OP Malhotra Class 10 Maths Solutions Chapter 10 Reflection Ex 10(b)

Interactive OP Malhotra Class 10 ICSE Solutions Chapter 10 Reflection Ex 10(b) engage students in active learning and exploration.

S Chand Class 10 ICSE Maths Solutions Chapter 10 Reflection Ex 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 (0 (2, 5), (ii) (1, -4), (iii) (-7, -13), (iv) (-6, 8) under reflection in’ the y-axis.
Solution:
(a) Points (i) (5, 7), (ii) (3, -4), (iii) (-8, 9), (iv) (-1, -2) under reflection in the x-axis will be (i) (5, -7), (ii) (3, 4), (iii) (-8, -9) and (iv) (-1, 2)

(b) Points (i) (2, 5), (ii) (1, -4), (iii) (-7, – 13), (iv) (-6, 8) under reflection in the y-axis will be (i) (-2, 5), (ii) (-1, -4), (iii) (7, -13) and (iv) (6, 8)

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.
Solution:
The vertices of a parallelogram ABCD are A (-3, 2) B (5, 2), C (7, -1), D (-1, -1)
The image of these points in the x-axis will be A’ (-3, -2), B’ (5, -2), C’ (7, 1) and D’ (-1, 1)

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 ?
Solution:
(a) Reflection of the point A (3, 4) in x-axis will be (3, -4) and then in y-axis, will be (-3, -4).

(b) Similarly the refection of B (-3, 2) in x-axis will be (-3, 2) and then in y-axis will be (3, 2).

(c) Reflection of C (5, 0) in x-axis will be (5,0) and then in y-axis will be (-5, 0).

(d) Reflection of D (-3, -3) in x-axis will be (-3, 3) and then in y-axis will be (3, 3)
Now (a) Reflection of A (3, 4) in y-axis will be (-3, 4) and then in x-axis, it will be (-3, -4) x’ which is same as in above (a).
Similarly we see that the reflection of each point first in y-axis and then in x-axis, the- result will same as in above.
So, we get the same points.

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.
Solution:
A point P (a, b) is reflected in y-axis, the image is P’ (- a, b) but the reflection P’ is given (- 3, 5)
∴ Comparing, we get – a = – 3 ⇒ a = 3
and b = 5, then coordinates of P will be (3, 5)
∵ P” is the image of P when reflected in x-aixs the coordinate of P” will be (3, – 5)
Again P”‘ is the image of P when reflected in the line parallel to x-axis such that its equation isy
= – 3
Whose coordinates will be 2b – y where b = – 3 and y = 5
∴ 2b – y = 2 (- 3) – 5 = – 6 – 5 = – 11
and abscissa, x = 3
∴ Coordinates will be (3, -11)

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. (Le. 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).
Solution:
(i) Point P (2, -4) is given.

(ii) The image of P’ when reflected to a line AB which is parallel to x-axis at a distance of 1 unit i.e. y= 1
∴ Co-ordinates of P’ will be (2, 2b – y) or [2, 2 x 1 – (- 4)]
or (2, 2 + 4), or (2, 6)

(iii) The co-ordinates of P”, which is the image of P’ when reflected in the line LM which is parallel to y-axis at a distance of 1 unit on the left side i.e. x = – 1

∴ Co-ordinates of P” will be (2a – x, y) or [2 x (-1) -2, 6], [(-2 -2), 6] or (-4, 6)

Question 6.
A point P is reflected in the jr-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)
Solution:
(i) Let P’ be the image of point P (x, y), whose co-ordinates are (8, -6) i.e. P’ (8, -6) when reflected in x-axis.
Co-ordinates of P will be (8, 6)
Let P” is the image of P when reflected iny-axis the co-ordinates of P” will be (-8, 6)

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.
Solution:
Let P’ be the image of point P (1, 5) when reflected a line LM, is (9, 5)
Here we see the co-ordinates of y are same in P and P’
∴ The line LM is parallel to y-axis (x = 0)
i.e. equation of the line will be = 2a – x
⇒ 9 = 2a – 1 ⇒ 2a = 9 + 1 = 10
⇒ a = 5
Equation of the line parallel to y-axis will be x = a ⇒ x = 5

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.
Solution:
Plot the points A (1, 2), B (3, 4) and C (6, 1) on the graph and join them to form a ∆ABC.
(i) The image of A, B and C under reflection in the origin are A’, B’ and C’ respectively and co-ordinate of A’, B’ and C’ will be A’ (-1, -2), B’ (-3, -4) and C’ (-6, -1) and are plotted on the graph as shown.

