Frank Brothers Solutions for ICSE Class 10 Mathematics Chapter 8 Reflection

Exercise 8.1

Answer 1.
Answer:
(i) (3,-9)
The co-ordinate of the given point under reflection in the x-axis is: (3,9).
(ii) (-7, 5)
The co-ordinate of the given point under reflection in the x-axis is: (-7, -5).
(iii) (0, 6)
The co-ordinate of the given point under reflection in the x-axis is: (0,-6).
(iv) (-4,-8)
The co-ordinate of the given point under reflection in the x-axis is: (-4, 8).
In simple words: When we reflect in x-axis, the x-value stays same but y-value changes sign. Positive becomes negative and negative becomes positive.

📝 Teacher's Note: Draw a coordinate plane and show how reflection in x-axis is like flipping the point up or down. The point moves to the opposite side of x-axis.

🎯 Exam Tip: In x-axis reflection, write (x,y) becomes (x,-y). Remember to change only the y-coordinate sign.

Answer 2.
Answer:
(i) (2, 8)
The co-ordinate of the given point under reflection in the y-axis is: (-2,8).
(ii) (-1,-3)
The co-ordinate of the given point under reflection in the y-axis is: (1,-3).
(iii) (5,-6)
The co-ordinate of the given point under reflection in the y-axis is: (-5,-6).
(iv) (-4, 7)
The co-ordinate of the given point under reflection in the y-axis is: (4, 7).
In simple words: When we reflect in y-axis, the y-value stays same but x-value changes sign. Left side points go to right side and right side points go to left side.

📝 Teacher's Note: Use a mirror placed on the y-axis to show reflection. Students can see how points on left go to right and vice versa.

🎯 Exam Tip: In y-axis reflection, write (x,y) becomes (-x,y). Remember to change only the x-coordinate sign.

Answer 3.
Answer:
(i) (-1,-4)
The co-ordinate of the given point under reflection in the origin is: (1, 4)
(ii) (2, 7)
The co-ordinate of the given point under reflection in the origin is: (-2,-7)
(iii) (0, 2)
The co-ordinate of the given point under reflection in the origin is: (0,-2)
(iv) (9,-9)
The co-ordinate of the given point under reflection in the origin is: (-9, 9)
In simple words: When we reflect in origin, both x and y values change sign. It is like rotating the point 180 degrees around the center.

📝 Teacher's Note: Show students that origin reflection is like doing both x-axis and y-axis reflections together. Both coordinates change sign.

🎯 Exam Tip: In origin reflection, write (x,y) becomes (-x,-y). Change the sign of both coordinates.

Answer 4.
Answer: P' = (2, 10). Therefore, the co-ordinates of P under reflection in the x-axis = (2,-10)
In simple words: We are given the reflected point and need to find the original point. Since x-axis reflection changes y-sign, we change the sign back.

📝 Teacher's Note: Explain that if P' is the reflection of P, then P is also the reflection of P'. Reflection works both ways.

🎯 Exam Tip: When finding original point from reflected point, apply the same reflection rule backwards.

Answer 5.
Answer: S= (2,-5). Therefore, the co-ordinates of S' under reflection in the origin = (-2, 5)
In simple words: We found the original point S and then reflected it in origin by changing both coordinate signs.

📝 Teacher's Note: Make students practice going from reflected point back to original point. This helps them understand reflection is reversible.

🎯 Exam Tip: Read the question carefully. Check if you need original point or reflected point.

Answer 6.
Answer: P' = (-3, 4).
Therefore, the co-ordinates of P under reflection in the x-axis = (-3,-4)
and the co-ordinates of P'' under reflection in the origin = (3,-4).
The single transformation = reflection in the y-axis.
In simple words: We did two reflections one after another. The final result is the same as doing one reflection in y-axis.

📝 Teacher's Note: Show that combining x-axis reflection and origin reflection gives the same result as y-axis reflection. Draw this step by step.

🎯 Exam Tip: When asked for single transformation, compare the original point with final point to see which reflection rule applies.

Answer 7.
Answer: P' = (8,-6).
Therefore, the co-ordinates of P under reflection in the x-axis = (8, 6)
and the co-ordinates of P'' under reflection in the y-axis = (-8, 6).
In simple words: First we reflected in x-axis, then in y-axis. This gave us a point in the second quadrant.

📝 Teacher's Note: Use coordinate plane to show how the point moves from one quadrant to another with each reflection.

🎯 Exam Tip: Do reflections step by step. Don't try to do both at once - you might make mistakes.

