Selina Concise Solutions for ICSE Class 10 Mathematics Chapter 12 Reflection

Exercise 12A

Question 1. Complete the following table:

Answer: The table shows different types of reflections and their results. When we reflect in the origin, both x and y change signs. When we reflect in x-axis, only y changes sign. When we reflect in y-axis, only x changes sign.
In simple words: A reflection is like looking at yourself in a mirror. The point moves to the opposite side of the line or origin. Each type of reflection has its own rule for changing the coordinates.

📝 Teacher's Note: Use a mirror to show students how reflection works. Place a point on one side and show where its reflection appears on the other side. This makes the concept very clear.

🎯 Exam Tip: Remember the rules: origin reflection changes both signs, x-axis reflection changes only y sign, y-axis reflection changes only x sign. Write the rule first, then apply it.

Question 2. A point P is its own image under the reflection in a line l. Describe the position of point P with respect to the line l.
Answer: Since the point P is its own image under the reflection in the line l, point P is an invariant point. Hence, the position of point P remains unaltered. This means point P lies on the line l itself.
In simple words: If a point looks exactly the same after reflection, it means the point is sitting right on the mirror line. Like when you put your finger on a mirror - the real finger and reflection finger are at the same place.

📝 Teacher's Note: Ask students to put their finger on a mirror and see that the real finger and reflected finger meet. This shows how a point on the line of reflection stays in the same place.

🎯 Exam Tip: Always write "The point lies on the line of reflection" when a point is its own image. Use the word "invariant point" to get full marks.

Question 3. State the co-ordinates of the following points under reflection in x-axis:
(i) (3, 2)
(ii) (-5, 4)
(iii) (0, 0)
Answer:
(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).
In simple words: When we reflect in the x-axis, the x-coordinate stays the same but the y-coordinate changes sign. It's like flipping the point up or down across the horizontal line.

📝 Teacher's Note: Draw the x-axis on the board. Show how points above the line go below and points below go above. The distance from the line stays the same.

🎯 Exam Tip: For x-axis reflection, keep x same and change y sign: (x, y) → (x, -y). Write this rule in your answer to show you understand.

Question 4. State the co-ordinates of the following points under reflection in y-axis:
(i) (6, -3)
(ii) (-1, 0)
(iii) (-8, -2)
Answer:
(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).
In simple words: When we reflect in the y-axis, the y-coordinate stays the same but the x-coordinate changes sign. It's like flipping the point left or right across the vertical line.

📝 Teacher's Note: Draw the y-axis on the board. Show how points on the right side go to the left side and vice versa. The distance from the line stays the same.

🎯 Exam Tip: For y-axis reflection, keep y same and change x sign: (x, y) → (-x, y). Remember this simple rule and you won't make mistakes.

Question 5. State the co-ordinates of the following points under reflection in origin:
(i) (-2, -4)
(ii) (-2, 7)
(iii) (0, 0)
Answer:
(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).
In simple words: When we reflect in the origin, both coordinates change signs. It's like rotating the point 180 degrees around the center point (0, 0).

📝 Teacher's Note: Mark the origin on the board. Show students how a point in one quadrant moves to the opposite quadrant when reflected in origin. Both coordinates flip signs.

🎯 Exam Tip: For origin reflection, change both signs: (x, y) → (-x, -y). The origin (0, 0) always stays at (0, 0) in any reflection.

Question 6. State the co-ordinates of the following points under reflection in the line x = 0:
(i) (-6, 4)
(ii) (0, 5)
(iii) (3, -4)
Answer:
(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).
In simple words: The line x = 0 is the same as the y-axis. So this is the same as reflecting in the y-axis. Only the x-coordinate changes sign.

📝 Teacher's Note: Explain that x = 0 means all points where x-coordinate is zero. This forms the y-axis. So reflection in line x = 0 is same as reflection in y-axis.

🎯 Exam Tip: Remember that x = 0 is the y-axis and y = 0 is the x-axis. Don't get confused by the notation. Apply the same reflection rules.

Question 7. State the co-ordinates of the following points under reflection in the line y = 0:
(i) (-3, 0)
(ii) (8, -5)
(iii) (-1, -3)
Answer:
(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).
In simple words: The line y = 0 is the same as the x-axis. So this is the same as reflecting in the x-axis. Only the y-coordinate changes sign.

