# ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 10

### ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 10 Reflection MCQS

Q1. The reflection of the point P ( – 2, 3) in the x-axis is
(a) (2, 3)
(b) (2, – 3)
(c) ( – 2, – 3)
(d) ( – 2, 3)
Solution:
Reflection of the point P (-2, 3) in x-axis is (-2, -3) (c)

Q2. The reflection of the point P ( – 2, 3) in the y- axis is
(a) (2, 3)
(b) (2, – 3)
(c) ( – 2, – 3)
(d) (0, 3)
Solution:
The reflection of the point P (-2, 3) under reflection in y-axis (2, 3) (a)

Q3. If the image of the point P under reflection in the x-axis is ( – 3, 2), then the coordinates of the point P are
(a) (3, 2)
(b) ( – 3, – 2)
(c) (3, – 2)
(d) ( – 3, 0)
Solution:
The image of the point P under reflection in the x-axis is (-3, 2),
then the co-ordinates of the point P will be (-3, -2) (b)

Q4. The reflection of the point P (1, – 2) in the line y = – 1 is
(a) ( – 3, – 2)
(b) (1, – 4)
(c) (1 , 4)
(d) (1, 0)
Solution:
The reflection of the point P (1, -2) in the line y = -1 is (1, 0) (d)

Q5. The reflection of the point A (4, -1) in the line x = 2 is
(a) (0, – 1)
(b) (8, – 1)
(c) (0, 1)
(d) none of these
Solution:
The reflection of A (4, -1) in the line x = 2 will be A’ (0, -1) (a)

Q6. The reflection of the point ( – 3, 0) in the origin is the point
(a) (0, – 3)
(b) (0, 3)
(c) (3, 0)
(d) none of these
Solution:
Reflection of the point (-3, 0) in origin will be (3, 0) (c)

Q7. Which of the following points is invariant with respect to the line y = – 2 ?
(a) (3, 2)
(b) (3, – 2)
(c) (2, 3)
(d) ( – 2, 3)
Solution:
The variant points are (3, -2) (b)

Q8. Find the co-ordinates of the images of the following points under reflection in the x- axis:
(i) (2, -5)
(ii) −32,−12
(iii) ( – 7, 0)
Solution:
Co-ordinates of the images of the points
under reflection in the x-axis will be
(i) Image of (2, -5) will be (2, 5)
(ii) Image of −32,−12 will be −32,12
(iii) Image of ( -7, 0) will be (-7, 0)

Q9. Find the co-ordinates of the images of the following points under reflection in the y-axis:
(i) (2, – 5)
(ii) −32,12
(iii) (0, – 7)
Solution:
Co-ordinates of the image of the points under reflection in the y-axis
(i) Image of (2, -5) will be ( -2, -5)
(ii) Image of −32,12 will be 32,12
(iii) Image of (0, -7) will be (0, -7)

Q10. Find the co-ordinates of the images of the following points under reflection in the origin:
(i) (2, – 5)
(ii) −32,−12
(iii) (0, 0)
Solution:
Co-ordinates of the image of the points under reflection in the y-axis
(i) Image of (2, -5) will be (-2, 5)
(ii) Image of −32,−12 will be 32,12
(iii) Image of (0, 0) will be (0, 0)

Q11. The image of a point P under reflection in the x-axis is (5, – 2). Write down the co-ordinates of P.
Solution:
As the image of a point (5, -2) under x – axis is P
∴ Co-ordinates of P will be (5, 2)

Q12. 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.
Solution:
The co-ordinates of image of P which is reflected in x-axis are (8, – 6), then
(0 Co-ordinates of P will be (8, 6)
(ii) Co-ordinates of image of P under reflection in the y-axis will be ( – 8, 6)

Q13. A point P is reflected in the origin. Co-ordinates of its image are (2, – 5). Find
(i) the co-ordinates of P.
(ii) the co-ordinates of the image of P in the x-axis.
Solution:
The co-ordinates of image of a point P which is reflected in origin are (2, – 5), then
(i) Co-ordinates of P will be ( – 2, 5)
(ii) Co-ordinates of the image of P in the x- axis will be ( – 2, – 5)

