OP Malhotra Class 10 Maths Solutions Chapter 10 Reflection Exercise 10 (B)

Question 1. State the coordinates of
(a) Points (i) (5, 7), (ii) (3, -4), (iii) (-8, 9), (iv) (-1, -2) under reflection in the x-axis.
(b) Points (i) (2, 5), (ii) (1, -4), (iii) (-7, -13), (iv) (-6, 8) under reflection in the y-axis.
Answer:
(a) When a point \( (x, y) \) is reflected in the x-axis, its image is \( (x, -y) \). We change the sign of the y-coordinate.
(i) Point \( (5, 7) \) becomes \( (5, -7) \).
(ii) Point \( (3, -4) \) becomes \( (3, -(-4)) = (3, 4) \).
(iii) Point \( (-8, 9) \) becomes \( (-8, -9) \).
(iv) Point \( (-1, -2) \) becomes \( (-1, -(-2)) = (-1, 2) \).
(b) When a point \( (x, y) \) is reflected in the y-axis, its image is \( (-x, y) \). We change the sign of the x-coordinate.
(i) Point \( (2, 5) \) becomes \( (-2, 5) \).
(ii) Point \( (1, -4) \) becomes \( (-1, -4) \).
(iii) Point \( (-7, -13) \) becomes \( (-(-7), -13) = (7, -13) \).
(iv) Point \( (-6, 8) \) becomes \( (-(-6), 8) = (6, 8) \).
In simple words: To reflect a point over the x-axis, you just flip the sign of the y-coordinate. To reflect over the y-axis, you flip the sign of the x-coordinate instead.

🎯 Exam Tip: Remember the rules for reflection: \( (x, y) \to (x, -y) \) for x-axis reflection and \( (x, y) \to (-x, y) \) for y-axis reflection. This is a fundamental concept in coordinate geometry.

Question 2. Parallelogram ABCD has vertices A (-3, 2), B (5, 2), C (7, -1), D (-1, -1). Determine the image A', B', C', D' of A, B, C, D respectively, under reflection in the x-axis.
Answer: The vertices of the parallelogram ABCD are given as \( A (-3, 2), B (5, 2), C (7, -1), D (-1, -1) \). To find the image of these points after reflection in the x-axis, we change the sign of the y-coordinate for each point.
The image of \( A (-3, 2) \) will be \( A' (-3, -2) \).
The image of \( B (5, 2) \) will be \( B' (5, -2) \).
The image of \( C (7, -1) \) will be \( C' (7, -(-1)) = C' (7, 1) \).
The image of \( D (-1, -1) \) will be \( D' (-1, -(-1)) = D' (-1, 1) \).
In simple words: When you reflect a shape across the x-axis, every point moves to a new spot where its x-value stays the same, but its y-value becomes the opposite. So, if a point was at 2 on the y-axis, it moves to -2. If it was at -1, it moves to 1.

🎯 Exam Tip: Always apply the reflection rule \( (x, y) \to (x, -y) \) consistently to all vertices of the figure. A common mistake is to change the x-coordinate or incorrectly flip the sign.

Question 3.
(i) Plot each of the given points on graph paper, reflect them in the x-axis and then reflect the points obtained in the y-axis. Write down the co-ordinates of the points obtained.
(a) (3, 4)
(b) (-3, 2)
(c) (5, 0)
(d) (-3, -3)
(ii) What happens if you reflect first in the y-axis and then in the x-axis ?
Answer:
(i) We need to reflect each point in the x-axis first, then reflect the new point in the y-axis.
(a) For point \( A (3, 4) \):
Reflection in x-axis: \( A' (3, -4) \).
Reflection of \( A' (3, -4) \) in y-axis: \( A'' (-3, -4) \).
(b) For point \( B (-3, 2) \):
Reflection in x-axis: \( B' (-3, -2) \).
Reflection of \( B' (-3, -2) \) in y-axis: \( B'' (3, -2) \).
(c) For point \( C (5, 0) \):
Reflection in x-axis: \( C' (5, 0) \). (The y-coordinate is 0, so it stays the same).
Reflection of \( C' (5, 0) \) in y-axis: \( C'' (-5, 0) \).
(d) For point \( D (-3, -3) \):
Reflection in x-axis: \( D' (-3, 3) \).
Reflection of \( D' (-3, 3) \) in y-axis: \( D'' (3, 3) \).

(ii) Now, we reflect first in the y-axis and then in the x-axis.
For point \( A (3, 4) \):
Reflection in y-axis: \( A''' (-3, 4) \).
Reflection of \( A''' (-3, 4) \) in x-axis: \( A'''' (-3, -4) \).
This result \( A'''' (-3, -4) \) is the same as \( A'' (-3, -4) \) obtained in part (i).
Similarly, for point \( B (-3, 2) \):
Reflection in y-axis: \( B''' (3, 2) \).
Reflection of \( B''' (3, 2) \) in x-axis: \( B'''' (3, -2) \).
This result \( B'''' (3, -2) \) is the same as \( B'' (3, -2) \) obtained in part (i).
The same pattern holds for points C and D. The order of reflection (x-axis then y-axis, or y-axis then x-axis) does not change the final coordinates. The final image is the same as reflecting the original point in the origin, \( (x, y) \to (-x, -y) \).
In simple words: First, for each point, we flip its y-sign, then its x-sign. Then, we try flipping the x-sign first, then the y-sign. We find that for every point, both ways of doing it lead to the exact same final position. This is like moving to the opposite corner of the graph for both x and y.

🎯 Exam Tip: Remember that two successive reflections across perpendicular lines (like the x and y axes) are equivalent to a single reflection through the origin. This can save time in multi-step reflection problems.

Question 4. A point P (a, b) is reflected in the y-axis to P' (-3, 5). Write down the values of a and b. P" is the image of P, when reflected in the x-axis. Write down the coordinates of P". Find the coordinates of P''', when P is reflected in the line, parallel to the x-axis, such that its equation is y = -3.
Answer:
1. To find the coordinates of P (a, b) from P' (-3, 5):
If P (a, b) is reflected in the y-axis, its image is \( P' (-a, b) \).
We are given \( P' (-3, 5) \).
Comparing the coordinates, we have \( -a = -3 \), which means \( a = 3 \).
And \( b = 5 \).
So, the coordinates of point P are \( (3, 5) \).

2. To find the coordinates of P" (image of P reflected in the x-axis):
Point P is \( (3, 5) \).
Reflection of P \( (3, 5) \) in the x-axis is \( P'' (3, -5) \). The x-coordinate stays the same, and the y-coordinate changes its sign.

3. To find the coordinates of P''' (image of P reflected in the line \( y = -3 \)):
Point P is \( (3, 5) \).
The line of reflection is parallel to the x-axis, with equation \( y = -3 \).
The formula for reflection of \( (x, y) \) in the line \( y = k \) is \( (x, 2k - y) \).
Here, \( x = 3, y = 5 \), and \( k = -3 \).
So, the x-coordinate of P''' will be \( 3 \).
The y-coordinate of P''' will be \( 2(-3) - 5 = -6 - 5 = -11 \).
Thus, the coordinates of P''' are \( (3, -11) \).
In simple words: First, we use the y-axis reflection rule to find the original point P. Then, we apply the x-axis reflection rule to P to get P''. Finally, we use a special reflection rule for a horizontal line \( y = -3 \) to find P'''. Each reflection changes the point's position in a specific way.

🎯 Exam Tip: Know the three basic reflection rules: x-axis \( (x, y) \to (x, -y) \), y-axis \( (x, y) \to (-x, y) \), and origin \( (x, y) \to (-x, -y) \). For reflection in a line \( y=k \) or \( x=k \), use the formulas \( (x, 2k-y) \) and \( (2k-x, y) \) respectively.

Question 5. Use graph paper for this question.
(i) Plot the points P (2, -4). Use 1 cm = 1 unit on both the axes.
(ii) P' is the image of P when reflected in AB, which is parallel to the x-axis and is at a distance 1 on the positive side of y-axis. (i.e. the line y = 1).
(iii) P" is the image of P' when reflected in the line LM which is parallel to the y-axis is at a distance 1 on the negative side of the x-axis. (i.e., the line x = -1).
Answer:
(i) The point P \( (2, -4) \) is plotted on the graph.

(ii) To find the image P' of P \( (2, -4) \) reflected in the line \( y = 1 \):
The line AB is \( y = 1 \). When a point \( (x, y) \) is reflected in the line \( y = k \), its image is \( (x, 2k - y) \).
Here, \( x = 2, y = -4 \), and \( k = 1 \).
The x-coordinate of P' will be \( 2 \).
The y-coordinate of P' will be \( 2(1) - (-4) = 2 + 4 = 6 \).
So, the coordinates of P' are \( (2, 6) \).