(ii) Under reflection of AABC in y-axis, the images of A, B and C will be A” (-1, 2), B” (-3,4) and C”(-6, 1)
And again under reflection it in the x-axis, we get the image of the ∆A”B”C” as ∆A’B’C’ which is the same as in (i).

(iii) So, we can say the results are same as in (i) and in (ii).

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 I, the point invariant under this reflection.
Solution:
Plot the vertices of ∆ABC whose vertices are A (3, 1), B (5, 0) and C (7, 4) .
Draw a line LM parallel to y-axis at a distance of 2 units on positive side which is x = 2 on reflection of AABC in the line x = 2, the image of A, B and C will be A’, B’ and C’ whose co-ordinates will be A’ (1, 1), B’ (-1, 0) and C’ (-3, 4) respectively.
Point is marked on the graph as the point in variant under this reflection whose co-ordinate are (2, 0).

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’.
Solution:
Plot the points B (3, 2) and C (0, 3) on the graph.
(i) The image B’ of B under reflection in the x-axis will be B’ (3, -2).
(ii) Join BB’.
(iii) The image C’ (0, 3) of C under reflection in the line BB’ is C’ (6, 3).

(iv) Join B’C’. Draw perpendicular from C’ and B’ meeting each other at D.
Now C’D = 5 units and B’D = 3 units
∴ In right angled AC’B’D,
(B’C’)² = (B’D)² + (C’D)² (Pythagoras Theorem)
= (5)² + (3)² = 25 + 9 = 34
B’C’ = \(\sqrt{34}\) units

Question 11.
The image of a point P under reflection in the x-axis is (5, -2). Write down the co-ordinates of P.
Solution:
Let co-ordinates of P be (x, y), then
the image of P under reflection in the x-axis will be P’ (x, -y)
But P’ is given (5, -2)
Comparing, we get
x = 5, – y = – 2 ⇒ y = 2
∴Co-ordinates of P are (5, 2)

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 ?
Solution:
Plot the points A (3, 3), B (5, 3), C (5, 5), D (3, 5) and join AB, BC, CD and DA forming a square. Draw the image of A, B, C and D under reflection in, y-axis. The co-ordinates of images are A’ (-3, 3), B’ (-5, 3), C (-5, 5) and D’ (-3, 5)
Again reflect the points A’, B’, C’ and D’ in origin
The images are A” (3, -3), B” (5, -3), C” (5, -5) and D” (3, -5) respectively.
We see that reflection A”B”C”D” is the image of ABCD under reflection in x-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”.
Solution:
The vertices of a AABC are A (1,2), B (4, 8) and C (6, 8) and are reflected in the x-axis to AA’B’C’.
Then the co-ordinates of A’, B’ and C’ will be (1, -2), (4, -8) and (6, -8).
The ∆A’B’C’ is again reflected in the origin to AA”B”C”. Then the co-ordinates of A”, B” and C” will be (-1, 2), (-4, 8) and (-6, 8) We see that the single transformation of A”, B” and C” are the images of ∆ABC when reflected in 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.
Solution:
Point P (a, b) is reflected under reflection in the origin
∴ Co-ordinates of the image of P will be P, (-a, -b)
Again P, (-a, -b) is reflected under reflection in the j-axis is to P’ whose co-ordinates will be (a, -b)
But co-ordinates of P’ are (3, – 4)
Then comparing, we get
a = 3, – b = – 4 ⇒ b = 4
∴ a = 3, b = 4

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”.
Solution:
P’ is the image of point P (-3, -2) in the x-axis
∴ Co-ordinates of P’ will be (-3, 2)
P’ is again reflected under reflection in the origin to P”
Then the co-ordinates of P” will be (3, -2)

Question 16.
Point A (5, -1) on reflection in the x-axis is mapped as A’. Also A on reflection in the 7-axis is mapped as A”. Write the co-ordinates of A’ and A”. Also calculate the distance AA”.
Solution:
Point A (5, -1) is reflected in the x-axis to A’
Then co-ordinates of A’ will be (5, 1)
Point A (5, -1) is reflected in the 7-axis to A”
The co-ordinates of A” will be (- 5, – 1)
Length of AA” = \(\sqrt{\left(x_1-x_2\right)^2+\left(y_1-y_2\right)^2}\)
= \(\sqrt{(5-(-5))^2+(-1-(-1))^2}\)
= \(\sqrt{(10)^2+(0)^2}=\sqrt{100+0}\)
= \(\sqrt{100}\)
= 10 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)
Solution:
Point A (2, – 4) is reflected in the origin to A’
Then co-ordinates of A’ will be (-2, 4)
Point B (-3, 2) is reflected in x-axis to B’
Then co-ordinates of B’ will be (-3, -2)
Now length of A’B’
= \(\sqrt{\left(x_1-x_2\right)^2+\left(y_1-y_2\right)^2}\)
= \(\sqrt{[-2-(-3)]^2+[4-(-2)]^2}\)
= \(\sqrt{(-2+3)^2+(4+2)^2}\)
= \(\sqrt{(1)^2+(6)^2}=\sqrt{1+36}\)
= \(\sqrt{37}\) = 6.1 units