📝 Teacher's Note: Show that y = 0 means all points where y-coordinate is zero. This forms the x-axis. Students often confuse x = 0 and y = 0 notation.

🎯 Exam Tip: When you see y = 0, think x-axis reflection. When you see x = 0, think y-axis reflection. This simple trick prevents confusion.

Question 8. 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.
Answer:
(i) Since M_x(-4, -5) = (-4, 5), the co-ordinates of P are (-4, -5).
(ii) Co-ordinates of the image of P under reflection in the y-axis are (4, -5).
In simple words: We work backwards to find the original point. If the image has coordinates (-4, 5) after x-axis reflection, the original point must be (-4, -5).

📝 Teacher's Note: Show students how to work backwards. If the image is (-4, 5) and we know it came from x-axis reflection, the original y-coordinate must be -5.

🎯 Exam Tip: To find the original point, reverse the reflection rule. If image is after x-axis reflection, change the sign of y-coordinate to get original point.

Question 9. 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.
Answer:
(i) Since M_O(2, -7) = (-2, 7), the co-ordinates of P are (2, -7).
(ii) Co-ordinates of the image of P under reflection in the x-axis are (2, 7).
In simple words: For origin reflection, we reverse both sign changes. If the image is (-2, 7), the original point is (2, -7).

📝 Teacher's Note: Emphasize that origin reflection changes both coordinates. To reverse it, change both signs again. Students should practice this type of backward working.

🎯 Exam Tip: For origin reflection questions, always change both signs when working backwards. Show your working clearly with the reflection notation.

Question 10. The point (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.
Answer:
M_O(a, b) = (-a, -b)
M_y(-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.
In simple words: We follow each step of reflection and work backwards from the final answer to find the original values.

📝 Teacher's Note: Break down multi-step transformations into separate steps. Show each reflection clearly before combining them. Students should practice step-by-step working.

🎯 Exam Tip: For multi-step transformations, show each step separately. Write the intermediate coordinates clearly before finding the final answer.

Question 11. The point P(x, y) is first reflected in the x-axis and reflected in the origin to P'. If P' has co-ordinates (-8, 5); evaluate x and y.
Answer:
M_x(x, y) = (x, -y)
M_O(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.
In simple words: First we reflect in x-axis, then in origin. We work step by step and compare with the final coordinates to find x and y.

📝 Teacher's Note: Make sure students write each transformation step clearly. Common mistake is to rush and combine steps without showing intermediate results.

🎯 Exam Tip: Always show the intermediate step between transformations. This helps avoid errors and gets you method marks even if final answer is wrong.

Question 12. 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".
Answer:
(i) The reflection in x-axis is given by M_x(x, y) = (x, -y).
A' = reflection of A(-3, 2) in the x-axis = (-3, -2).
The reflection in origin is given by M_O(x, y) = (-x, -y).
A" = reflection of A'(-3, -2) in the origin = (3, 2)

(ii) The reflection in y-axis is given by M_y(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".
In simple words: Two reflections can often be replaced by one single reflection. We find what single transformation gives the same result.

📝 Teacher's Note: Show students that combinations of reflections can be simplified. Help them see the pattern and find equivalent single transformations.

🎯 Exam Tip: For single transformation questions, check your answer by applying it directly to the original point. It should give the same result as the multi-step process.

Question 13. The point A(4, 6) is first reflected in the origin to point A'. Point A' is then reflected in the y-axis to the point A".
(i) Write down the co-ordinates of A".
(ii) Write down a single transformation that maps A onto A".
Answer:
(i) The reflection in origin is given by M_O(x, y) = (-x, -y).
A' = reflection of A(4, 6) in the origin = (-4, -6)
The reflection in y-axis is given by M_y(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 M_x(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".
In simple words: Again, we can replace two reflections with one single reflection that gives the same final result.

📝 Teacher's Note: Practice different combinations of reflections with students. Help them recognize patterns in how multiple reflections can be simplified.

🎯 Exam Tip: Double-check your single transformation by testing it on the original coordinates. The result should match your multi-step calculation.

Question 14. 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".
Answer:
(i) Reflection in y-axis is given by M_y(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 M_O(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.
In simple words: Even with triangles, we can find one single reflection that does the same job as two separate reflections.

📝 Teacher's Note: Show that transformations work the same way for shapes as for individual points. Each vertex follows the same reflection rules.