(iii) To find the image P" of P' \( (2, 6) \) reflected in the line \( x = -1 \):
The line LM is \( x = -1 \). When a point \( (x, y) \) is reflected in the line \( x = k \), its image is \( (2k - x, y) \).
Here, \( x = 2, y = 6 \), and \( k = -1 \).
The x-coordinate of P" will be \( 2(-1) - 2 = -2 - 2 = -4 \).
The y-coordinate of P" will be \( 6 \).
So, the coordinates of P" are \( (-4, 6) \).
In simple words: We start with point P. First, we reflect P across a horizontal line \( y=1 \), which gives us P'. Next, we take P' and reflect it across a vertical line \( x=-1 \), which gives us P''. Each reflection creates a new point based on its distance to the reflection line.

🎯 Exam Tip: When reflecting across a line \( y=k \) or \( x=k \), visualizing the line and the point helps. The reflected point will be the same perpendicular distance from the line but on the opposite side. This is correctly captured by the formulas \( (x, 2k-y) \) and \( (2k-x, y) \).

Question 6. A point P is reflected in the x-axis. Co-ordinates of its image are (8, -6).
(i) Find the co-ordinates of P
(ii) Find the co-ordinates of the image of P under reflection in the y-axis. (ICSE)
Answer:
(i) Let the coordinates of point P be \( (x, y) \).
When P \( (x, y) \) is reflected in the x-axis, its image is \( P' (x, -y) \).
We are given that the image \( P' \) has coordinates \( (8, -6) \).
Comparing these, we get \( x = 8 \) and \( -y = -6 \), which means \( y = 6 \).
Therefore, the coordinates of P are \( (8, 6) \).

(ii) Now we need to find the image of P \( (8, 6) \) under reflection in the y-axis.
When a point \( (x, y) \) is reflected in the y-axis, its image is \( (-x, y) \).
So, for P \( (8, 6) \), the image \( P'' \) will be \( (-8, 6) \).
In simple words: First, we use the rule for x-axis reflection backward to find the original point P. Then, we take that point P and apply the rule for y-axis reflection to find its new position, P''.

🎯 Exam Tip: Understanding inverse reflections is crucial. If \( P' \) is the reflection of \( P \) in a line, then \( P \) is also the reflection of \( P' \) in the same line. Always work backwards carefully if the image is given.

Question 7. The image of the point (1, 5) when reflected in a line LM is (9, 5). Write down the equation of the line LM.
Answer:
Let the original point be \( P (1, 5) \).
Let its image be \( P' (9, 5) \).
We observe that the y-coordinate for both the original point P and its image P' is the same (which is 5). This tells us that the line of reflection must be a vertical line, parallel to the y-axis, and of the form \( x = k \).
The line of reflection is always the perpendicular bisector of the segment connecting the point and its image.
To find the value of \( k \), we find the midpoint of \( PP' \).
Midpoint \( M = \left( \frac{1 + 9}{2}, \frac{5 + 5}{2} \right) \)
\( M = \left( \frac{10}{2}, \frac{10}{2} \right) \)
\( M = (5, 5) \)
Since the line of reflection is vertical and passes through the midpoint \( (5, 5) \), its equation is \( x = 5 \).
In simple words: We have a point and its reflection. Because their y-coordinates are the same, the mirror line must be straight up and down. To find exactly where this line is, we just find the middle point between the original point and its reflection. The x-value of this middle point tells us the equation of the mirror line.

🎯 Exam Tip: If the y-coordinates are the same, the reflection line is \( x = k \). If the x-coordinates are the same, the reflection line is \( y = k \). The value of \( k \) is always the midpoint of the non-invariant coordinate.

Question 8. Use graph paper to plot the triangle ABC where A is (1, 2), B is (3, 4) and C is (6,1). On the same graph paper plot.
(i) the image A, B, C of the triangle ABC under reflection in the origin O (0, 0);
(ii) the image of triangle ABC under reflection in the y-axis followed by a reflection in the x-axis.
Compare your results for (i) and (ii) above and make a statement connecting the two results.
Answer:
First, plot the points A \( (1, 2) \), B \( (3, 4) \), and C \( (6, 1) \) on the graph paper and join them to form triangle ABC.

(i) To find the image of A, B, C under reflection in the origin O \( (0, 0) \):
The rule for reflection in the origin is \( (x, y) \to (-x, -y) \).
For A \( (1, 2) \), the image is \( A' (-1, -2) \).
For B \( (3, 4) \), the image is \( B' (-3, -4) \).
For C \( (6, 1) \), the image is \( C' (-6, -1) \).
Plot these points \( A', B', C' \) and join them to form triangle A'B'C'.

(ii) To find the image of triangle ABC under reflection in the y-axis, followed by reflection in the x-axis:
First reflection (y-axis): \( (x, y) \to (-x, y) \)
For A \( (1, 2) \), image \( A'' (-1, 2) \).
For B \( (3, 4) \), image \( B'' (-3, 4) \).
For C \( (6, 1) \), image \( C'' (-6, 1) \).

Second reflection (x-axis) on \( A'', B'', C'' \): \( (-x, y) \to (-x, -y) \)
For \( A'' (-1, 2) \), image \( A''' (-1, -2) \).
For \( B'' (-3, 4) \), image \( B''' (-3, -4) \).
For \( C'' (-6, 1) \), image \( C''' (-6, -1) \).
Plot these points \( A''', B''', C''' \) and join them to form triangle A'''B'''C'''.

Comparison of results:
Comparing the coordinates from part (i) \( (A'(-1, -2), B'(-3, -4), C'(-6, -1)) \) with the coordinates from part (ii) \( (A'''(-1, -2), B'''(-3, -4), C'''(-6, -1)) \), we observe that the results are exactly the same.
This means that reflection in the origin is equivalent to reflection in the y-axis followed by reflection in the x-axis (or vice versa).
In simple words: We draw a triangle. For the first part, we flip the triangle across the center point (the origin). For the second part, we flip it across the y-axis first, then flip the result across the x-axis. We see that both methods end up with the triangle in the exact same final position. So, flipping it over the y-axis then the x-axis is like flipping it over the center.

🎯 Exam Tip: This question highlights an important property of transformations: two successive reflections across perpendicular axes (like x and y) are equivalent to a single rotation of 180 degrees about the intersection of the axes (the origin).

Question 9. On the graph paper, taking 1 cm = 1 unit, plot the triangle ABC whose vertices are at the points A (3, 1), B (5, 0) and C (7, 4). On the same diagram, draw the image of the AABC under reflection in the line x = 2. Mark the point invariant under this reflection.
Answer:
1. Plot the vertices of triangle ABC:
\( A (3, 1) \), \( B (5, 0) \), and \( C (7, 4) \). Join them to form triangle ABC.

2. Draw the line of reflection:
The line of reflection is \( x = 2 \), which is a vertical line parallel to the y-axis, located 2 units to the right of the y-axis.

3. Reflect each vertex of triangle ABC in the line \( x = 2 \):
The rule for reflection in the line \( x = k \) is \( (x, y) \to (2k - x, y) \). Here \( k = 2 \).
For A \( (3, 1) \):
\( x' = 2(2) - 3 = 4 - 3 = 1 \)
\( y' = 1 \)
So, \( A' (1, 1) \).

For B \( (5, 0) \):
\( x' = 2(2) - 5 = 4 - 5 = -1 \)
\( y' = 0 \)
So, \( B' (-1, 0) \).

For C \( (7, 4) \):
\( x' = 2(2) - 7 = 4 - 7 = -3 \)
\( y' = 4 \)
So, \( C' (-3, 4) \).
Plot the points \( A' (1, 1) \), \( B' (-1, 0) \), and \( C' (-3, 4) \) and join them to form the image triangle A'B'C'.

4. Mark the invariant point:
An invariant point is a point that does not change its position after the reflection. This happens if the point lies on the line of reflection.
Since the reflection line is \( x = 2 \), any point with an x-coordinate of 2 will be invariant.
Looking at the graph, the point \( (2, 0) \) lies on the line \( x = 2 \). Thus, \( (2, 0) \) is the invariant point under this reflection. For instance, the intersection of the x-axis and the line \( x=2 \) is \( (2,0) \).
In simple words: First, we draw the triangle. Then, we draw a vertical mirror line at \( x=2 \). For each corner of the triangle, we find its mirror image by calculating its new position based on the reflection rule. Any point that sits directly on the mirror line itself will not move at all, and that's our "invariant point."

🎯 Exam Tip: An invariant point in a reflection is any point that lies on the line of reflection. For a line \( x=k \), any point \( (k, y) \) is invariant. For a line \( y=k \), any point \( (x, k) \) is invariant.

Question 10. B, C have co-ordinates (3, 2) and (0, 3). Find
(i) the image B' of B under the reflection in the x-axis;
(ii) the image C' of C under reflection in the line BB';
(iii) Calculate the length of B'C'.
Answer:
First, plot the points B \( (3, 2) \) and C \( (0, 3) \) on the graph.

(i) To find the image B' of B \( (3, 2) \) under reflection in the x-axis:
The rule for reflection in the x-axis is \( (x, y) \to (x, -y) \).
So, the image B' will be \( (3, -2) \).