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.
Solution:
Point A (4, -1) is reflected in y-axis as A’
∴ Co-ordinates of A’ will be (-4, -1)
Point B is reflected in x-axis as B’ (-2, 5)
Let co-ordinates of B’ be (x, y) then B’ will be (x, -y)
Comparing, we get
x = – 2, – y = 5 or x = – 2, y = – 5
∴ Co-ordinates of B are (- 2, – 5)
Let P (x₁, y₂) be the middle point of line segment A’B
∴ Co-ordinates of P will be
\(\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)\)
or \(\left(\frac{-4-2}{2}, \frac{-1-5}{2}\right)\)
or \(\left(\frac{-6}{2}, \frac{-6}{2}\right)\) or (- 3, – 3)

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’.
Solution:
Point A(1, – 5) is mapped as A’ on reflection in the x-axis
∴ Co-ordinates of A’ will be (1, 5)
Point B (3, 2) is mapped as B’ on reflection in origin
∴ Co-ordinates of B’ will be (-3, -2)
Length of AB’ = \(\sqrt{[1-(-3)]^2+[-5-(-2)]^2}\)
= \(\sqrt{(1+3)^2+(-5+2)^2}\)
= \(\sqrt{16+9}\)
= \(\sqrt{25}\) = 5

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 image of A” of A, when A is reflection in the origin,
(e) the single transformation that maps A’ onto A”.
Solution:
(i) Plot the point A (3, 2) and B (5, 4) on the graph.

(ii) A’ and B’ are the reflection under reflection in x-axis
∴ Co-ordinates of A’ (3, -2) and of B (5, -4)

(iii) (a) Joining AB, BA’, A’B’ and B’A, the figure ABB’A’ is formed.
The figure so formed is of an isosceles trapezium.
(b) The axis of symmetry of the figure trapezium is the x-axis.
(c) The measure of ∠ABB’ is 45°.
(d) The A” is the image of A when reflected in origin.
The co-ordinates of A” will be (-3, -2).
(e) The single transformation that maps A’ onto A” is the reflection of A’ in y-axis.

Self Evaluation And Revision
(LATEST ICSE QUESTIONS)

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.
(i) Name or write equations for the lines L₁ and L₂.
(ii) Write down the images of P (3, 4) and Q (-5, -2) on reflection in 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”.
Solution:
(i) ∵ Points (3,0) and (-1,0) are invariant points under reflection L₁
Then the equation of L₁ will be x-axis is or y = 0
∵ Points (0, -3) and (0, 1) are invariant points under reflection L₂
Then the equaiton ofL, will bey-axis or x = 0

(ii) The images of point P (3, 4) and Q (-5, -2) on reflection is L, or in x-axis is will be P’ (3, -4) and Q’ (-5, 2)

(iii) The image of P and Q on reflection is L₂ i.e. y-axis will be P” (-3, 4) and Q” (5, -2)

(iv) Single transformation of P and Q that maps P’ onto P” is reflection in origin.

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”.
Solution:
(i) The reflection of point P (a, b) in x-axis will be P’ (a, -b)
But P’ is (5, -2)
∴ Comparing, we get a = 5, – b = – 2 or b = 2
∴ Co-ordinates of P are (5, 2)

(ii) P” is the image of P when reflected in y-axis
∴ Co-ordinates of P” will be (-5, 2)

(iii) Single transformation that maps P’ to P” is in origin

Question 3.
A point P (a, b) is reflected in the x-axis to P’ (2, -3). Write down the value of a and b. P” is the image of P, when reflected in the y-axis. Write down the coordinates of P”. Find the coordinates of P'” when P is reflected in the line parallel to the y- axis, such that x = 4.
Solution:
(i) A point P (a, b) is reflected in x-axis to P’ (a, -b) but P’ is given (2, -3)
∴ Comparing, we get
a = 2, – b = – 3 or b = 3
∴ a = 2, b = 3
∴ Co-ordinates of P will be (2, 3)

