reflection.

First, reflect point A over the y-axis:

• A(2, 6) becomes A'(-2, 6).

Next, reflect A’ over the origin:

• A'(-2, 6) turns into A”(2, -6).

For point B, reflect over the y-axis:

• B(-3, 5) becomes B'(3, 5).

Then, reflect B’ over the origin:

• B'(3, 5) changes to B”(-3, -5).

For point C, reflect over the y-axis:

• C(4, 7) becomes C'(-4, 7).

Finally, reflect C’ over the origin:

• C'(-4, 7) results in C”(4, -7).

Therefore, the coordinates of A” = (2, -6), B” = (-3, -5), C” = (4, -7).

(ii) Now, consider the overall transformation from triangle ABC to triangle A”B”C”.

• A(2, 6) becomes A”(2, -6),
• B(-3, 5) becomes B”(-3, -5),
• C(4, 7) becomes C”(4, -7).

This transformation is equivalent to reflecting the entire triangle ABC across the x-axis.

Thus, reflecting in the x-axis maps triangle ABC onto triangle A”B”C”.

Question 9

Attempt this question on graph paper.

(a) Plot A (3, 2) and B (5, 4) on graph paper. Take 2 cm = 1 unit on both the axes.

(b) Reflect A and B in the x-axis to A’ and B’ respectively. Plot these points also on the same graph paper.

(c) Write down :

(i) the geometrical name of the figure ABB’A’;

(ii) the measure of angle ABB’;

(iii) the image A” of A, when A is reflected in the origin.

(iv) the single transformation that maps A’ to A”.

The graph is depicted as follows:

(c) (i) Observing the figure, we can identify that ABB’A’ is an isosceles trapezium.

(ii) Upon measuring the angle, we find that ∠ABB’ is 45°.

Thus, ∠ABB’ = 45°.

(iii) Looking at the reflection of point A in the origin, we see that A(3, 2) transforms to A”(-3, -2).

Therefore, the coordinates of A” are (-3, -2).

(iv) Examining the figure, when A’ is reflected across the y-axis, it maps to A”.

Thus, the reflection of A’ in the y-axis maps A’ to A”.

Question 10

Points (3, 0) and (-1, 0) are invariant points under reflection in the line L~1; points (0, -3) and (0, 1) are invariant points on reflection in line L~2.

(i) Name and write equations for the lines L~1 and L~2.

(ii) Write down the images of points P(3, 4) and Q(-5, -2) on reflection in L~1. Name the images as P’ and Q’ respectively.

(iii) Write down the images of P and Q on reflection in L~2. Name the images as P” and Q” respectively.

(iv) State or describe a single transformation that maps P’ onto P”.

(i) Points that lie on a line remain unchanged when reflected over the same line.

Notice that (3, 0) and (-1, 0) are on the x-axis. ∴ they are unchanged when reflected over the x-axis.

Thus, L~1 is the x-axis.

Similarly, (0, -3) and (0, 1) are on the y-axis. ∴ they are unchanged when reflected over the y-axis.

Thus, L~2 is the y-axis.

Hence, L~1 = x-axis whose equation is y = 0 and L~2 = y-axis whose equation is x = 0.

(ii) Since L~1 is the x-axis, reflecting a point over the x-axis involves changing the sign of the y-coordinate.

The formula for reflection over the x-axis is:

M~x(x, y) = (x, -y)

∴ Reflecting P(3, 4) over L~1 gives P'(3, -4).

Similarly, reflecting Q(-5, -2) over L~1 gives Q'(-5, 2).

Hence, co-ordinates of P’ = (3, -4) and Q’ = (-5, 2).

(iii) Since L~2 is the y-axis, reflecting a point over the y-axis involves changing the sign of the x-coordinate.

The formula for reflection over the y-axis is:

M~y(x, y) = (-x, y)

∴ Reflecting P(3, 4) over L~2 gives P”(-3, 4).

Similarly, reflecting Q(-5, -2) over L~2 gives Q”(5, -2).

Hence, co-ordinates of P” = (-3, 4) and Q” = (5, -2).