(ii) To find the image C' of C \( (0, 3) \) under reflection in the line BB':
First, join B \( (3, 2) \) and B' \( (3, -2) \) to form the line BB'. This line is a vertical line passing through \( x = 3 \). So, the equation of line BB' is \( x = 3 \).
Now, reflect C \( (0, 3) \) in the line \( x = 3 \).
The rule for reflection in the line \( x = k \) is \( (x, y) \to (2k - x, y) \). Here \( k = 3 \).
For C \( (0, 3) \):
\( x' = 2(3) - 0 = 6 - 0 = 6 \)
\( y' = 3 \)
So, the image C' will be \( (6, 3) \).

(iii) To calculate the length of B'C':
We have B' \( (3, -2) \) and C' \( (6, 3) \).
We can use the distance formula: \( \text{Distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \)
Let \( (x_1, y_1) = (3, -2) \) and \( (x_2, y_2) = (6, 3) \).
\( B'C' = \sqrt{(6 - 3)^2 + (3 - (-2))^2} \)
\( B'C' = \sqrt{(3)^2 + (3 + 2)^2} \)
\( B'C' = \sqrt{3^2 + 5^2} \)
\( B'C' = \sqrt{9 + 25} \)
\( B'C' = \sqrt{34} \)
Alternatively, from the graph, draw a perpendicular from C' to the line extended from B'. Let D be the point \( (6, -2) \). Then triangle B'DC' is a right-angled triangle.
\( C'D = |6 - 3| = 3 \) units (horizontal distance between x-coordinates of C' and B'). This is incorrect. It should be \( C'D \) is the horizontal distance from C' to the vertical line passing through B'. So, \( C'D = |6-3| = 3 \). No. Let D be the point (3,3). No. Let's use the points B' (3, -2) and C'(6,3). Drop a perpendicular from C' to the line y = -2. The point of intersection is (6, -2). So, B' \( (3,-2) \), D \( (6,-2) \), C' \( (6,3) \). The horizontal distance B'D = \( |6-3| = 3 \) units. The vertical distance C'D = \( |3-(-2)| = 5 \) units. Using Pythagoras theorem in \( \triangle B'DC' \):
\( (B'C')^2 = (B'D)^2 + (C'D)^2 \)
\( (B'C')^2 = 3^2 + 5^2 \)
\( (B'C')^2 = 9 + 25 \)
\( (B'C')^2 = 34 \)
\( B'C' = \sqrt{34} \) units.
In simple words: First, we find where point B goes when reflected across the x-axis, calling it B'. Then, we figure out the line that connects B and B'. This line acts as a new mirror for point C, creating C'. Finally, we measure the straight distance between B' and C' using the distance formula or by forming a right triangle and using Pythagoras.

🎯 Exam Tip: When calculating distance, the distance formula is reliable. Alternatively, drawing a right-angled triangle between the two points and using the Pythagorean theorem can be quicker for points easily plotted on a grid. Ensure the line of reflection for C is correctly identified as a vertical line through B and B'.

Question 11. The image of a point P under reflection in the x-axis is (5, -2). Write down the co-ordinates of P.
Answer:
Let the original point be P \( (x, y) \).
When P \( (x, y) \) is reflected in the x-axis, its image is \( P' (x, -y) \).
We are given that the image \( P' \) has coordinates \( (5, -2) \).
Comparing the coordinates, we get:
\( x = 5 \)
\( -y = -2 \implies y = 2 \)
Therefore, the coordinates of the original point P are \( (5, 2) \).
In simple words: We know where a point landed after being reflected over the x-axis. To find out where it started, we just do the opposite of the x-axis reflection rule: keep the x-value the same and flip the sign of the y-value back.

🎯 Exam Tip: Reflection is a reversible transformation. If you know the image and the line of reflection, you can easily find the original point by applying the inverse reflection rule. For x-axis reflection, the x-coordinate is invariant.

Question 12. Draw a square whose vertices are (3, 3), (5, 3), (5, 5) and (3, 5). Reflect the square in the y-axis and then reflect the image in the origin. What single transformation would give the same result ?
Answer:
1. Plot the vertices of the square ABCD:
\( A (3, 3) \), \( B (5, 3) \), \( C (5, 5) \), and \( D (3, 5) \). Join these points to form square ABCD.

2. Reflect square ABCD in the y-axis:
The rule for reflection in the y-axis is \( (x, y) \to (-x, y) \).
\( A (3, 3) \to A' (-3, 3) \)
\( B (5, 3) \to B' (-5, 3) \)
\( C (5, 5) \to C' (-5, 5) \)
\( D (3, 5) \to D' (-3, 5) \)
Plot \( A', B', C', D' \) to form the image square A'B'C'D'.

3. Reflect square A'B'C'D' in the origin:
The rule for reflection in the origin is \( (x, y) \to (-x, -y) \).
\( A' (-3, 3) \to A'' (-(-3), -3) = A'' (3, -3) \)
\( B' (-5, 3) \to B'' (-(-5), -3) = B'' (5, -3) \)
\( C' (-5, 5) \to C'' (-(-5), -5) = C'' (5, -5) \)
\( D' (-3, 5) \to D'' (-(-3), -5) = D'' (3, -5) \)
Plot \( A'', B'', C'', D'' \) to form the final image square A''B''C''D''.

4. Determine the single transformation:
Compare the coordinates of the original square ABCD with the final image A''B''C''D'':
A \( (3, 3) \) maps to \( A'' (3, -3) \)
B \( (5, 3) \) maps to \( B'' (5, -3) \)
C \( (5, 5) \) maps to \( C'' (5, -5) \)
D \( (3, 5) \) maps to \( D'' (3, -5) \)
We observe that for each point \( (x, y) \) in ABCD, the corresponding point in A''B''C''D'' is \( (x, -y) \). This is the rule for reflection in the x-axis.
Therefore, a single reflection in the x-axis would give the same result as reflecting in the y-axis followed by reflecting in the origin.
In simple words: We draw a square. First, we flip it across the y-axis. Then, we take that new square and flip it across the center point (the origin). When we look at the final square, we notice that it looks exactly like what we would get if we had just flipped the original square directly across the x-axis.

🎯 Exam Tip: Successive transformations can often be replaced by a single equivalent transformation. In this case, reflection in y-axis followed by reflection in origin is equivalent to reflection in x-axis. Similarly, reflection in x-axis followed by reflection in origin is equivalent to reflection in y-axis.

Question 13. The triangle ABC, where A (1, 2), B (4, 8), C (6, 8), is reflected in the x-axis to triangle A'B'C'. Triangle A'B'C' is then reflected in the origin to triangle A”B”C”. Write down the coordinates of A”, B”, C”. Write down a single transformation that maps ABC onto A”B”C”.
Answer:
Given vertices of triangle ABC: \( A (1, 2), B (4, 8), C (6, 8) \).

1. Reflection of triangle ABC in the x-axis to get A'B'C':
The rule for reflection in the x-axis is \( (x, y) \to (x, -y) \).
For A \( (1, 2) \), the image is \( A' (1, -2) \).
For B \( (4, 8) \), the image is \( B' (4, -8) \).
For C \( (6, 8) \), the image is \( C' (6, -8) \).

2. Reflection of triangle A'B'C' in the origin to get A”B”C”:
The rule for reflection in the origin is \( (x, y) \to (-x, -y) \).
For \( A' (1, -2) \), the image is \( A'' (-1, -(-2)) = A'' (-1, 2) \).
For \( B' (4, -8) \), the image is \( B'' (-4, -(-8)) = B'' (-4, 8) \).
For \( C' (6, -8) \), the image is \( C'' (-6, -(-8)) = C'' (-6, 8) \).
So, the coordinates of A”, B”, C” are \( A'' (-1, 2), B'' (-4, 8), C'' (-6, 8) \).

3. Single transformation that maps ABC onto A”B”C”:
Compare the original points A, B, C with their final images A'', B'', C'':
A \( (1, 2) \to A'' (-1, 2) \)
B \( (4, 8) \to B'' (-4, 8) \)
C \( (6, 8) \to C'' (-6, 8) \)
We can see that for each point \( (x, y) \), the corresponding image is \( (-x, y) \). This is the rule for reflection in the y-axis.
Therefore, the single transformation that maps triangle ABC onto triangle A”B”C” is a reflection in the y-axis.
In simple words: We start with a triangle. First, we flip it across the x-axis. Then, we take that flipped triangle and flip it across the center point (the origin). When we compare the starting triangle to the final one, we notice that the final position is exactly what we would get if we had just flipped the original triangle across the y-axis in one go.

🎯 Exam Tip: When dealing with multiple transformations, always write down the coordinates after each step. Comparing the initial and final coordinates helps in identifying the single equivalent transformation. This specific sequence of reflection (x-axis then origin) is equivalent to reflection in the y-axis.

Question 14. The point P (a, b) is first reflected in the origin, and then reflected in the y-axis to P'. If P' has coordinates (3, -4), evaluate a, b.
Answer:
Let the original point be P \( (a, b) \).

1. Reflection of P \( (a, b) \) in the origin:
The rule for reflection in the origin is \( (x, y) \to (-x, -y) \).
So, the image \( P_1 \) after reflection in the origin is \( P_1 (-a, -b) \).