🎯 Exam Tip: For shapes, apply the transformation to each vertex separately. Keep track of which vertex becomes which in your working.

Question 15. P and Q have co-ordinates (-2, 3) and (5, 4) respectively. Reflect P in the x-axis to P' and Q in the y-axis to Q'. State the co-ordinates of P' and Q'.
Answer:
Reflection in x-axis is given by M_x(x, y) = (x, -y)
P' = Reflection of P(-2, 3) in x-axis = (-2, -3)
Reflection in y-axis is given by M_y(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.
In simple words: We can reflect different points in different lines. Each point follows the rule for its own reflection.

📝 Teacher's Note: This question shows that different points can have different transformations applied to them. Make sure students read carefully which point gets which transformation.

🎯 Exam Tip: Read the question carefully to see which transformation applies to which point. Don't apply the same transformation to all points unless stated.

Question 16. On a graph paper, plot the triangle ABC, whose vertices are at points A (3, 1), B (5, 0) and C (7, 4). On the same diagram, draw the image of the triangle ABC under reflection in the origin O (0, 0).
Answer: The graph shows triangle ABC and triangle A'B'C' which is obtained when ABC is reflected in the origin.
[Diagram: This shows a coordinate plane with triangle ABC in the first quadrant with vertices at A(3,1), B(5,0), and C(7,4), and its reflection A'B'C' in the third quadrant with vertices at A'(-3,-1), B'(-5,0), and C'(-7,-4)]
In simple words: When we reflect in the origin, each point (x, y) becomes (-x, -y). So A(3,1) becomes A'(-3,-1). The new triangle is on the opposite side of the origin.

📝 Teacher's Note: Show students that reflection in origin is like rotating the shape 180 degrees around the center point. Every coordinate changes sign.

🎯 Exam Tip: Always write the coordinates of all reflected points clearly. Check that both x and y coordinates change signs for origin reflection.

Question 17. Find the image of point (4, -6) under the following operations:
(i) \( M_x \cdot M_y \) (ii) \( M_y \cdot M_x \)
(iii) \( M_O \cdot M_x \) (iv) \( M_x \cdot M_O \)
(v) \( M_O \cdot M_y \) (vi) \( M_y \cdot M_O \)
Write down a single transformation equivalent to each operation given above. State whether:
(a) \( M_O \cdot M_x = M_x \cdot M_O \)
(b) \( M_y \cdot M_O = M_O \cdot M_y \)
Answer:
(i) \( M_x \cdot M_y \) (4, -6) = \( M_x \) (-4, -6) = (-4, 6)
Single transformation equivalent to \( M_x \cdot M_y \) is \( M_O \).

(ii) \( M_y \cdot M_x \) (4, -6) = \( M_y \) (4, 6) = (-4, 6)
Single transformation equivalent to \( M_y \cdot M_x \) is \( M_O \).

(iii) \( M_O \cdot M_x \) (4, -6) = \( M_O \) (4, 6) = (-4, -6)
Single transformation equivalent to \( M_O \cdot M_x \) is \( M_y \).

(iv) \( M_x \cdot M_O \) (4, -6) = \( M_x \) (-4, 6) = (-4, -6)
Single transformation equivalent to \( M_x \cdot M_O \) is \( M_y \).

(v) \( M_O \cdot M_y \) (4, -6) = \( M_O \) (-4, -6) = (4, 6)
Single transformation equivalent to \( M_O \cdot M_y \) is \( M_x \).

(vi) \( M_y \cdot M_O \) (4, -6) = \( M_y \) (-4, 6) = (4, 6)
Single transformation equivalent to \( M_y \cdot M_O \) is \( M_x \).

From (iii) and (iv), it is clear that \( M_O \cdot M_x = M_x \cdot M_O \).
From (v) and (vi), it is clear that \( M_y \cdot M_O = M_O \cdot M_y \).
In simple words: We apply transformations one by one from right to left. \( M_x \) reflects across x-axis, \( M_y \) across y-axis, \( M_O \) across origin. Some combinations give the same result.

📝 Teacher's Note: Remind students that transformations are done from right to left. Use simple examples like (1,1) to show each step clearly.

🎯 Exam Tip: Show all working steps. Write coordinates after each transformation. Remember that order matters for some combinations but not others.