(iv) Consider P’ = (3, -4) and P” = (-3, 4).

To transform P'(3, -4) to P”(-3, 4), we change the signs of both coordinates.

This transformation is achieved by reflecting over the origin.

Hence, reflection in origin maps P’ onto P”.

Question 11

The point (-2, 0) on reflection in a line is mapped to (2, 0) and the point (5, -6) on reflection in the same line is mapped to (-5, -6).

(i) State the name of mirror line and write its equation.

(ii) State the co-ordinates of the image of (-8, -5) in the mirror line.

(i) Consider the transformations:

(-2, 0) becomes (2, 0) and (5, -6) becomes (-5, -6).

Observe how the sign of the x-coordinate changes in each case. This indicates reflection across the y-axis.

Thus, the mirror line is the y-axis, represented by the equation x = 0.

(ii) The reflection formula for a point across the y-axis is:

M~y(x, y) = (-x, y)

∴ The image of (-8, -5) when reflected over the y-axis (mirror line) is (8, -5).

Therefore, the co-ordinates of the image of (-8, -5) in the mirror line are (8, -5).

Question 12

The points P(4, 1) and Q(-2, 4) are reflected in line y = 3. Find the co-ordinates of P’, the image of P and Q’, the image of Q.

The line given, y = 3, is a horizontal line parallel to the x-axis, situated 3 units above it. In the diagram, this line is represented by AB.

Steps for finding the reflection:

1. Plot the points P(4, 1) and Q(-2, 4) on the coordinate plane.
2. For point P, draw a line perpendicular to AB (the line y = 3) and extend it.
3. Identify point P’ on this line such that it is equidistant from AB as P(4, 1) is, but on the opposite side.
4. Similarly, from point Q, draw a line perpendicular to AB and extend it.
5. Locate point Q’ on this line so that it is at the same distance from AB as Q(-2, 4), but on the opposite side.

Here’s a look at the graph:

From the graph, we determine:

P’ = (4, 5) and Q’ = (-2, 2)

Thus, the coordinates of P’ are (4, 5) and those of Q’ are (-2, 2).

Question 13

A point P (-2, 3) is reflected in the line x = 2 to point P’. Find the co-ordinates of P’.

The line x = 2 is a vertical line parallel to the y-axis, located 2 units away from it. In the diagram, this line is denoted as AB.

To find the image of point P(-2, 3) when reflected across this line, follow these steps:

1. Plot the point P(-2, 3) on the coordinate plane.
2. Draw a line from P that is perpendicular to the line AB and extend it.
3. Identify point P’ on this extended line such that it is equidistant from AB as point P is, but on the opposite side.

According to the graph:

P’ = (6, 3).

Therefore, the coordinates of P’ are (6, 3).

Question 14

A point P(a, b) is reflected in the x-axis to P'(2, -3). Write down the values of a and b. P” is the image of P, reflected in the y-axis. Write down the co-ordinates of P”. Find the co-ordinates of P”’, when P is reflected in the line, parallel to y-axis, such that x = 4.

When a point is reflected over the x-axis, the y-coordinate changes its sign, while the x-coordinate remains the same.

Given that the reflection of point P(a, b) in the x-axis is P'(2, -3), we can compare:

⇒ a = 2 and -b = -3

⇒ a = 2 and b = 3.

Thus, the original point P(a, b) is (2, 3) and its reflection in the x-axis is P'(2, -3).

For reflection in the y-axis, the x-coordinate changes its sign, while the y-coordinate remains unchanged.

The reflection of point P(2, 3) in the y-axis gives us P''(-2, 3).

Now, considering the reflection in a line parallel to the y-axis, specifically x = 4, note that this line is 4 units away from the y-axis. To find the reflection of P(2, 3) across this line:

1. Plot P(2, 3) on the graph.
2. Draw a perpendicular line from P to the line x = 4 and extend it.
3. Mark the point P''' on this line such that the distance from x = 4 to P''' is equal to the distance from x = 4 to P(2, 3).