2. Reflection of \( P_1 (-a, -b) \) in the y-axis to get P':
The rule for reflection in the y-axis is \( (x, y) \to (-x, y) \).
So, the image P' will be \( P' (-(-a), -b) = P' (a, -b) \).

We are given that P' has coordinates \( (3, -4) \).
Comparing our derived P' \( (a, -b) \) with the given P' \( (3, -4) \):
\( a = 3 \)
\( -b = -4 \implies b = 4 \)
Therefore, the values are \( a = 3 \) and \( b = 4 \).
In simple words: We imagine a point P at \( (a, b) \). We first flip it across the center (origin), then flip that new point across the y-axis. This gives us a final point P' at \( (a, -b) \). Since we know P' is \( (3, -4) \), we can easily figure out that \( a \) must be 3 and \( b \) must be 4.

🎯 Exam Tip: When working backwards through multiple transformations, carefully apply the inverse of each rule. In this specific sequence (origin then y-axis), an initial point \( (a,b) \) becomes \( (a,-b) \). If the final image is \( (x',y') \), then \( a=x' \) and \( b=-y' \).

Question 15. The point P (-3, -2) on reflection in the x-axis is mapped as P'. Then P' on reflection in the origin is mapped as P”. Find the coordinates of P' and P".
Answer:
Given point P \( (-3, -2) \).

1. Reflection of P \( (-3, -2) \) in the x-axis to get P':
The rule for reflection in the x-axis is \( (x, y) \to (x, -y) \).
For P \( (-3, -2) \), the image P' will be \( (-3, -(-2)) = (-3, 2) \).
So, the coordinates of P' are \( (-3, 2) \).

2. Reflection of P' \( (-3, 2) \) in the origin to get P”:
The rule for reflection in the origin is \( (x, y) \to (-x, -y) \).
For P' \( (-3, 2) \), the image P” will be \( (-(-3), -2) = (3, -2) \).
So, the coordinates of P” are \( (3, -2) \).
In simple words: We start with point P. First, we flip P over the x-axis to get P'. Then, we take P' and flip it over the center point (the origin) to get P''. We just apply each reflection rule step-by-step.

🎯 Exam Tip: Always show each step of the transformation clearly, writing down the coordinates of the intermediate images (like P' in this case). This helps avoid errors in complex sequences of reflections.

Question 16. Point A (5, -1) on reflection in the x-axis is mapped as A'. Also A on reflection in the y-axis is mapped as A”. Write the co-ordinates of A' and A”. Also calculate the distance AA”.
Answer:
Given point A \( (5, -1) \).

1. To find the coordinates of A' (reflection of A in the x-axis):
The rule for reflection in the x-axis is \( (x, y) \to (x, -y) \).
For A \( (5, -1) \), the image A' will be \( (5, -(-1)) = (5, 1) \).
So, the coordinates of A' are \( (5, 1) \).

2. To find the coordinates of A” (reflection of A in the y-axis):
The rule for reflection in the y-axis is \( (x, y) \to (-x, y) \).
For A \( (5, -1) \), the image A” will be \( (-5, -1) \).
So, the coordinates of A” are \( (-5, -1) \).

3. To calculate the distance AA”:
We have A \( (5, -1) \) and A” \( (-5, -1) \).
We use the distance formula: \( \text{Distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \)
Let \( (x_1, y_1) = (5, -1) \) and \( (x_2, y_2) = (-5, -1) \).
\( AA'' = \sqrt{(-5 - 5)^2 + (-1 - (-1))^2} \)
\( AA'' = \sqrt{(-10)^2 + (-1 + 1)^2} \)
\( AA'' = \sqrt{(-10)^2 + (0)^2} \)
\( AA'' = \sqrt{100 + 0} \)
\( AA'' = \sqrt{100} \)
\( AA'' = 10 \) units.
In simple words: We start with point A. We find its reflection across the x-axis, which is A'. Then, we find its reflection across the y-axis, which is A''. Finally, we measure the straight line distance between the original point A and its second reflection A''.

🎯 Exam Tip: The distance formula \( \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2} \) is essential for calculating lengths between points. If the y-coordinates are the same, the distance is simply the absolute difference of the x-coordinates, and similarly for x-coordinates. This is a common application of coordinate geometry.

Question 17. Point A (2, -4) is reflected in origin as A'. Point B (-3, 2) is reflected in x-axis as B'. Write the coordinates of A' and B'. Calculate the distance A'B'. Give your answer correct to 1 decimal place. (Do not consult tables)
Answer:
Given points: A \( (2, -4) \) and B \( (-3, 2) \).

1. To find the coordinates of A' (reflection of A in the origin):
The rule for reflection in the origin is \( (x, y) \to (-x, -y) \).
For A \( (2, -4) \), the image A' will be \( (-2, -(-4)) = (-2, 4) \).
So, the coordinates of A' are \( (-2, 4) \).

2. To find the coordinates of B' (reflection of B in the x-axis):
The rule for reflection in the x-axis is \( (x, y) \to (x, -y) \).
For B \( (-3, 2) \), the image B' will be \( (-3, -2) \).
So, the coordinates of B' are \( (-3, -2) \).

3. To calculate the distance A'B':
We have A' \( (-2, 4) \) and B' \( (-3, -2) \).
Using the distance formula: \( \text{Distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \)
Let \( (x_1, y_1) = (-2, 4) \) and \( (x_2, y_2) = (-3, -2) \).
\( A'B' = \sqrt{(-3 - (-2))^2 + (-2 - 4)^2} \)
\( A'B' = \sqrt{(-3 + 2)^2 + (-6)^2} \)
\( A'B' = \sqrt{(-1)^2 + (-6)^2} \)
\( A'B' = \sqrt{1 + 36} \)
\( A'B' = \sqrt{37} \)
To give the answer correct to 1 decimal place, we estimate \( \sqrt{37} \).
Since \( 6^2 = 36 \), \( \sqrt{37} \) is slightly more than 6.
\( \sqrt{37} \approx 6.08 \)
Rounding to 1 decimal place, \( A'B' \approx 6.1 \) units.
In simple words: We take point A and flip it over the center to get A'. Then, we take point B and flip it over the x-axis to get B'. After finding both new points, we measure the straight line distance between A' and B' and round the answer.

🎯 Exam Tip: Remember to carry out each transformation correctly before calculating the distance. When rounding, look at the second decimal place: if 5 or higher, round up the first decimal place; otherwise, keep it the same.

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 the coordinates of A' and B. Write the coordinates of the middle point of the line segment A'B.
Answer:
Given: Point A \( (4, -1) \). Point B' \( (-2, 5) \) is the image of B reflected in the x-axis.

1. To find the coordinates of A' (reflection of A in the y-axis):
The rule for reflection in the y-axis is \( (x, y) \to (-x, y) \).
For A \( (4, -1) \), the image A' will be \( (-4, -1) \).
So, the coordinates of A' are \( (-4, -1) \).

2. To find the coordinates of B (original point from B'):
B' \( (-2, 5) \) is the image of B \( (x, y) \) reflected in the x-axis.
The x-axis reflection rule is \( (x, y) \to (x, -y) \).
So, for B \( (x, y) \), its image would be \( (x, -y) \).
Comparing this with B' \( (-2, 5) \):
\( x = -2 \)
\( -y = 5 \implies y = -5 \)
So, the coordinates of the original point B are \( (-2, -5) \).

3. To find the coordinates of the middle point of the line segment A'B:
We have A' \( (-4, -1) \) and B \( (-2, -5) \).
The midpoint formula is \( M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) \).
Let \( (x_1, y_1) = (-4, -1) \) and \( (x_2, y_2) = (-2, -5) \).
\( M = \left( \frac{-4 + (-2)}{2}, \frac{-1 + (-5)}{2} \right) \)
\( M = \left( \frac{-4 - 2}{2}, \frac{-1 - 5}{2} \right) \)
\( M = \left( \frac{-6}{2}, \frac{-6}{2} \right) \)
\( M = (-3, -3) \)
So, the coordinates of the middle point of A'B are \( (-3, -3) \).
In simple words: First, we find where point A goes when reflected across the y-axis, calling it A'. Then, we use the x-axis reflection rule in reverse to find the original point B from its reflection B'. Finally, we calculate the exact middle point between A' and B.

🎯 Exam Tip: When given an image and reflection line to find the original point, apply the inverse transformation: for reflection in x-axis \( (x, -y) \to (x, y) \), for reflection in y-axis \( (-x, y) \to (x, y) \). The midpoint formula is also critical for finding the center of a line segment.

Question 19. Point A (1, -5) is mapped as A' on reflection in the x-axis. Point B (3, 2) is mapped as B' on reflection in the origin. Write the coordinates of A' and B'. Calculate AB'.
Answer:
Given points: A \( (1, -5) \) and B \( (3, 2) \).

1. To find the coordinates of A' (reflection of A in the x-axis):
The rule for reflection in the x-axis is \( (x, y) \to (x, -y) \).
For A \( (1, -5) \), the image A' will be \( (1, -(-5)) = (1, 5) \).
So, the coordinates of A' are \( (1, 5) \).