Question 18. 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 down the co-ordinates of A' and B.
Answer:
Reflection in y-axis is given by \( M_y \) (x, y) = (-x, y)
A' = Reflection of A(4, -1) in y-axis = (-4, -1)

Reflection in x-axis is given by \( M_x \) (x, y) = (x, -y)
B' = Reflection of B in x-axis = (-2, 5)
Thus, B = (-2, -5)
In simple words: For y-axis reflection, x-coordinate changes sign. For x-axis reflection, y-coordinate changes sign. We work backwards from B' to find B.

📝 Teacher's Note: Draw both axes and show students that y-axis reflection flips left-right, x-axis reflection flips up-down.

🎯 Exam Tip: For y-axis reflection, only x changes sign. For x-axis reflection, only y changes sign. Check your answer by reflecting it back.

Question 19. 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 down the co-ordinates of the image of (5, -8) in the line obtained in (a).
Answer:
(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 \( M_y \) (x, y) = (-x, y)
Co-ordinates of the image of (5, -8) in the line x = 0 are (-5, -8).
In simple words: The line x = 0 is the y-axis. When we reflect in y-axis, only the x-coordinate changes sign. The y-coordinate stays the same.

📝 Teacher's Note: Point out that x = 0 means the y-axis. Students often confuse this. Show examples on the coordinate plane.

🎯 Exam Tip: Remember x = 0 is the y-axis and y = 0 is the x-axis. For y-axis reflection, x changes sign but y stays same.

Exercise 12B

Question 1. 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 of A" of A, when A is reflected in the origin.
(iv) the single transformation that maps A' to A".
Answer:
[Diagram: This shows a coordinate plane with points A(3,2), B(5,4), A'(3,-2), B'(5,-4), and A"(-3,-2) plotted, forming an isosceles trapezium ABB'A']

(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.
In simple words: When we reflect points across x-axis, we get a trapezium shape. The angle is 45° because of the special positions of the points.

📝 Teacher's Note: Help students see that isosceles trapezium has parallel sides and equal angles. Use a ruler to measure the 45° angle.

🎯 Exam Tip: Write "isosceles trapezium" clearly. Show that A' to A" needs y-axis reflection because only x-coordinate changes.

Question 2. Points (3, 0) and (-1, 0) are invariant points under reflection in the line L₁; points (0, -3) and (0, 1) are invariant points on reflection in line L₂.
(i) Name or write equations for the lines L₁ and L₂.
(ii) Write down the images of the points P (3, 4) and Q (-5, -2) on reflection in line L₁. Name the images as P' and Q' respectively.
(iii) Write down the images of P and Q on reflection in L₂. Name the images as P" and Q" respectively.
(iv) State or describe a single transformation that maps P' onto P".
Answer:
(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.
In simple words: Invariant points don't move when reflected. Points on x-axis stay on x-axis when reflected in x-axis. Same for y-axis.

📝 Teacher's Note: Explain that invariant means "doesn't change." If a point is on the mirror line, it stays in the same place when reflected.

🎯 Exam Tip: Write y = 0 for x-axis and x = 0 for y-axis. Check that P'(3,-4) to P"(-3,4) needs origin reflection (both coordinates change sign).

Question 3.
(i) Point P (a, b) is reflected in the x-axis to P' (5, -2). Write down the values of a and b.
(ii) P" is the image of P when reflected in the y-axis. Write down the co-ordinates of P".
(iii) Name a single transformation that maps P' to P".
Answer:
(i) We know \( M_x \) (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.
In simple words: We work backwards from P' to find P. Then we reflect P in y-axis to get P". From P' to P" needs origin reflection.

📝 Teacher's Note: Show students how to work backwards. If P' is (5,-2) from x-axis reflection, then P must be (5,2).

🎯 Exam Tip: Always check your answer. Reflect P(5,2) in x-axis to get (5,-2) which matches P'. Show all steps clearly.

Question 4. 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 the mirror line and write its equation.
(ii) State the co-ordinates of the image of (-8, -5) in the mirror line.
Answer:
(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).
In simple words: Both examples show that only x-coordinate changes sign. This happens in y-axis reflection. The mirror line is x = 0.

📝 Teacher's Note: Show students that y-coordinate stays same in both examples. This confirms it's y-axis reflection.

🎯 Exam Tip: Write "y-axis" and "x = 0" clearly. For y-axis reflection, change sign of x only.