From the graph, the coordinates of P''' are (6, 3).

Hence, a = 2, b = 3, P'' = (-2, 3) and P''' = (6, 3).

Question 15

Points A and B have co-ordinates (3, 4) and (0, 2) respectively. Find the image :

(a) A’ of A under reflection in the x-axis.

(b) B’ of B under reflection in the line AA’

(c) A” of A under reflection in the y-axis.

(d) B” of B under reflection in the line AA”.

The graph is shown below:

To find the images of points A and B under various reflections, follow these steps:

(i) When reflecting point A with coordinates (3, 4) in the x-axis, the y-coordinate changes sign. Thus, the image A’ has coordinates (3, -4).

(ii) For point B with coordinates (0, 2), reflecting it in the line AA’ (which connects A and A’) involves finding the perpendicular bisector of AA’. The reflection results in the image B’ having coordinates (6, 2).

(iii) Reflecting point A in the y-axis involves changing the sign of the x-coordinate. Therefore, the image A” has coordinates (-3, 4).

(iv) Finally, reflecting point B in the line AA” (which connects A and A”), we use the perpendicular bisector of AA”. This reflection gives the image B” with coordinates (0, 6).

Question 16

(i) Plot the points A(3, 5) and B(-2, -4). Use 1 cm = 1 unit on both the axes.

(ii) A’ is the image of A when reflected in the x-axis. Write down the co-ordinates of A’ and plot it on the graph paper.

(iii) B’ is the image of B when reflected in the y-axis, followed by reflection in the origin. Write down the co-ordinates of B’ and plot it on the graph paper.

(iv) Write down the geometrical name of the figure AA’BB’.

(v) Name two invariant points under reflection in the x-axis.

(i) The graph is depicted below:

(ii) Observing the graph, the coordinates of A’ after reflecting A over the x-axis are (3, -5).

(iii) Consider B” as the intermediate reflection of B over the y-axis. From the graph, the coordinates of B’ after reflecting B” over the origin are (-2, 4).

(iv) The shape formed by connecting the points AA’BB’ is identified as an isosceles trapezium.

(v) Points that remain unchanged under reflection over the x-axis are called invariant points. Such points lie directly on the x-axis, hence their ordinate is zero. Two examples of such invariant points are (5, 0) and (-17, 0).

Question 17

The point P(5, 3) was reflected in the origin to get the image P’.

(a) Write down the co-ordinates of P’.

(b) If M is the foot of the perpendicular from P to the x-axis, find the co-ordinates of M.

(c) If N is the foot of the perpendicular from P’ to the x-axis, find the co-ordinates of N.

(d) Name the figure PMP’N.

(e) Find the area of the figure PMP’N.

The diagram illustrates the reflection of the point P(5, 3) at the origin, resulting in the image P'.

(a) Observing the graph,

The coordinates of P' are (-5, -3).

(b) From the graph, the foot of the perpendicular from P to the x-axis is located at:

Coordinates of M are (5, 0).

(c) Similarly, the foot of the perpendicular from P' to the x-axis is:

Coordinates of N are (-5, 0).

(d) Examining the graph, the shape formed by the points P, M, P', and N is:

The figure PMP'N is a parallelogram.

(e) The diagonal NM splits the parallelogram into two congruent right-angled triangles.

∴ The area of △NMP is equal to the area of △NMP'.

Given that each block represents 1 unit, we find:

NM = 10 and NP = 3.

The area of △NMP is calculated as:

∴ The area of △NMP' also equals 15 sq. units.

Thus, the total area of the parallelogram PMP'N is:

Hence, the area of the parallelogram PMP'N is 30 sq. units.

Test Yourself

Question 1(a)

Point (5, 6) is reflected in a line to get the point (-5, 6). The line of reflection is :

• (a) x-axis
• (b) x = 5
• (c) y = 0
• (d) y-axis

When a point is reflected across the y-axis, the x-coordinate changes its sign while the y-coordinate remains the same.

∴ The point (5, 6) transforms into (-5, 6) when reflected over the y-axis.