2. To find the coordinates of B' (reflection of B in the origin):
The rule for reflection in the origin is \( (x, y) \to (-x, -y) \).
For B \( (3, 2) \), the image B' will be \( (-3, -2) \).
So, the coordinates of B' are \( (-3, -2) \).

3. To calculate the distance A'B':
We have A' \( (1, 5) \) and B' \( (-3, -2) \).
Using the distance formula: \( \text{Distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \)
Let \( (x_1, y_1) = (1, 5) \) and \( (x_2, y_2) = (-3, -2) \).
\( A'B' = \sqrt{(-3 - 1)^2 + (-2 - 5)^2} \)
\( A'B' = \sqrt{(-4)^2 + (-7)^2} \)
\( A'B' = \sqrt{16 + 49} \)
\( A'B' = \sqrt{65} \)
In simple words: We find A' by flipping A across the x-axis. Then, we find B' by flipping B across the center point (the origin). Finally, we use a formula to calculate how far apart A' and B' are.

🎯 Exam Tip: Be careful with the signs when applying reflection rules, especially when dealing with negative coordinates. The distance calculation involves squaring, so negative differences become positive squares, but always double-check subtraction of negative numbers.

Question 20. Attempt this question on the graph paper.
(i) Plot A (3, 2) and B (5, 4) on the graph paper. Take 2 cm = 1 unit on both axes.
(ii) Reflect A and B in the x-axis to A' and B'. Plot these on the same graph paper.
(iii) Write down
(a) the geometrical name of the figure ABBA',
(b) the axis of symmetry of ABB'A,
(c) the measure of angle ABB',
(d) The A" is the image of A when reflected in origin.
(e) The single transformation that maps A' onto A".
Answer:
(i) Plot the points A \( (3, 2) \) and B \( (5, 4) \) on the graph.

(ii) Reflect A and B in the x-axis to get A' and B':
The rule for reflection in the x-axis is \( (x, y) \to (x, -y) \).
For A \( (3, 2) \), the image A' is \( (3, -2) \).
For B \( (5, 4) \), the image B' is \( (5, -4) \).
Plot A' \( (3, -2) \) and B' \( (5, -4) \) on the same graph paper.

(iii) Write down:
(a) Joining A, B, B', and A' forms the figure ABB'A'. This figure has two parallel sides (AB and A'B') and two non-parallel sides of equal length (AA' and BB'). Therefore, it is an isosceles trapezium.
(b) The axis of symmetry for the figure ABB'A' is the x-axis itself. This is because the figure is symmetric about the line where the reflection occurred.
(c) The measure of angle ABB': To find \( \angle ABB' \), we consider the line segment AB and the line segment BB'. B is \( (5, 4) \) and B' is \( (5, -4) \). The line BB' is a vertical line along \( x=5 \). A is \( (3, 2) \). The slope of AB is \( (4-2)/(5-3) = 2/2 = 1 \). The line BB' is vertical. The angle between a line with slope 1 and a vertical line can be found. The line AB makes an angle of 45 degrees with the positive x-axis. The line BB' is perpendicular to the x-axis. Therefore, the angle between AB and BB' is \( 90 - 45 = 45^\circ \). So, \( \angle ABB' = 45^\circ \).

(d) A" is the image of A when reflected in the origin.
Point A is \( (3, 2) \). The rule for reflection in the origin is \( (x, y) \to (-x, -y) \).
So, the coordinates of A" will be \( (-3, -2) \).

(e) The single transformation that maps A' onto A":
A' is \( (3, -2) \). A" is \( (-3, -2) \).
Comparing \( (3, -2) \) with \( (-3, -2) \), we see that the y-coordinate remains the same, but the x-coordinate changes its sign. This corresponds to a reflection in the y-axis.
Therefore, the single transformation that maps A' onto A" is reflection in the y-axis.
In simple words: We plot two points A and B. Then, we find their mirror images A' and B' by reflecting them across the x-axis. We then connect these points to form a shape, which we identify as an isosceles trapezium. We also find its mirror line. Next, we find A'', which is A reflected across the center point. Finally, we figure out how to get from A' to A'' in just one step, which turns out to be a reflection across the y-axis.

🎯 Exam Tip: Pay close attention to the scale (2 cm = 1 unit) if drawing a graph. For identifying quadrilaterals, check properties like parallel sides and equal side lengths. For angles with vertical/horizontal lines, slopes or basic trigonometry can quickly give the answer.

Self Evaluation And Revision

Question 1. Points (3, 0) and (-1, 0) are invariant points under reflection L₁ ; points (0, -3) and (0, 1) are invariant points under reflection in line L2. Write the equations for the lines L₁ and L2. Write down the images of P (3, 4) and Q (-5, -2) on reflection in L₁. Name the images as P' and Q' respectively. Write down the images of P and Q on reflection in L2. Name the images as P” and Q” respectively. State or describe a single transformation that maps P' onto P”.
Answer:
1. Equations for the lines L₁ and L₂:
For line L₁: Points \( (3, 0) \) and \( (-1, 0) \) are invariant. Since both points have a y-coordinate of 0, the line of reflection L₁ must be the x-axis. The equation for L₁ is \( y = 0 \).
For line L₂: Points \( (0, -3) \) and \( (0, 1) \) are invariant. Since both points have an x-coordinate of 0, the line of reflection L₂ must be the y-axis. The equation for L₂ is \( x = 0 \).

2. Images of P \( (3, 4) \) and Q \( (-5, -2) \) on reflection in L₁ (x-axis):
The rule for reflection in the x-axis is \( (x, y) \to (x, -y) \).
For P \( (3, 4) \), the image P' is \( (3, -4) \).
For Q \( (-5, -2) \), the image Q' is \( (-5, -(-2)) = (-5, 2) \).

3. Images of P \( (3, 4) \) and Q \( (-5, -2) \) on reflection in L₂ (y-axis):
The rule for reflection in the y-axis is \( (x, y) \to (-x, y) \).
For P \( (3, 4) \), the image P” is \( (-3, 4) \).
For Q \( (-5, -2) \), the image Q” is \( (-(-5), -2) = (5, -2) \).

4. Single transformation that maps P' onto P”:
We have P' \( (3, -4) \) and P” \( (-3, 4) \).
Comparing these two points, we observe that both the x and y coordinates have changed their signs:
\( (3, -4) \to (-3, 4) \).
This transformation \( (x, y) \to (-x, -y) \) is a reflection in the origin.
Therefore, the single transformation that maps P' onto P” is reflection in the origin.
In simple words: First, we identify the mirror lines L1 and L2 because certain points don't move when reflected. Then, we find where points P and Q land when reflected in L1 (calling them P' and Q'). Next, we find where P and Q land when reflected in L2 (calling them P'' and Q''). Finally, we figure out how to get directly from P' to P'' with just one transformation, which turns out to be a flip across the center.

🎯 Exam Tip: Invariant points are points that lie on the line of reflection. If two points with the same x-coordinate are invariant, the line of reflection is the y-axis. If two points with the same y-coordinate are invariant, the line of reflection is the x-axis. This is a quick way to find the equation of the reflection line.

Question 2.
(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 coordinate of P".
(iii) Name a single transformation that maps P' to P".
Answer:
(i) To find the values of a and b for point P \( (a, b) \):
If P \( (a, b) \) is reflected in the x-axis, its image is \( P' (a, -b) \).
We are given \( P' (5, -2) \).
Comparing the coordinates: \( a = 5 \) and \( -b = -2 \), which means \( b = 2 \).
So, the coordinates of point P are \( (5, 2) \).

(ii) To find the coordinates of P” (image of P reflected in the y-axis):
Point P is \( (5, 2) \).
The rule for reflection in the y-axis is \( (x, y) \to (-x, y) \).
For P \( (5, 2) \), the image P” will be \( (-5, 2) \).
So, the coordinates of P” are \( (-5, 2) \).

(iii) To name a single transformation that maps P' to P”:
We have P' \( (5, -2) \) and P” \( (-5, 2) \).
Comparing the coordinates, we see that both the x and y coordinates have changed their signs from P' to P”:
\( (5, -2) \to (-5, 2) \).
This transformation \( (x, y) \to (-x, -y) \) is a reflection in the origin.
Therefore, a single transformation that maps P' to P” is reflection in the origin.
In simple words: First, we use the x-axis reflection rule backwards to find the original point P. Then, we take P and reflect it across the y-axis to find P''. Finally, we look at P' and P'' and see that P'' is what you get if you flip P' across the center point.

🎯 Exam Tip: Remember to clearly state the coordinates of intermediate points (like P) before proceeding to the next step. Also, practice identifying common single transformations (reflection in x-axis, y-axis, origin, or specific lines) from the coordinate changes.