Answer 8.
Answer: R = (3,-2). Therefore, reflection of R in the origin is R' = (-3, 2)
Q = (-7, 1). Therefore, reflection of Q in the x-axis is Q' = (-7,-1)
Distance between R' Q' = \(\sqrt{(-7 - (-3))^2 + (-1 - 2)^2}\)
= \(\sqrt{(-4)^2 + (-3)^2}\)
= \(\sqrt{16 + 9}\)
= \(\sqrt{25}\)
= 5 units
In simple words: We reflected both points, then used distance formula to find how far apart the new points are.

📝 Teacher's Note: Teach distance formula clearly: \(\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}\). Make students practice this formula separately.

🎯 Exam Tip: Always show the reflection coordinates first, then apply distance formula. Write each step clearly.

Answer 9.
Answer: B = (3, 2). Therefore, reflection of B in the x-axis is B' = (3,-2)
C = (0, 3). Therefore, reflection of C in the line B' is C' = (6, 3).
In simple words: We reflected point B in x-axis first, then used that to reflect point C.

📝 Teacher's Note: When reflecting in a line like "line B'", find the equation of that line first. Here it appears to be a vertical line.

🎯 Exam Tip: Identify what type of line you are reflecting in - is it x-axis, y-axis, or some other line?

Answer 10.
Answer: P' = (5,-2), therefore, co-ordinates of P' = (-5, 2) and hence the coordinates of P = (-5,-2)
Single transformation = reflection in the y-axis
In simple words: We found the original point P, then checked which single reflection gives us P'. It was y-axis reflection.

📝 Teacher's Note: Students should check their answer by applying the reflection rule to see if they get the given reflected point.

🎯 Exam Tip: Always verify your answer. Apply your reflection rule to the original point to check if you get the reflected point.

Answer 11.
Answer: Let P be the point = (-2, 4).
Image under reflection in the origin P' = (2,-4)
Image under reflection in the y-axis P'' = (2, 4)
Distance between points of reflection = \(\sqrt{(4 - (-4))^2 + (2 - 2)^2}\)
= \(\sqrt{8^2}\)
= \(\sqrt{64}\)
= 8 units
In simple words: We reflected the point in two different ways, then found the distance between the two reflected points.

📝 Teacher's Note: Draw both reflected points on coordinate plane. Students can see that they are on opposite sides of the original point.

🎯 Exam Tip: Be careful with calculations. Check your arithmetic, especially when squaring negative numbers.

Answer 12.
Answer: A = (2, 3); B = (4,-4); C = (6,-7)
Co-ordinates of \(\triangle A'B'C'\) under reflection in the line y=0:
A' = (2,-3); B' = (4, 4); C' = (6, 7)
Co-ordinates of \(\triangle A''B''C''\) under reflection in the origin:
A'' = (-2, 3); B'' = (-4,-4); C'' = (-6,-7)
In simple words: We reflected the triangle first in x-axis (line y=0), then reflected the new triangle in origin.

📝 Teacher's Note: Line y=0 means the x-axis. Make this clear to students. They should know that y=0 is another way to write x-axis.

🎯 Exam Tip: Write coordinates of all vertices clearly. Use proper notation like A', A'' to show the sequence of reflections.

Answer 13.
Answer: P = (-8, 1), therefore co-ordinates of P' under reflection in the x-axis = (-8,-1).
Hence, the co-ordinates of P'' under reflection in the origin = (8, 1).
The single transformation = reflection in the y-axis.
In simple words: After two reflections, the final result is same as doing one y-axis reflection from the original point.

📝 Teacher's Note: This shows that x-axis reflection followed by origin reflection equals y-axis reflection. Students can remember this pattern.

🎯 Exam Tip: Learn the combination patterns. X-axis + Origin = Y-axis. This saves time in exams.

Answer 14.
Answer:
(i) \(M_x M_y\) on P (2,-5)
\(M_x M_y\) on P (2,-5)
= \(M_x M_y\) (2,-5)
= \(M_x\) (-2,-5)
= (-2, 5); reflection in the origin
(ii) \(M_y M_x\) on A (-7, 3)
\(M_y M_x\) on A (-7, 3)
=\(M_y M_x\) (-7, 3)
= \(M_y\) (7,-3)
= (-7,-3); reflection in the x-axis
(iii) \(M_x M_y\) on B (4, 6)
\(M_x M_y\) on B (4, 6)
=\(M_x M_y\) (4, 6)
= \(M_x\) (-4, 6)
= (4,-6); reflection in the x-axis
(iv) \(M_x M_x\) on P (-1,-3)
\(M_x M_x\) on P (-1,-3)
=\(M_x M_x\) (-1,-3)
= \(M_x\) (-1, 3)
= (-1,-3); reflection in the y-axis
In simple words: \(M_x\) means reflect in x-axis, \(M_y\) means reflect in y-axis. We do them in order from right to left.