Hence, Option 4 is the correct option.

Question 1(b)

(i) Point A is reflected in y-axis to get point B.

(ii) Point B is reflected in origin to get point C.

(iii) Point C is reflected in y = 0 to get point P. Now, which of the following coincides with point P.

• (a) Point A
• (b) Point B
• (c) Points B and C
• (d) Points A and B

Consider point A as ((x, y)).

Notice that when reflecting across the y-axis, the x-coordinate changes its sign. Thus, if point A is reflected over the y-axis to get point B, we have:

∴ (A(x, y) = B(-x, y))

Next, reflecting in the origin alters the signs of both coordinates. Hence, reflecting point B over the origin to get point C gives:

∴ (B(-x, y) = C(x, -y))

Now, reflecting across the x-axis (or y = 0) changes the sign of the y-coordinate. Thus, reflecting point C over y = 0 results in point P:

∴ (C(x, -y) = P(x, y))

As a result, point P coincides with point A.

Hence, Option 1 is the correct option.

Question 1(c)

The point P(3, 4) is reflected in the line y = x to point (a, b). Then:

• (a) a = b
• (b) a-b = 1
• (c) b – a = 1
• (d) none of these

Reflecting a point (x, y) over the line y = x involves swapping its coordinates, giving us the point (y, x).

For point P(3, 4), its reflection becomes (4, 3). Thus, we have a = 4 and b = 3.

Let’s examine each option:

1. Option (a): a = b implies 4 = 3, which is incorrect.

2. Option (b): a - b = 1 implies 4 \times 3 = 12, which is not equal to 1. So, this option is incorrect.

3. Option (c): b - a = 1 implies 3 - 4 = -1, which does not equal 1.

Hence, option 4 is the correct option.

Question 1(d)

The point P(5, 7) is reflected to P’ in x-axis and O’ is the image of O (origin) in the line PP’. The co-ordinates of O’ are :

• (a) (10, 0)
• (b) (0, 10)
• (c) (10, 10)
• (d) (5, 5)

When a point is reflected across the x-axis, the x-coordinate remains the same, while the y-coordinate changes its sign. Given point P(5, 7), its reflection P’ will have coordinates (5, -7).

To find the image of the origin O in the line formed by points P and P’, follow these steps:

1. Plot points P and P’ on a coordinate plane.
2. Connect P and P’ with a straight line, forming line PP’.
3. Draw a perpendicular from the origin O to line PP’ and extend it. The point O’ will lie on this line, equidistant from O.

From the diagram, the coordinates of O’ are determined to be (10, 0).

Hence, Option 1 is the correct option.

Question 1(e)

The point P is reflected in x = 0 to get the point P’ and the point P’ is reflected in y = 0 to get the point P”. Which two points out of P, P’ and P” are invariant under this reflection.

• (a) P” = P
• (b) P” = P’
• (c) P’ = P
• (d) no-one

Consider the coordinates of point P as ((x, y)).

Reflecting point P across the y-axis, which is the line x = 0, alters the x-coordinate’s sign. Thus, point P transforms into (P'(-x, y)).

Next, reflecting point P’ across the x-axis, which is the line y = 0, changes the y-coordinate’s sign. This results in point (P'(-x, y)) becoming (P”(-x, -y)).

Notice that none of the points P, P’, or P” share identical coordinates, indicating that there are no invariant points in this sequence of reflections.

Hence, Option 4 is the correct option.

Question 1(f)

A triangle ABC is reflected in y-axis to get triangle A’B’C’. Triangle A’B’C’ is reflected in line y = 0, to get △A”B”C”. Then which of the following is not true ?

• (a) △A’B’C’ ~ △A”B”C”
• (b) △A’B’C’ ≅ △A”B”C”
• (c) △ABC ≅ △A”B”C”
• (d) △ABC ≠ △A”B”C”

Reflections preserve the shape and size of figures, meaning they are isometries.

∴ The triangles involved are congruent.

Since congruent triangles maintain their shape and size, they are inherently similar.