Question 3.
(i) Plot each of the given points on graph paper, reflect them in the x-axis and then reflect the points obtained in the y-axis. Write down the co-ordinates of the points obtained.
(a) (3, 4)
(b) (-3, 2)
(c) (5, 0)
(d) (-3, -3)
(ii) What happens if you reflect first in the y-axis and then in the x-axis ?
Answer:
(i) We apply the reflection rule: (x, y) reflected in the x-axis becomes (x, -y), and then reflected in the y-axis becomes (-x, -y).
(a) For point (3, 4):
Reflection in x-axis: (3, -4)
Reflection in y-axis: (-3, -4)
(b) For point (-3, 2):
Reflection in x-axis: (-3, -2)
Reflection in y-axis: (3, -2)
(c) For point (5, 0):
Reflection in x-axis: (5, 0) (This point is on the x-axis, so it is invariant)
Reflection in y-axis: (-5, 0)
(d) For point (-3, -3):
Reflection in x-axis: (-3, -(-3)) = (-3, 3)
Reflection in y-axis: (-(-3), 3) = (3, 3)

(ii) If we reflect first in the y-axis and then in the x-axis, the final coordinates are the same as reflecting first in the x-axis and then in the y-axis. The result is equivalent to reflecting the original point in the origin.
For example, taking point (3, 4):
Reflection in y-axis: (-3, 4)
Reflection in x-axis: (-3, -4)
This matches the result from part (i)(a). This means the order of reflection across perpendicular axes does not change the final image.
In simple words: To find the new spots, first flip the point over the x-axis, then flip that new spot over the y-axis. If you do it the other way around – first flip over the y-axis, then over the x-axis – you will end up in the exact same final place. It's like a double flip that always gives the same ending position.

🎯 Exam Tip: Remember that reflection in the x-axis changes the sign of the y-coordinate, reflection in the y-axis changes the sign of the x-coordinate, and reflection in the origin changes the signs of both coordinates. If a point lies on the axis of reflection, it remains unchanged.

Question 4.
A point P (a, b) is reflected in the y-axis to P' (-3, 5). Write down the values of a and b.
P" is the image of P, when reflected in the x-axis. Write down the coordinates of P".
Find the coordinates of P'", when P is reflected in the line, parallel to the x-axis, such that its equation is y = -3.
Answer:
A point P(a, b) reflected in the y-axis results in the image P'(-a, b).
Given that P' is (-3, 5).
Comparing the coordinates, we get:
\( -a = -3 \implies a = 3 \)
\( b = 5 \)
So, the coordinates of point P are (3, 5).

Next, P" is the image of P when reflected in the x-axis.
Reflection of P(3, 5) in the x-axis means (x, y) becomes (x, -y).
So, P"(3, -5).
The coordinates of P" are (3, -5).

Finally, P''' is the image of P when reflected in the line y = -3.
The rule for reflection in the line y = k is (x, y) becomes (x, 2k-y). Here, k = -3.
For P(3, 5) reflected in y = -3:
P'''(3, \( 2 \times (-3) - 5 \))
P'''(3, \( -6 - 5 \))
P'''(3, -11)
The coordinates of P''' are (3, -11).
In simple words: First, we use the y-axis reflection rule to find point P. Then, we flip point P over the x-axis to get P". After that, we flip point P over the horizontal line y=-3 to find P'''. Each flip has a specific rule to find the new coordinates.

🎯 Exam Tip: When reflecting a point, carefully identify the line of reflection (x-axis, y-axis, origin, or a specific line) and apply the correct coordinate transformation rule. Mistakes often happen with signs or using the wrong coordinate for the line of reflection.

Question 5.
Use graph paper for this question.
(i) Plot the points P (2, -4). Use 1 cm = 1 unit on both the axes.
(ii) P' is the image of P when reflected in AB, which is parallel to the x-axis and is at a distance 1 on the positive side of y-axis. (i.e. the line y = 1).
(iii) P" is the image of P' when reflected in the line LM which is parallel to the y-axis is at a distance 1 on the negative side of the x-axis. (i.e., the line x = -1).
Answer:
(i) The point P (2, -4) is plotted as shown in the graph below.

(ii) P' is the image of P (2, -4) reflected in the line y = 1.
The rule for reflection in the line y = k is (x, y) becomes (x, 2k-y). Here, k = 1.
P'(2, \( 2 \times 1 - (-4) \))
P'(2, \( 2 + 4 \))
P'(2, 6)
The coordinates of P' are (2, 6).

(iii) P" is the image of P' (2, 6) reflected in the line x = -1.
The rule for reflection in the line x = k is (x, y) becomes (2k-x, y). Here, k = -1.
P''( \( 2 \times (-1) - 2 \), 6)
P''( \( -2 - 2 \), 6)
P''(-4, 6)
The coordinates of P" are (-4, 6).

🎯 Exam Tip: When reflecting across a line like y=k or x=k, remember the formula: for y=k, (x, y) becomes (x, 2k-y); for x=k, (x, y) becomes (2k-x, y). Draw the lines of reflection on your graph paper to visualize the transformations correctly.

Question 6.
A point P is reflected in the x-axis. Co-ordinates of its image are (8, -6).
(i) Find the co-ordinates of P
(ii) Find the co-ordinates of the image of P under reflection in the y-axis. (ICSE)
Answer:
(i) Let the coordinates of point P be (x, y).
The rule for reflection in the x-axis is (x, y) becomes (x, -y).
Given that the image of P is (8, -6).
Comparing (x, -y) with (8, -6), we get:
\( x = 8 \)
\( -y = -6 \implies y = 6 \)
Therefore, the coordinates of P are (8, 6).

(ii) Now, we need to find the image of P (8, 6) under reflection in the y-axis.
The rule for reflection in the y-axis is (x, y) becomes (-x, y).
So, the image of P(8, 6) will be (-8, 6).
In simple words: If a point is flipped over the x-axis, only its y-value changes sign. We use this to find the original point. Then, if we flip that original point over the y-axis, only its x-value changes sign. This helps us find its new reflected position.

🎯 Exam Tip: Understand that reflection is a reversible transformation. If you know the image and the line of reflection, you can find the original point by applying the reflection rule in reverse or simply applying the rule again if it's across a coordinate axis.

Question 7.
The image of the point (1, 5) when reflected in a line LM is (9, 5). Write down the equation of the line LM.
Answer:
Let the original point be A(1, 5) and its image be A'(9, 5).
We observe that the y-coordinates of A and A' are the same (which is 5). This tells us that the line of reflection, LM, must be a vertical line, parallel to the y-axis. The equation of such a line is of the form x = k.
For a reflection in the line x = k, the rule is (x, y) becomes (2k-x, y).
Applying this rule to A(1, 5) to get A'(9, 5):
\( 2k - 1 = 9 \)
\( 2k = 9 + 1 \)
\( 2k = 10 \)
\( k = 5 \)
Thus, the equation of the line LM is x = 5. This line acts like a mirror, perfectly centered between the original point and its reflection.
In simple words: We have a point and its reflection. Since their 'up and down' position (y-coordinate) did not change, the mirror line must be a straight line going 'up and down' (a vertical line). We find the middle of the 'left and right' positions (x-coordinates) of the point and its reflection to get the exact location of this vertical mirror line.

🎯 Exam Tip: If the y-coordinates of a point and its image are the same, the line of reflection is a vertical line (x=k). If the x-coordinates are the same, the line of reflection is a horizontal line (y=k). The line of reflection is always the perpendicular bisector of the segment connecting the point and its image.

Question 8.
Use graph paper to plot the triangle ABC where A is (1, 2), B is (3, 4) and C is (6,1). On the same graph paper plot.
(i) the image A', B', C' of the triangle ABC under reflection in the origin O (0, 0);
(ii) the image of triangle ABC under reflection in the y-axis followed by a reflection in the x-axis.
Compare your results for (i) and (ii) above and make a statement connecting the two results.
Answer:
The vertices of triangle ABC are A(1, 2), B(3, 4), and C(6, 1).

(i) Reflection in the origin O(0, 0): The rule is (x, y) becomes (-x, -y).
A(1, 2) \( \implies \) A'(-1, -2)
B(3, 4) \( \implies \) B'(-3, -4)
C(6, 1) \( \implies \) C'(-6, -1)
The coordinates of the image triangle A'B'C' are A'(-1, -2), B'(-3, -4), and C'(-6, -1).

(ii) First, reflection in the y-axis: The rule is (x, y) becomes (-x, y).
A(1, 2) \( \implies \) A''(-1, 2)
B(3, 4) \( \implies \) B''(-3, 4)
C(6, 1) \( \implies \) C''(-6, 1)

Next, reflection of A''B''C'' in the x-axis: The rule is (x, y) becomes (x, -y).
A''(-1, 2) \( \implies \) A'''(-1, -2)
B''(-3, 4) \( \implies \) B'''(-3, -4)
C''(-6, 1) \( \implies \) C'''(-6, -1)
The coordinates of the image triangle A'''B'''C''' are A'''(-1, -2), B'''(-3, -4), and C'''(-6, -1).

Comparison: The coordinates obtained in (i) for A'B'C' are A'(-1, -2), B'(-3, -4), C'(-6, -1). The coordinates obtained in (ii) for A'''B'''C''' are A'''(-1, -2), B'''(-3, -4), C'''(-6, -1).
Both sets of coordinates are identical. This shows that reflecting a point in the y-axis followed by a reflection in the x-axis is equivalent to a single reflection in the origin. This sequential reflection creates the same image as a direct reflection through the center point (0,0).
In simple words: We start with a triangle and find its reflection if we flip it through the very center point (origin). Then, we find its reflection by first flipping it over the 'up-down' line (y-axis), and then flipping that result over the 'left-right' line (x-axis). We notice that both ways give us the same final triangle. So, flipping over the y-axis and then the x-axis is the same as just flipping through the origin.