Question 5. 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.
Answer:
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).
In simple words: Line y = 3 is horizontal. P is 2 units below this line, so P' is 2 units above it. Q is 1 unit above, so Q' is 1 unit below.

📝 Teacher's Note: Draw y = 3 as a horizontal line. Show students how to count units above and below this line to find reflected points.

🎯 Exam Tip: For line y = k, the x-coordinate stays same. Find distance from point to line, then go same distance on other side.

Question 6. A point P (-2, 3) is reflected in line x = 2 to point P'. Find the coordinates of P'.
Answer: 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).
In simple words: When we reflect a point in a vertical line, the y-coordinate stays the same. The x-coordinate changes by moving the same distance on the other side of the line.

[Diagram: This shows a coordinate plane with point P at (-2, 3), the reflection line x = 2 shown as a vertical line, and point P' at (6, 3). The diagram shows how P is 4 units left of the line, so P' is 4 units right of the line.]

📝 Teacher's Note: Show students how to count units from the point to the line. Then count the same number of units on the other side. Use folding paper to show reflection.

🎯 Exam Tip: Always keep the y-coordinate same for vertical line reflections. Count distance carefully from original point to the line.

Question 7. 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 coordinates of P"', when P is reflected in the line, parallel to y-axis, such that x = 4.
Answer: 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).
In simple words: When reflecting in x-axis, x stays same but y changes sign. When reflecting in y-axis, y stays same but x changes sign. For vertical line reflection, y stays same.

[Diagram: This shows multiple reflections of point P(2,3) - P'(2,-3) reflected in x-axis, P"(-2,3) reflected in y-axis, and P"'(6,3) reflected in line x=4.]

📝 Teacher's Note: Make students write the reflection rules clearly. x-axis reflection changes y sign only. y-axis reflection changes x sign only.

🎯 Exam Tip: Remember the reflection formulas: x-axis gives (x,-y), y-axis gives (-x,y). For line x=k, use distance formula.

Question 8. 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".
Answer:
(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)
In simple words: We use different reflection rules for each part. Some reflections are in axes, some are in lines we create by joining points.

[Diagram: This shows points A(3,4), B(0,2), and all their reflections A'(3,-4), B'(6,2), A"(-3,4), and B"(0,6) plotted on a coordinate plane.]

📝 Teacher's Note: Draw each reflection step by step. Show students how line AA' is vertical and line AA" is horizontal. This makes the reflections easier to find.

🎯 Exam Tip: Always find the equation of the reflection line first. Then apply the correct reflection rule for that type of line.

Question 9. (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 the invariant points under reflection in the x-axis.
Answer:
(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).
In simple words: We do step by step reflections. First reflect B in y-axis, then reflect that result in origin. Invariant points are points that do not move when reflected.

[Diagram: This shows all four points A(3,5), A'(3,-5), B(-2,-4), and B'(-2,4) plotted and connected to form an isosceles trapezium.]

📝 Teacher's Note: Show students that invariant points lie on the line of reflection. For x-axis reflection, any point with y=0 stays fixed.

🎯 Exam Tip: For double reflections, do them one by one. Do not try to combine steps. Invariant points always lie on the reflection line.

Question 10. 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 if the perpendicular from P to the x-axis, find the co-ordinates of M. (c) If N is the foot if 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.
Answer:
(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) Area = base × height = 10 × 3 = 30 square units
In simple words: Reflection in origin changes both signs. Foot of perpendicular to x-axis means the y-coordinate becomes 0. The four points make a parallelogram.

[Diagram: This shows points P(5,3), P'(-5,-3), M(5,0), and N(-5,0) connected to form a parallelogram with the origin at the center.]

📝 Teacher's Note: Show that foot of perpendicular means dropping a vertical line down to the axis. The x-coordinate stays same, y becomes 0.

🎯 Exam Tip: For origin reflection, both coordinates change signs: (x,y) becomes (-x,-y). Always show area calculation clearly with formula.

(e) Area of PMP'N = 2 (Area of △ PMN)
= 2 × ½ × 10 × 3
= 30 sq. units

Question 11. 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'.
Answer:
(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' = \( \sqrt{(4)^2 + (3)^2} = 5 \) units
So, perimeter of quadrilateral POP'O' = 4 PO' = 4 × 5 units = 20 units
(iv) Quadrilateral POP'O' is a rhombus.
In simple words: P' is just P flipped over the x-axis. O' is O flipped over the line connecting P and P'. All four sides of this shape are equal, making it a rhombus.