Thus, the assertion △ABC ≠ △A”B”C” is incorrect.

Hence, Option 4 is the correct option.

Question 1(g)

Point M(x, y) is reflected in line AB, the reflection of M(x, y) in AB is the point M itself.

Assertion (A) : The reflection is called invariant transformation.

Reason (R) : In case of invariant transformation, the point is its own image.

• (a) A is true, R is false.
• (b) A is false, R is true.
• (c) Both A and R are true and R is correct reason for R.
• (d) Both A and R are true and R is incorrect reason for R.

When point M(x, y) is reflected over line AB and ends up as M itself, this indicates that M is unchanged by the reflection.

Such a reflection, where the point remains identical to its image, is termed an invariant transformation.

∴ Both the assertion and the reason are accurate, and the reason correctly explains the assertion.

Hence, option 3 is the correct option.

Question 1(h)

Δ ABC is reflected in origin to get Δ A’B’C’.

Statement 1: Δ ABC is congruent to Δ A’B’C’.

Statement 2: The two triangles are similar to each other.

• (a) Both the statement are true.
• (b) Both the statement are false.
• (c) Statement 1 is true, and statement 2 is false.
• (d) Statement 1 is false, and statement 2 is true.

Reflecting a point (x, y) through the origin results in the point (-x, -y). This transformation maintains all side lengths and angles, ensuring that the triangles’ size and shape remain unchanged, although the orientation is flipped.

For Statement 1, Δ ABC is indeed congruent to Δ A’B’C’ because the reflection preserves both the lengths and angles of the triangle. Thus, Statement 1 holds true.

For Statement 2, Δ ABC is similar to Δ A’B’C’ as well. This is because congruent triangles, by definition, are always similar, sharing identical angles and proportional sides.

∴ Both statements are accurate. Hence, option 1 is the correct option.

Question 1(i)

Points (-5, 1) and (4, 1) are invariant points under reflection in the line L.

Statement 1: The equation of the line L is x = 1.

Statement 2: A point P is called an invariant point with respect to a given line L, if its image in the line L is the point P itself.

• (a) Both the statement are true.
• (b) Both the statement are false.
• (c) Statement 1 is true, and statement 2 is false.
• (d) Statement 1 is false, and statement 2 is true.

The points given, (-5, 1) and (4, 1), are described as invariant under reflection across line L. Notice that both these points share the same y-coordinate, 1. This means they lie on the line y = 1. When reflected over y = 1, they remain unchanged, confirming their invariance. Thus, statement 1, which claims the line is x = 1, is incorrect.

Statement 2 correctly defines an invariant point as one that remains unchanged after reflection over line L. This is indeed true by definition.

∴ Statement 1 is false, and statement 2 is true. Hence, option 4 is the correct option.

Question 2

Point A (4, -1) is reflected as A’ in the y-axis. Point B on reflection in the x-axis is mapped as B’ (-2, 5). Write the co-ordinates of A’ and B.

When reflecting a point across the y-axis, the transformation is given by:

M~y(x, y) = (-x, y)

∴ The reflection of point A (4, -1) over the y-axis results in the image A'(-4, -1).

Now, let’s determine the original coordinates of point B. Assume B = (a, b).

For reflection across the x-axis, the transformation is:

M~x(x, y) = (x, -y)

∴ The reflection of point B(a, b) over the x-axis gives the image B'(a, -b).

Given that B’ = (-2, 5), we have:

a = -2 and -b = 5

⇒ a = -2 and b = -5.

Hence, A’ = (-4, -1) and B = (-2, -5).

Question 3

The point (-5, 0) on reflection in a line is mapped as (5, 0) and the point (-2, -6) on reflection in the same line is mapped as (2, -6).

(a) Name the line of reflection.

(b) Write the co-ordinates of the image of (5, -8) in the line obtained in (a).

(a) When reflecting a point across the y-axis, the transformation is given by:

Observe that the point (-5, 0) becomes (5, 0), which matches this transformation.

∴ The reflection is across the y-axis.

Hence, y-axis is the line of reflection.