🎯 Exam Tip: When comparing transformations, visually plot the points if possible. This helps confirm your calculations. Remember that a reflection in the y-axis followed by a reflection in the x-axis is a common combination that simplifies to a single reflection in the origin.

Question 9.
On the graph paper, taking 1 cm = 1 unit, plot the triangle ABC whose vertices are at the points A (3, 1), B (5, 0) and C (7, 4). On the same diagram, draw the image of the AABC under reflection in the line x = 2. Mark the point invariant under this reflection.
Answer:
The vertices of triangle ABC are A(3, 1), B(5, 0), and C(7, 4).

Reflection in the line x = 2: The rule for reflection in the line x = k is (x, y) becomes (2k-x, y). Here, k = 2.
A(3, 1) \( \implies \) A'( \( 2 \times 2 - 3 \), 1) = A'(1, 1)
B(5, 0) \( \implies \) B'( \( 2 \times 2 - 5 \), 0) = B'(-1, 0)
C(7, 4) \( \implies \) C'( \( 2 \times 2 - 7 \), 4) = C'(-3, 4)
The image triangle A'B'C' has vertices A'(1, 1), B'(-1, 0), and C'(-3, 4).

An invariant point under reflection is any point that lies on the line of reflection itself. The line of reflection is x = 2. Therefore, any point with an x-coordinate of 2 is an invariant point. A common example to mark is (2, 0), where the line x=2 crosses the x-axis. This point stays in its place after the reflection.
In simple words: First, we plot the triangle ABC. Then, we flip each corner of the triangle over the vertical line x=2 to get a new triangle A'B'C'. Any point that is exactly on the line x=2 will not move during this flip; this is called an invariant point, like (2, 0).

🎯 Exam Tip: When reflecting across a vertical line x=k, the y-coordinate remains unchanged. The new x-coordinate is found by \( 2k-x \). Always mark the line of reflection clearly on your graph to avoid errors and show invariant points if asked.

Question 10.
B, C have co-ordinates (3, 2) and (0, 3). Find
(i) the image B' of B under the reflection in the x-axis;
(ii) the image C' of C under reflection in the line BB';
(iii) Calculate the length of B'C'.
Answer:
The given points are B(3, 2) and C(0, 3).

(i) Reflection of B(3, 2) in the x-axis: The rule is (x, y) becomes (x, -y).
B(3, 2) \( \implies \) B'(3, -2)
The coordinates of B' are (3, -2).

(ii) Reflection of C(0, 3) in the line BB'.
First, we need to find the equation of the line BB'. B is (3, 2) and B' is (3, -2).
Since both B and B' have the same x-coordinate (3), the line BB' is a vertical line with the equation x = 3.
Now, reflect C(0, 3) in the line x = 3. The rule for reflection in the line x = k is (x, y) becomes (2k-x, y). Here, k = 3.
C(0, 3) \( \implies \) C'( \( 2 \times 3 - 0 \), 3) = C'(6, 3)
The coordinates of C' are (6, 3).

(iii) Calculate the length of B'C'.
B' is (3, -2) and C' is (6, 3).
Using the distance formula \( d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2} \):
\( B'C' = \sqrt{(6-3)^2 + (3-(-2))^2} \)
\( B'C' = \sqrt{(3)^2 + (3+2)^2} \)
\( B'C' = \sqrt{3^2 + 5^2} \)
\( B'C' = \sqrt{9 + 25} \)
\( B'C' = \sqrt{34} \) units. For distance calculation, it's often helpful to imagine a right triangle formed by the points and the axes.
In simple words: First, we flip point B over the x-axis to get B'. Then, we use B and B' to find a straight vertical line. We flip point C over this vertical line to get C'. Finally, we measure the distance between B' and C' using a special ruler formula.

🎯 Exam Tip: When a point is reflected in a line formed by two other points, first find the equation of that line. In this case, since the x-coordinates of B and B' are the same, the line is a vertical line. Then apply the appropriate reflection rule.

Question 11.
The image of a point P under reflection in the x-axis is (5, -2). Write down the co-ordinates of P.
Answer:
Let the coordinates of point P be (x, y).
The rule for reflection in the x-axis is (x, y) becomes (x, -y).
Given that the image of P is (5, -2).
Comparing (x, -y) with (5, -2), we get:
\( x = 5 \)
\( -y = -2 \implies y = 2 \)
Therefore, the coordinates of P are (5, 2). The x-axis acts as a mirror, and the original point is the "pre-image" of the given reflection.
In simple words: When you flip a point over the x-axis, its 'left-right' number stays the same, but its 'up-down' number changes sign. Since we know the flipped point, we can just change the sign of its 'up-down' number back to find the original point.

🎯 Exam Tip: Reflection in the x-axis only affects the y-coordinate by changing its sign. The x-coordinate remains unchanged. Use this property to quickly find the original coordinates from an image.

Question 12.
Draw a square whose vertices are (3, 3), (5, 3), (5, 5) and (3, 5). Reflect the square in the y-axis and then reflect the image in the origin. What single transformation would give the same result ?
Answer:
The vertices of the square are A(3, 3), B(5, 3), C(5, 5), and D(3, 5).

1. Reflection in the y-axis: The rule is (x, y) becomes (-x, y).
A(3, 3) \( \implies \) A'(-3, 3)
B(5, 3) \( \implies \) B'(-5, 3)
C(5, 5) \( \implies \) C'(-5, 5)
D(3, 5) \( \implies \) D'(-3, 5)
The image square A'B'C'D' has vertices A'(-3, 3), B'(-5, 3), C'(-5, 5), and D'(-3, 5).

2. Reflection of A'B'C'D' in the origin: The rule is (x, y) becomes (-x, -y).
A'(-3, 3) \( \implies \) A''(-(-3), -3) = A''(3, -3)
B'(-5, 3) \( \implies \) B''(-(-5), -3) = B''(5, -3)
C'(-5, 5) \( \implies \) C''(-(-5), -5) = C''(5, -5)
D'(-3, 5) \( \implies \) D''(-(-3), -5) = D''(3, -5)
The final image square A''B''C''D'' has vertices A''(3, -3), B''(5, -3), C''(5, -5), and D''(3, -5).

Comparing the original coordinates (A(3, 3), B(5, 3), C(5, 5), D(3, 5)) with the final coordinates (A''(3, -3), B''(5, -3), C''(5, -5), D''(3, -5)), we observe that the x-coordinates remain the same, while the y-coordinates have changed their signs.
This type of transformation, where (x, y) becomes (x, -y), is a reflection in the x-axis.
Therefore, a single transformation that would give the same result is a reflection in the x-axis. This shows how multiple transformations can sometimes be represented by a simpler one.
In simple words: We start with a square and first flip it over the 'up-down' line (y-axis). Then, we flip that new square through the center point (origin). After all these flips, we look at the final square and realize we could have gotten to the same spot by just flipping the very first square over the 'left-right' line (x-axis) in one go.

🎯 Exam Tip: When performing multiple reflections, keep track of the coordinates after each step. If the final image's coordinates are a simple transformation (like only x-sign change or only y-sign change) of the original, it often means there's a single equivalent reflection.

Question 13.
The triangle ABC, where A (1, 2), B (4, 8), C (6, 8), is reflected in the x-axis to triangle A'B'C'. Triangle A'B'C' is then reflected in the origin to triangle A"B"C". Write down the coordinates of A", B", C". Write down a single transformation that maps ABC onto A"B"C".
Answer:
The vertices of triangle ABC are A(1, 2), B(4, 8), and C(6, 8).

1. Reflection in the x-axis to get A'B'C': The rule is (x, y) becomes (x, -y).
A(1, 2) \( \implies \) A'(1, -2)
B(4, 8) \( \implies \) B'(4, -8)
C(6, 8) \( \implies \) C'(6, -8)
The coordinates of triangle A'B'C' are A'(1, -2), B'(4, -8), and C'(6, -8).

2. Reflection of A'B'C' in the origin to get A''B''C'': The rule is (x, y) becomes (-x, -y).
A'(1, -2) \( \implies \) A''(-1, -(-2)) = A''(-1, 2)
B'(4, -8) \( \implies \) B''(-4, -(-8)) = B''(-4, 8)
C'(6, -8) \( \implies \) C''(-6, -(-8)) = C''(-6, 8)
The coordinates of A'', B'', C'' are A''(-1, 2), B''(-4, 8), and C''(-6, 8).