(ii) P” is the image of P when reflected in y-axis
∴ Co-ordinates of P” will be (-2, 3)

(iii) When P is reflected in a line parallel to y-axis
such that x = 4, to P”
∴ Co-ordinates of P” will be (2a – x, y)
or (2 x 4 – 2, 3) or (8 – 2, 3) or (6, 3)

Question 4.
Use graph paper for this question.
(i) Plot the points A (3, 5) and B (-2, -4). Use 1 cm – 1 unit on both axes.
(ii) A’ is the image of A when reflected in the x-axis. Write down the coordinates 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 coordinates of B’ and plot it on the graph paper.
(iv) Write down the geometrical name of AA’ BB’.
(v) Name two invariant points under reflection in the x-axis.
Solution:
(i) Plot the points A (3, 5) and B (-2, -4) on the graph.
(ii) ∵ A’ is the image of A (3, 5) when reflected in x-axis
∴ Co-ordinates of A’ will be (3, -5) which has been plotted.

(iii) ∵ B’ is the image of B (-2, -4) in the y-axis
∴ Co-ordinates of B’ will be (2, -4) which has been plotted.

(iv) By joining AA’ BB’, the figure so formed is of an isosceles trapezium.

(v) Points which are invariant under reflection in x-axis are C (3, 0) and D (-2, 0).

Question 5.
Write down the coordinates of the image of the point (3, -2) when :
(i) reflected in the x-axis;
(ii) reflected in the y-axis;
(iii) reflected in the x-axis followed by the reflection in the y-axis;
(iv) reflected in the origin;
(v) reflected in the line y = 5.
Solution:
Let point A (3, -2) be given.
(i) The image of A (3, -2) when reflected in x- axis will be A’ and its co-ordinates will be (3, 2).

(ii) The image of A (3, -2) when reflected in y- axis will be A” and its co-ordinates will be (-3, -2).

(iii) The image of A (3, -2) when reflected in x- axis followed by the reflection in y-axis be A'” whose co-ordinates will be (-3, 2).

(iv) The image of point A (3, -2) when reflected in origin will be (-3, 2).

(v) The image of point A (3, -2) when reflected in a line y = 5 The co-ordinates will be (x, 2y-y)
or [3, 2 x 5 – (-2)] or [3, (10+ 2)] or (3, 12).

Question 6.
Use graph paper for this question.
The points P (5, 3) was reflected in the origin to get the image P’.
(a) Write down the coordinates of P’.
(b) If M is the foot of the perpendicular from P to the x-axis, find the coordinates of M.
(c) If N is the foot of the perpendicular from P’ to the x-axis. Find the coordinates of N.
(d) Name the figure PMP’N.
(e) Find the area of the figure PMP’N.
Solution:
(a) The point P (5,3) has been plotted on the graph.
P’ is the image of P (5, 3) under reflected in origin
∴ Co-ordinates of P’ will be (-5, -3)

(b) PM is perpendicular on x-axis
Co-ordinates of M are (5, 0)

(c) P’N is perpendicular on y-axis
∴ Co-ordinates of N will be (-5, 0) x1

(d) By joining PM, MP’, P’N and NP, the figure PMP’N so found is a parallelogram whose opposite sides of are parallel and equal.

(e) Area of parallelogram PMP’N’ = 2 x area of ∆PMN
= 2 x \(\frac { 1 }{ 2 }\) x MN x PM = MN x PM
= 10 x 3 = 30 sq. units

Question 7.
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’. Using graph paper, give
(i) The coordinates of P’ and O’.
(ii) The lengths of the segments PP’ and OO’.
(iii) The perimeter of the quadrilateral POP’O’.
(iv) The geometrical name of the figure POP’O’.
Solution:
(i) P’ is the image of point P (3, 4) under reflection in x-axis, then the co-ordinates of P’ will be (3, – 4)
O’ is the image of origin O (0, 0) when reflected in the line PP’.
Then the co-ordinates of O’ will be (6, 0)

(ii) Length of PP = 4 + 4 = 8
and length of OO’ = 3 + 3 = 6

(iii) Join OP, PO’, O’P’ and P’O forming a rhombus
Area of figure OPO’P’ = \(\frac { 1 }{ 2 }\) PP’ x OO’
= \(\frac { 1 }{ 2 }\) x 8 x 6 = 24 sq. units.