(b) Using the same reflection rule across the y-axis:

∴ The image of the point (5, -8) when reflected over the y-axis is (-5, -8).

Hence, image of (5, -8) in line of reflection = (-5, -8).

Question 4

The point P(3, 4) is reflected to P’ in the x-axis; and O’ is the image of O (the origin) when reflected in the line PP’. Write :

(i) the co-ordinates of P’ and O’,

(ii) the length of the segments PP’ and OO’,

(iii) the perimeter of the quadrilateral POP’O’,

(iv) the geometrical name of the figure POP’O’.

(i) When reflecting a point across the x-axis, the y-coordinate changes sign. Thus, the coordinates of P' become (3, -4). For the origin O, reflecting across the line PP' results in O' being at (6, 0).

(ii) To find the length of PP', apply the distance formula:

PP' = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}.

Substitute P(3, 4) and P'(3, -4) into the formula:

Now, for OO', substitute O(0, 0) and O'(6, 0):

Thus, PP' = 8 units and OO' = 6 units.

(iii) Calculate PO using the distance formula:

PO = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}.

Substitute P(3, 4) and O(0, 0):

Similarly, for P'O with P'(3, -4) and O(0, 0):

Since adjacent sides are equal, POP'O' forms a rhombus.

∴ Perimeter = PO + OP' + P'O' + O'P = 5 + 5 + 5 + 5 = 20.

Therefore, the perimeter of POP'O' is 20 units.

(iv) The quadrilateral POP'O' is classified as a rhombus.

Question 5

A(1, 1), B(5, 1), C(4, 2) and D(2, 2) are vertices of a quadrilateral. Name the quadrilateral ABCD. A, B, C, and D are reflected in the origin on to A’, B’, C’ and D’ respectively. Locate A’, B’, C’ and D’ on the graph sheet and write their co-ordinates. Are D, A, A’ and D’ collinear ?

Observing the graph, we see that quadrilateral ABCD forms an isosceles trapezium.

When we reflect each vertex across the origin, the new coordinates are:

• A' becomes (-1, -1)
• B' becomes (-5, -1)
• C' becomes (-4, -2)
• D' becomes (-2, -2)

Notice that the points D, A, A', and D' all lie on a straight line, confirming that they are collinear.

Question 6

P and Q have co-ordinates (0, 5) and (-2, 4).

(a) P is invariant when reflected in an axis. Name the axis.

(b) Find the image of Q on reflection in the axis found in (a).

(c) (0, k) on reflection in the origin is invariant. Write the value of k.

(d) Write the co-ordinates of the image of Q, obtained by reflecting it in the origin followed by reflection in x-axis.

(a) Notice that point P is located on the y-axis. A point remains unchanged when reflected over the axis on which it lies.

Hence, P is invariant on y-axis.

(b) Observing the graph, the reflection of point Q across the y-axis results in a new point, denoted as Q’.

Hence, coordinates of Q’ = (2, 4).

(c) For a point to be invariant under reflection through the origin, it must be its own image.

Therefore, the point (0, k) must coincide with the origin.

Hence, k = 0.

(d) Referring to the graph, reflecting Q first through the origin and then through the x-axis gives us a new point, Q’.

Hence, coordinates of the image after reflection in origin followed by reflection in x-axis is (2, 4).

Question 7

(a) The point P(2, -4) is reflected about the line x = 0 to get the image Q. Find the co-ordinates of Q.

(b) The point Q is reflected about the line y = 0 to get the image R. Find the co-ordinates of R.

(c) Name the figure PQR.

(d) Find the area of figure PQR.

(a) Reflecting point P(2, -4) across the line x = 0 (the y-axis), we find the image Q at (-2, -4).

(b) Next, reflecting point Q(-2, -4) across the line y = 0 (the x-axis), the coordinates of R become (-2, 4).

(c) The shape formed by points P, Q, and R is a right-angled triangle.

(d) Observing the graph, each block represents 1 unit. Thus, the lengths are:

• QR = 8 units
• PQ = 4 units

The area of a right-angled triangle is calculated using the formula:

Substituting the values:

Hence, area of PQR = 16 sq. unit.