Now, to find a single transformation that maps ABC onto A''B''C'':
Compare original A(1, 2) with final A''(-1, 2). The x-coordinate changed sign, but the y-coordinate remained the same.
This corresponds to a reflection in the y-axis.
Similarly, for B(4, 8) to B''(-4, 8) and C(6, 8) to C''(-6, 8), the x-coordinate changes sign and y-coordinate remains the same.
Therefore, the single transformation that maps triangle ABC onto triangle A''B''C'' is a reflection in the y-axis. This combines two steps into one simpler operation.
In simple words: First, we flip the triangle over the x-axis. Then, we flip that new triangle through the center point (origin). When we compare the final triangle to the very first one, we see that it's the same as if we had just flipped the original triangle over the y-axis in one single move.

🎯 Exam Tip: When a problem involves a sequence of reflections, calculating the coordinates after each step helps identify the final positions. Often, a combination of two reflections (like x-axis then origin) simplifies to a single basic reflection (like y-axis), which is a key concept to remember.

Question 14.
The point P (a, b) is first reflected in the origin, and then reflected in the y-axis to P'. If P' has coordinates (3, -4), evaluate a, b.
Answer:
Let the original point be P(a, b).

1. Reflection in the origin: The rule is (x, y) becomes (-x, -y).
P(a, b) \( \implies \) P1(-a, -b)

2. Reflection of P1(-a, -b) in the y-axis: The rule is (x, y) becomes (-x, y).
P1(-a, -b) \( \implies \) P'(-(-a), -b) = P'(a, -b)

Given that the final coordinates of P' are (3, -4).
Comparing P'(a, -b) with P'(3, -4), we get:
\( a = 3 \)
\( -b = -4 \implies b = 4 \)
Therefore, the values are a = 3 and b = 4. Working backward from the final image, step-by-step, helps reveal the original point.
In simple words: We have a starting point P. First, we flip P through the center (origin). Then, we flip that new point over the 'up-down' line (y-axis) to get P'. We are told what P' is. By doing the flip rules backward, we can figure out what the original P point was.

🎯 Exam Tip: When working backward from a transformed image to the original point, apply the inverse of each transformation in reverse order. For reflections, the inverse is simply the same reflection again. This ensures you undo the changes correctly.

Question 15.
The point P (-3, -2) on reflection in the x-axis is mapped as P'. Then P' on reflection in the origin is mapped as P". Find the coordinates of P' and P".
Answer:
The original point is P(-3, -2).

1. Reflection in the x-axis to get P': The rule is (x, y) becomes (x, -y).
P(-3, -2) \( \implies \) P'(-3, -(-2)) = P'(-3, 2)
The coordinates of P' are (-3, 2).

2. Reflection of P'(-3, 2) in the origin to get P": The rule is (x, y) becomes (-x, -y).
P'(-3, 2) \( \implies \) P''(-(-3), -2) = P''(3, -2)
The coordinates of P" are (3, -2). Each reflection applies a distinct rule to the coordinates, leading to a new position.
In simple words: We start with point P. First, we flip P over the 'left-right' line (x-axis) to get P'. Then, we take P' and flip it through the center point (origin) to find P". We write down the coordinates for both P' and P".

🎯 Exam Tip: Always perform transformations step-by-step as given in the problem. A common error is to apply the second transformation to the original point instead of the already transformed image.

Question 16.
Point A (5, -1) on reflection in the x-axis is mapped as A'. Also A on reflection in the y-axis is mapped as A". Write the co-ordinates of A' and A". Also calculate the distance AA".
Answer:
The original point is A(5, -1).

1. Reflection in the x-axis to get A': The rule is (x, y) becomes (x, -y).
A(5, -1) \( \implies \) A'(5, -(-1)) = A'(5, 1)
The coordinates of A' are (5, 1).

2. Reflection in the y-axis to get A": The rule is (x, y) becomes (-x, y).
A(5, -1) \( \implies \) A"(-5, -1)
The coordinates of A" are (-5, -1).

Now, calculate the distance AA" using A'(5, 1) and A"(-5, -1).
Using the distance formula \( d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2} \):
\( AA" = \sqrt{(-5-5)^2 + (-1-1)^2} \)
\( AA" = \sqrt{(-10)^2 + (-2)^2} \)
\( AA" = \sqrt{100 + 4} \)
\( AA" = \sqrt{104} \) units. This distance represents the shortest path between the two reflected images, A' and A''.
In simple words: We start with point A. First, we flip A over the x-axis to get A'. Then, we flip A over the y-axis to get A". After finding where A' and A" are, we measure the straight line distance between them.

🎯 Exam Tip: Be careful with the signs, especially when subtracting negative numbers in the distance formula. A simple calculation error can lead to an incorrect final answer. If asked for distance, always provide units.

Question 17.
Point A (2, -4) is reflected in origin as A'. Point B (-3, 2) is reflected in x-axis as B'. Write the coordinates of A' and B'. Calculate the distance A'B'. Give your answer correct to 1 decimal place. (Do not consult tables)
Answer:
The original point A is (2, -4).

1. Reflection of A(2, -4) in the origin to get A': The rule is (x, y) becomes (-x, -y).
A(2, -4) \( \implies \) A'(-2, -(-4)) = A'(-2, 4)
The coordinates of A' are (-2, 4).

The original point B is (-3, 2).

2. Reflection of B(-3, 2) in the x-axis to get B': The rule is (x, y) becomes (x, -y).
B(-3, 2) \( \implies \) B'(-3, -2)
The coordinates of B' are (-3, -2).

Now, calculate the distance A'B' using A'(-2, 4) and B'(-3, -2).
Using the distance formula \( d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2} \):
\( A'B' = \sqrt{(-3-(-2))^2 + (-2-4)^2} \)
\( A'B' = \sqrt{(-3+2)^2 + (-6)^2} \)
\( A'B' = \sqrt{(-1)^2 + 36} \)
\( A'B' = \sqrt{1 + 36} \)
\( A'B' = \sqrt{37} \)
To 1 decimal place, \( \sqrt{37} \approx 6.0827 \) which is 6.1 units. The distance formula is useful for measuring the separation between any two points in the coordinate plane.
In simple words: First, we flip point A through the center (origin) to get A'. Then, we flip point B over the x-axis to get B'. After we find A' and B', we use a special math rule to measure the straight line distance between them, rounding the answer to one decimal place.

🎯 Exam Tip: When calculating distances, it's crucial to correctly identify the coordinates of the points after all transformations. Use parentheses carefully in the distance formula to avoid sign errors, especially when dealing with negative coordinates.

Question 19. Use a graph paper for this question taking 1 cm = 1 unit along both the x-axis and j-axis years:
(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 j-axis and name them respectively as B', C', D' and Eʹ.
(iii) Write the coordinates of B', C', D' and Eʹ.
(iv) Name the figure formed by BCDEE'D'C'B'.
(v) Name a line of symmetry for the figure formed.
Answer:
(i) The points A (0, 5), B (2, 5), C (5, 2), D (5, -2), E (2, -5), and F (0, -5) are plotted on the graph paper as specified.
(ii) To reflect points B, C, D, and E on the j-axis (which is the y-axis), we change the sign of their x-coordinates. The reflected points are named B', C', D', and E' respectively.
(iii) The coordinates of the reflected points are: B' (-2, 5), C' (-5, 2), D' (-5, -2), and E' (-2, -5). Reflection across the y-axis keeps the y-coordinate the same while reversing the x-coordinate's sign.
(iv) The figure formed by BCDEE'D'C'B' is an octagon. This eight-sided polygon is symmetric due to the reflections.
(v) The lines of symmetry for this octagon are the x-axis, the y-axis, the line \( y = x \), and the line \( y = -x \). There are also four other lines of symmetry that pass through the midpoints of opposite sides, making a total of eight lines of symmetry for a regular octagon.
In simple words: First, plot the given points on a graph. Then, reflect some of these points across the y-axis to get new points. The shape you get by joining these points will be an octagon, and it has eight lines where you can fold it perfectly in half.

🎯 Exam Tip: When reflecting a point across an axis, only one coordinate changes sign. For the x-axis, \( (x, y) \rightarrow (x, -y) \); for the y-axis, \( (x, y) \rightarrow (-x, y) \). For the origin, \( (x, y) \rightarrow (-x, -y) \).

Question 20. 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 the points A and B on they-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 the figure.
Answer:
The points O (0, 0), A (-4, 4), B (-3, 0), and C (0, -3) are plotted on the graph paper with the specified scale.
(i) When reflecting point A (-4, 4) on the y-axis, its image A' will have coordinates (4, 4). Similarly, reflecting point B (-3, 0) on the y-axis gives its image B' with coordinates (3, 0). Reflection across the y-axis means only the x-coordinate changes its sign.
(ii) The figure OABCB'A' formed by connecting these points is an irregular hexagon. A hexagon is any polygon with six sides.
(iii) The line of symmetry of the figure OABCB'A' is the y-axis. The entire figure is perfectly balanced across this vertical line.
In simple words: After plotting the points, reflect A and B across the y-axis to find A' and B'. The shape you get by joining O, A, B, C, B', A' is a hexagon. If you fold this shape along the y-axis, both sides will match up perfectly.

🎯 Exam Tip: Always double-check the coordinates of reflected points, especially when the line of reflection is one of the axes. The y-axis acts as a mirror for the x-coordinates.