[Diagram: A coordinate grid showing points P(3,4), P'(3,-4), O at origin, and O'(6,0) forming a diamond-shaped quadrilateral]

📝 Teacher's Note: Draw the reflection step by step. Show students that P(3,4) becomes P'(3,-4) when flipped over x-axis. The y-coordinate changes sign, x stays same.

🎯 Exam Tip: Always write coordinates clearly in brackets. For perimeter, first find one side length, then multiply by 4 since all sides are equal in a rhombus.

Question 12. 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?
Answer:
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.
In simple words: When we reflect points in the origin, both x and y coordinates change their signs. The shape ABCD is like a trapezium with equal slant sides. The four points D, A, A', D' all lie on the same straight line.

[Diagram: A coordinate grid showing quadrilateral ABCD in the first quadrant and its reflection A'B'C'D' in the third quadrant, with points D, A, A', D' forming a straight line through the origin]

📝 Teacher's Note: To reflect in origin, change signs of both coordinates. So (x,y) becomes (-x,-y). Show this pattern clearly to students with examples.

🎯 Exam Tip: For reflection in origin, write the rule: (x,y) → (-x,-y). For collinear points, check if they lie on the same straight line by plotting or using slope.

Question 13. 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 (i).
(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.
Answer:
(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).
In simple words: P stays the same when reflected in y-axis because it lies on y-axis. When we reflect Q in y-axis, only x-coordinate changes sign. For origin reflection, only (0,0) stays unchanged. Double reflection gives us the final position.

📝 Teacher's Note: Invariant means "does not change". Points on y-axis stay same when reflected in y-axis. Only origin stays same when reflected in origin.

🎯 Exam Tip: Write "invariant" clearly and explain what it means. For y-axis reflection, only x-coordinate changes sign. For origin reflection, both coordinates change signs.

Question PQ. The points P (1, 2), Q (3, 4) and R (6, 1) are the vertices of PQR.
(a) Write down the co-ordinates of P', Q' and R', if P'Q'R' is the image of PQR, when reflected in the origin.
(b) Write down the co-ordinates of P", Q" and R", if P"Q"R" is the image of PQR, when reflected in the x-axis.
(c) Mention the special name of the quadrilateral QRR"Q" and find its area.
Answer:
(a) The co-ordinates of P', Q' and R' are (-1, -2), (-3, -4) and (-6, -1) respectively.
(b) The co-ordinates of P", Q" and R" are (1, -2), (3, -4) and (6, -1) respectively.
(c) The quadrilateral QRR"Q" is an isosceles trapezium.
Area of QRR"Q" = ½ (RR"+ QQ") × Height
= ½ (2 + 8) × 3 = 15 sq units
In simple words: When reflected in origin, both coordinates change signs. When reflected in x-axis, only y-coordinate changes sign. The quadrilateral formed has two parallel sides of different lengths, making it a trapezium.

[Diagram: A coordinate grid showing triangle PQR and its reflections, forming quadrilateral QRR"Q"]

📝 Teacher's Note: Show students the trapezium area formula clearly. Draw parallel sides and mark the height. Students often forget to multiply by ½ in area calculations.

🎯 Exam Tip: For origin reflection: (x,y) → (-x,-y). For x-axis reflection: (x,y) → (x,-y). Write area formula clearly and show all steps.

Question 14.
(i) The point P (2, -4) is reflected about the line x = 0 to get the image Q. Find the coordinates of Q.
(ii) The point Q is reflected about the line y = 0 to get the image R. Find the co-ordinates of R.
(iii) Name the figure PQR.
(iv) Find the area of figure PQR.
Answer:
(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 △PQR = ½ × PQ × QR = ½ × 4 × 8 = 16 sq units
In simple words: Line x = 0 is the y-axis. Line y = 0 is the x-axis. The three points form a right-angled triangle because two sides are parallel to the axes.

[Diagram: A coordinate grid showing right triangle PQR with P(2,-4), Q(-2,-4), and R(-2,4)]

📝 Teacher's Note: Explain that x = 0 means y-axis and y = 0 means x-axis. Show students how PQ is horizontal and QR is vertical, making a right angle.

🎯 Exam Tip: Write x = 0 as "y-axis" and y = 0 as "x-axis" in your answer. For right triangle area, use ½ × base × height formula.