📝 Teacher's Note: Explain that reflections are done from right to left, like reading backwards. Make students practice this order.

🎯 Exam Tip: Always work from right to left when doing multiple reflections. Write each step clearly to avoid confusion.

Answer 15.
Answer:
(i) y = 0
Co-ordinates of image = (-5, 2x0-4)= (-5,-4)
(ii) y = 4
Co-ordinates of image = (-5, 2x4-4)= (-5,4)
In simple words: When reflecting in line y = k, the x-coordinate stays same. The y-coordinate changes using the formula \(y' = 2k - y\).

📝 Teacher's Note: Teach the reflection formula for horizontal lines: when reflecting in y = k, new y = 2k - old y. X stays same.

🎯 Exam Tip: Remember the formula for reflection in y = k. Write "x stays same, y becomes 2k - y".

Answer 16.
Answer:
(i) x = 0
Co-ordinates of image =(2x0-4,-1)= (-4,-1)
(ii) y = 5
Co-ordinates of image = (4, 2x5-(-1)) = (4, 11)
In simple words: When reflecting in line x = k, the y-coordinate stays same. The x-coordinate changes using the formula \(x' = 2k - x\).

📝 Teacher's Note: Teach the reflection formula for vertical lines: when reflecting in x = k, new x = 2k - old x. Y stays same.

🎯 Exam Tip: Remember the formula for reflection in x = k. Write "y stays same, x becomes 2k - x".

Answer 17.
Answer:
(i) Find the co-ordinates of P'Q'R', Q' and R'
The co-ordinates are: P'Q'R' = (-1,-1); Q' = (-4,-1); R' = (-4, 3)
(ii) What kind of figure is formed by RR' Q'Q?
A rectangle is formed by RR' Q'Q.
(iii) Find the perimeter of the figure P'Q'R'

[Diagram: This diagram shows a coordinate plane with points plotted and a rectangle formed by the original and reflected points]

The figure is a right angled triangle with sides 4 units, 3 units and 5 units.
Here, height = 4 units, base = 3 units and
Hypotenuse = \(\sqrt{4^2 + 3^2}\)
= \(\sqrt{16 + 9}\)
= \(\sqrt{25}\)
= 5 units
Perimeter = height + base + hypotenuse
= 4 + 3 + 5
= 12 units
In simple words: We reflected the triangle and found it forms a right triangle. We calculated all three sides and added them for perimeter.

📝 Teacher's Note: Draw the reflected triangle clearly. Show students how to identify if it's a right triangle by checking if sides follow Pythagorean theorem.

🎯 Exam Tip: Always check if reflected shape is a right triangle. Use Pythagorean theorem to find missing side lengths.

Answer 18.
Answer: A (1,-5), the co-ordinates of A' = (1, 2x1-(-5)) = (1, 7)
B (-5, 1), the co-ordinates of B' = (-5, 2x4-(1)) = (-5, 7)
The distance AB' = \(\sqrt{(-5 - 1)^2 + (7 - (-5))^2}\)
= \(\sqrt{(-6)^2 + 12^2}\)
= \(\sqrt{36 + 144}\)
= \(\sqrt{180}\)
= 13.41 units
In simple words: We reflected both points in the line y = 4, then found distance between the original point A and reflected point B'.

📝 Teacher's Note: Make sure students understand they need distance from A to B', not A to A' or B to B'.

🎯 Exam Tip: Read carefully which distance is asked - between original and reflected points, or between two reflected points.

Answer 19.
Answer: A (4,-1), the co-ordinates of A' = (2x1-4,-1) = (-2,-1)
A' (6,-1), the co-ordinates of B = (6, 2x3-(-1)) = (6, 7)
The distance between A' B' = \(\sqrt{(6 - (-2))^2 + (-1 - (-1))^2}\)
= \(\sqrt{8^2 + 0}\)
= 8 units
Distance till midpoint = 4 units
Co-ordinates of mid-point = (-2+4, -1+4) = (2, 3)
In simple words: We reflected points in different lines, found distance between them, then calculated the midpoint coordinates.

📝 Teacher's Note: Midpoint formula is \(\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)\). Distance to midpoint is half the total distance.

🎯 Exam Tip: Midpoint is exactly halfway between two points. Its distance from each endpoint is half the total distance.