Question 8.
Use a graph paper for this question. (Take 10 small division = 1 unit on both axes.) Plot the points P (3, 2) and Q (-3, -2). From P and Q draw perpendiculars PM and QN on the x-axis.
(a) Name the image of P on reflection in the origin.
(b) Assign the special name to the geometrical figure PMQN and find its area.
(c) Write the coordinates of the point to which M is mapped on reflection in (i) x- axis; (ii) y-axis, (iii) origin.
Solution:
Plot points P (3, 2) and Q (-3, -2) on the graph.
Draw PM and QN perpendicular on x-axis Co-ordinates of M are (3, 0) and of N are (-3, 0)
(a) The image of P on reflection in origin is Q (-3, -2)

(b) The figure so formed by joining PN and QM is a parallelogram
Area = 2 x area of ∆PMN = 2 x \(\frac { 1 }{ 2 }\) x MN x PM
= MN x PM = 6 x 2 = 12 sq. units

(c) (i) The image of M when reflected in x-axis is M itself.
(ii) The image of M when reflected in y-axis is N(-3, 0).
(iii) The image of M when reflected in origin is N (-3, 0).

Question 9.
Use a graph paper for this question.
A (1,1), B (5,1), C (4, 2) and D (2, 2) are the vertices of a quadrilateral.
Name the quadrilateral ABCD. A, B, C and D are reflected in the origin onto 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?
Solution:
Plot points A (1, 1), B, (5, 1) C (4, 2) and D (2, 2) on the graph and join then to form a quadrilateral ABCD, which is an isosceles trapezium.
The images of A, B, C, D are A’, B’, C’, and D’ when reflected in origin.
∴ The co-ordinates of A’ are (-1, -1), of B’ are (-5, -1), of C are (-4, -2) and of D’ are (-2, -2). The figure A’B ’C’D’ is the image of the figure ABCD in origin.
We see that D, A, A’ and D’ are collinear when joined.

Question 10.
Use a graph paper for this question (take 10 small division = 1 unit on both axes). P and Q have co-ordinates (0, 5) and (-2, 4).
(i) P is invariant when reflected in an axis. Name the axis.
(ii) Find the image of Q on reflection in the axis found in (i).
(iii) (0, k) on reflection in the origin is invariant. Write the value of k.
(iv) Write the co-ordinates of the image of Q obtained by reflecting it in the origin followed by reflection in the x-axis.
Solution:

(i) Plot points P (0, 5) is invariant on y-axis and Q (-2, 4) on the graph.

(ii) Q’ is the image of Q on reflection in the line y-axis (x = 0)
The co-ordinates of Q’ are (2, 4)

(iii) ∵ (0, k) is invariant in the origin (0, 0)
∴ k = 0

(iv) The image of Q on reflection in the origin followed by the reflection in x-axis is Q”
The co-ordinates of Q” will be (2, 4)

Question 11.
Use graph paper for this question.
The points A (2, 3), B (4, 5) and C (7, 2) are the vertices of AABC.
(i) Write down the co-ordinates of A’, B’, C’, if ∆A’B’C’ is the image of ∆ABC, when reflected in the origin.
(ii) Write down the co-ordinates of A”, B”, C” if ∆A”B”C” is the image of ∆A’B’C’, when reflected in the x-axis.
(iii) Mention the special name of the quadrilateral BC C”B” and find its area.
Solution:
Plot the points A (2, 3), B (4, 5) and C (7, 2) and join them to form the AABC
(i) A’, B’, C’ are the images of A, B, C respectively
Therefore ∆A’B’C’ is the image of ∆ABC when reflected in origin
∴ Co-ordinates of A’ are are (-2, -3), of B’ are (-4, -5) and of C’ are (-7, -2).

(ii) A”, B” and C” are the images of A, B, C under reflection in x-axis
∴ Co-ordinates of A” (2, -3), of B” (4, -5) and C” (7, -2)
Thus ∆A”B”C” is the image of AABC reflected in x-axis.

(iii) By joining C, C” and B, B”, the figure so formed is an isosceles trapezium BCC”B”.

Question 12.
Use a graph paper for this question.
(i) The point P (2, -4) is reflected about the line x = 0 to get the image Q. Find the co-ordinates of Q.
(ii) 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 the figure PQR.
Solution:

(i) Plot point P (2, – 4) on the graph,
Q is the image of P under reflection iny-axis i.e. x = 0
∴ Co-ordinates of Q will be (-2, -4)

(ii) Point Q (-2, -4) is reflected about the line = 0, i.e. x-axis we get the image R of Q, whose co-ordinates are (-2, 4).