Question 8

Using a graph paper, plot the points A(6, 4) and B(0, 4).

(a) Reflect A and B in the origin to get the images A’ and B’.

(b) Write the co-ordinates of A’ and B’.

(c) State the geometrical name for the figure ABA’B’.

(d) Find its perimeter.

(b) Observing the graph, the coordinates of the reflected points are A’ = (-6, -4) and B’ = (0, -4).

(c) From the graph, the shape formed by connecting the points ABA’B’ is a parallelogram.

(d) According to the graph, where 1 block represents 1 unit:

∴ A’B’ = AB = 6 units

∴ BB’ = 8 units

Using the Pythagorean theorem in the right triangle △A’B’B:

⇒ A’B^2 = A’B’^2 + BB’^2

⇒ A’B^2 = 6^2 + 8^2

⇒ A’B^2 = 36 + 64

⇒ A’B^2 = 100

⇒ A’B = 10 units.

Since ABA’B’ is a parallelogram, the opposite sides are equal:

∴ AB’ = A’B = 10 units

Thus, the perimeter of the parallelogram is calculated as follows:

Perimeter = A’B + BA + AB’ + B’A’ = 10 + 6 + 10 + 6 = 32 units.

Therefore, the perimeter is 32 units.

Question 9

Use graph paper for this question.

Plot the points O(0, 0), A(-4, 4), B(-3, 0) and C(0, -3)

(i) Reflect points A and B on the y-axis and name them A’ and B’ respectively. Write down their co-ordinates.

(ii) Name the figure OABCB’A’.

(iii) State the line of symmetry of this figure.

(i) Upon reflecting the points on the graph:

The coordinates for A’ become (4, 4), and for B’ they are (3, 0).

(ii) Observing the plotted points:

The shape formed by OABCB’A’ resembles an arrow head.

(iii) Analyzing the symmetry:

The y-axis acts as the line of symmetry, dividing the shape into two mirror-image halves.

Hence, y-axis is the line of symmetry.

Question 10

Use a graph paper for this question.

(i) Plot the following points :

A(0, 4), B(2, 3), C(1, 1) and D(2, 0).

(ii) Reflect the points B, C, D on the y-axis and write down their coordinates. Name the images as B’, C’, D’ respectively.

(iii) Join the points A, B, C, D, D’, C’, B’ and A in order, so as to form a closed figure. Write down the equation of the line about which if this closed figure obtained is folded, the two parts of the figure exactly coincide.

(i) Begin by plotting the points A(0, 4), B(2, 3), C(1, 1), and D(2, 0) on graph paper. You will see these points clearly marked on the coordinate plane.

(ii) Next, reflect the points B, C, and D across the y-axis. The coordinates of the reflected points are as follows: B’ = (-2, 3), C’ = (-1, 1), and D’ = (-2, 0). These new positions are the mirror images of the original points with respect to the y-axis.

(iii) Proceed to join the points in this sequence: A, B, C, D, D’, C’, B’, and finally back to A. This will create a closed figure on your graph. Notice that the y-axis acts as a line of symmetry for this shape.

Therefore, the equation of the line that serves as the axis of symmetry is x = 0. Folding the figure along this line results in perfect overlap of the two halves.

Question 11

Use a graph paper for this question taking 1 cm = 1 unit along both the x-axis and the y axis.

(i) Plot the points A(0, 5), B(2, 5), C(5, 2), D(5, -2), E(2, -5) and F(0, -5).

(ii) Reflect the points B, C, D and E on the y-axis and name them B’, C’, D’ and E’ respectively.

(iii) Write the co-ordinates of B’, C’, D’ and E’.

(iv) Name the close figure formed.

Upon examining the graph, the coordinates after reflecting the points on the y-axis are as follows: B’ becomes (-2, 5), C’ shifts to (-5, 2), D’ changes to (-5, -2), and E’ moves to (-2, -5).

The resulting closed figure is an octagon, characterized by having equal opposite sides.