Question PQ. A' and B' are images of A (-3, 5) and B (-5, 3) respectively on reflection in y-axis. Find:
(a) the co-ordinates of A' and B'.
(b) Assign special name of quadrilateral AA'B'B.
(c) Are AB' and BA' equal in length?
Answer:
(a) A (-3, 5) reflected in y-axis gives A' (3, 5)
B (-5, 3) reflected in y-axis gives B' (5, 3)
(b) The quadrilateral AA'B'B is an isosceles trapezium.
(c) Yes, AB' and BA' are equal in length because the quadrilateral is symmetric about the y-axis.
In simple words: When we reflect in y-axis, x-coordinate changes sign but y stays same. The four points make a trapezium with equal slant sides. The diagonal lines AB' and BA' have the same length due to symmetry.

📝 Teacher's Note: Draw the quadrilateral on board and show students the symmetry. Mark the equal sides and equal diagonals to make it clear.

🎯 Exam Tip: For y-axis reflection, write (x,y) → (-x,y). State clearly that diagonals are equal due to symmetry of the figure.

Question 15. Using a graph paper, plot the point A (6, 4) and B (0, 4).
(a) Reflect A and B in the origin to get the image 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.
Answer:
(a) When we reflect A (6, 4) and B (0, 4) in the origin, both coordinates change sign.
(b) Co-ordinates of A' = (-6, -4) and Co-ordinates of B' = (0, -4)
(c) ABA'B' is a parallelogram.
(d) In triangle ABA'B', BB' = 8 units, A'B' = 6 units
\( BA' = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10 \) units
Therefore B'A = 10 units
AB = A'B' = 6 units
Perimeter of ABA'B' = AB + BA' + A'B' + B'A = 6 + 10 + 6 + 10 = 32 units
In simple words: When we reflect in origin, both x and y change sign. The four points make a parallelogram shape. We add all four sides to get perimeter.

📝 Teacher's Note: Show students that reflecting in origin means changing (x,y) to (-x,-y). Draw the shape on graph paper to show it looks like a kite or diamond.

🎯 Exam Tip: Always write coordinates clearly. Use distance formula for diagonal lengths. Write "parallelogram" as the shape name - this gets you marks.

Question 16. Use graph paper for this question. (Take 2 cm = 1 unit along both x and y axis. 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 coordinates.
(ii) Name the figure OABCB'A'.
(iii) State the line of symmetry of this figure.
Answer:
(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'.
In simple words: When we reflect on y-axis, only x-coordinate changes sign. The six points joined together look like an arrow pointing up. The y-axis is the mirror line.

[Diagram: This diagram shows a coordinate plane with an arrow-head shaped figure. Points A(-4,4), B(-3,0), O(0,0), C(0,-3), B'(3,0), and A'(4,4) are connected to form an arrow pointing upward. The y-axis acts as the line of symmetry.]

📝 Teacher's Note: Tell students that reflecting on y-axis means x becomes -x, but y stays same. The arrow shape is easy to remember - like pointing direction.

🎯 Exam Tip: Write coordinates in brackets with comma. Say "arrow head" as the shape name. Always mention "y-axis" or "x = 0" for the symmetry line.

Question 17.
(i) Plotting A(0, 4), B(2, 3), C(1, 1) and D(2, 0).
(ii) Reflected points B'(-2, 3), C'(-1, 1) and D'(-2, 0).
(iii) The figure is symmetrical about x = 0
Answer:
This question shows the plotting of points and their reflections on the y-axis. Point A stays on the y-axis so it does not move when reflected. The other points B, C, D reflect to B', C', D' across the y-axis. The complete figure is symmetric about the line x = 0 (the y-axis).
In simple words: We plot some points and flip them across the y-axis like a mirror. The whole shape looks the same on both sides of the y-axis.

[Diagram: This diagram shows a coordinate plane with points plotted to form a symmetric figure. Original points A(0,4), B(2,3), C(1,1), D(2,0) are on the right side, and their reflections B'(-2,3), C'(-1,1), D'(-2,0) are on the left side, with the y-axis as the line of symmetry.]

📝 Teacher's Note: Explain that points on the y-axis do not move when reflected across y-axis. Other points jump to opposite side at same distance.

🎯 Exam Tip: When reflecting across y-axis, x changes sign but y stays same. Always check your reflected coordinates are correct distances from the y-axis.