(iii) Join PQ, QR and RP, the figure so formed PQR is a right angled triangle.

(iv) Area of ∆PQR = \(\frac { 1 }{ 2 }\) x base x height
= \(\frac { 1 }{ 2 }\) PQ x QR
= \(\frac { 1 }{ 2 }\) x 4 x 8 = 16 sq. units

Question 13.
Use graph paper to answer this question:
(i) Plot the points A (4, 6) and B (1, 2).
(ii) A’ is the image of A when reflected in ac-axis.
(iii) B’ is the image of B when B is reflected in the line AA’.
(iv) Give the geometrical name for the figure AB A’B’.
Solution:
(i), (ii), (iii)
(iv) Figure ABA’B’ is a kite.

Question 14.
Use a graph paper for this question. Y’
A (0, 3), B (3, -2) and O (0, 0) are the vertices of a ∆ABO.
(i) Plot the triangle on the graph paper taking 2 cm = 1 unit on both the axes.
(ii) Plot D the reflection of B in the y-axis, and write its coordinates.
(iii) Give the geometrical name of the figure ABOD.
(iv) Write the equation of the line of symmetry of the figure ABOD.
Solution:
(i)

(ii) Coordinates of D are (-3, -2)
(iii) Arrow head.
(iv) Equation of line of symmetry of figure ∆BOD is x = 0 i.e., y-axis.

Question 15.
Use a graph paper to answer the following questions. (Take 1 cm = 1 unit on both axes)
(i) Plot A (4, 4), B (4, -6) and C (8, 0) the vertices of a triangle AABC.
(ii) Reflect ABC on the y-axis and name it as A’B’C’.
(iii) Write the coordinates of the images A’, B’ and C’.
(iv) Give a geometrical name for the figure AA’C’B’BC.
(v) Identify the line of symmetry of AA’C’ B’BC.
Solution:

(iii) Coordinates of A’ (-4, 4), B’ (-4, -6) and C’ (-8, 0)
(iv) Hexagon
(v) YY’ is line of symmetry

Question 16.
Using graph paper and taking 1 cm = 1 unit along both xr-axis and y-axis :
(i) Plot the points A (-4, 4) and B (2, 2)
(ii) Reflect A and B in the origin to get the images A’ and B’ respectively.
(iii) Write down the co-ordinates of A’ and B’.
(iv) Give the geometrical name for the figure ABA’B’.
(v) Draw and name its lines of symmetry.
Solution:
(i) The points are plotted in the graph.

(ii) The points A and B have been reflected in the origin to get the image A’ and B’ respectively.

(iii) Coordinates of A’ and B’ are (4, -4) and (-2, -2) respectively.

(iv) In the figure, AB = BA’ = A’B’ = B’A. i.e., ABA’B’ is a rhombus.

(v) As we know a rhombus has two lines of symmetry namely its two diagonals. So, these lines of symmetry have been drawn.

Question 17.
Using a graph paper, plot the points A (6, 4) and B (0, 4).
(i) Reflect A and B in the origin to get the images A’ and B’.
(ii) Write the co-ordinates of A’ and B’.
(iii) State the geometrical name for the figure ABA’B’.
(iv) Find its perimeter.
Solution:

Question 18.
Use a graph paper to answer the following questions. (Take 2 cm = 1 unit on both axis)
(i) Plot the points A (-4, 2) and B (4, 2)
(ii) A’ is the image of A when reflected in the y-axis. Plot it on the graph paper and write the coordinates of A’.
(iii) B’ is the image of B when reflected in the line AA’. Write the coordinates of B’.
(iv) Write the geometric name of the figure ABA’B’.
(v) Name a line of symmetry of the figure formed.
Solution:

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 E’.
(iii) Write the coordinates of B’, C’, D’ and E’.
(iv) Name the figure formed by BCDEE’D’C’B’.
(v) Name a line of symmetry for the figure formed.
Solution:
(i)

(ii) Reflection of points on the y-axis will result in the change of the x-coordinate.
(iii) Points will be B’ (-2, 5), C’ (-5, 2), D’ (-5, -2), E’ (-2, -5).
(iv) The figure BCDEE’D’C’B’ is a octagon.
(v) The lines of symmetry are x-axis, y-axis, Y = X and Y = -X, total 8.

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.
Solution:
(i)

(ii) The figure OABCB’A’ is an irregular hexagon.
(iii) They-axis is the line of symmetry of figure OABCB’A’.