Q14. (i) The point P (2, 3) is reflected in the line x = 4 to the point P’. Find the co-ordinates of the point P’.
(ii) Find the image of the point P (1, – 2) in the line x = – 1.
Solution:
(i) (a) Draw axis XOX’ and YOY’ and take 1 cm = 1 unit
(b) Plot point P (2, 3) on it.
(c) Draw a line x = 4 which is parallel to y-axis.
(d) From P, draw a perpendicular on x = 4, which intersects x = 4 at Q.
(e) Produce PQ to P’, such that QP’ = QP.
∴ P’ is the reflection of P in the line x = 4
Co-ordinates of P’ are (6, 3) (ii) (a) Draw axis XOX’ and YOY’ and take 1 cm = 1 unit.
(b) Plot the point P (1, -2) on it.
(c) Draw a line x = -1 which is parallel toy-axis.
(d) From P, draw a perpendicular on the line x = -1, which meets it at Q.
(e) Produce PQ to P’ such that PQ = QP’
P’ is the image or reflection of P in the line x = -1
Co-ordinates of P’ are (-3, -2)
Co-ordinates of P’ are (-3, -2) Q15. (i) The point P (2, 4) on reflection in the line y = 1 is mapped onto P’ Find the co-ordinates of P’.
(ii) Find the image of the point P ( – 3, – 5) in the line y = – 2.
Solution:
(i) (a) Draw axis XOX’ and YOY’ and take 1 cm = 1 unit.
(b) Plot point P (2, 4) on it.
(c) Draw a line y = 1, which is parallel to x-axis.
(d) From P, draw a perpendicular on y = 1 meeting it at Q.
(e) Produce PQ to P’ such that QP’ = PQ.
P’ is the reflection of P whose co-ordinates are (2, -2) (ii) (a) Draw axis XOX’ and YOY’ and take 1 cm = 1 unit.
(b) Plot point P (-3, -5) on it.
(c) Draw a line y = -2 which is parallel to the x-axis.
(d) From P, draw a perpendicular on y = -2 which meets it at Q.
(e) Produce PQ to P’ such that QP’ = PQ.
Then P’ is the image of P, whose co-ordinates are (-3, 1). Q16. The point P ( – 4, – 5) on reflection in y-axis is mapped on P’. The point P’ on reflection in the origin is mapped on P”. Find the co-ordinates of P’ and P”. Write down a single transformation that maps P onto P”.
Solution:
P’ is the image of point P (-4, -5) in y-axis
∴Co-ordinates of P’ will be (4, -5)
Again P” is the image of P’ under reflection in origin will be (-4, 5).
The single transformation that maps P onto P” is the x-axis

Q17. Write down the co-ordinates 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 reflection in the y-axis
(iv) reflected in the origin. (2000)
Solution:
Co-ordinates of the given points are (3, -2).
(i) Co-ordinates of the image reflected in x- axis will be (3, 2)
(ii) Co-ordinates of the image reflected in y- axis will be (-3, -2)
(iii) Co-ordinates of the point reflected in x- axis followed by reflection in the y-axis will be (-3, 2)
(iv) Co-ordinates of the point reflected in the origin will be (-3, 2)

Q18. Find the co-ordinates of the image of (3, 1) under reflection in x-axis followed by a reflection in the line x = 1.
Solution:
(i) Draw axis XOX’ and YOY’ taking 1 cm = 1 unit.
(ii) Plot a point P (3, 1).
(iii) Draw a line x = 1, which is parallel to y-axis.
(iv) From P, draw a perpendicular on x-axis meeting it at Q.
(v) Produce PQ to P’ such that QP’ = PQ, then
P’ is the image of P is x-axis. Then co-ordinates of P’ will be (3, -1)
(vi) From P’, draw a perpendicular on x = 1 meeting it at R.
(vii) Produce P’R to P” such that RP” = P’R
∴P” is the image of P’ in the line x = 1
Co-ordinates of P” are (-1, -1) Q19. If P’ ( – 4, – 3) is the image of a point P under reflection in the origin, find
(i) the co-ordinates of P.
(ii) the co-ordinates of the image of P under reflection in the line y = – 2.
Solution:
(i) Reflection of P is P’ (-4, -3) in the origin
∴ Co-ordinates of P will be (4, 3)
Draw a line y = -2, which is parallel to x-axis
(ii) From P, draw a perpendicular on y = -2 meetings it at Q
Produce PQ to P” such that QP” = PQ
∴P” will the image of P in the line y = -2
∴Co-ordinates of P” will be (4, -7) Q20. A Point P (a, b) is reflected in the x-axis to P’ (2, – 3), write down the values of a and b. P” is the image of P, when reflected in the y-axis. Write down the co-ordinates of P”. Find the co-ordinates of P”’, when P is reflected in the line parallel to y-axis such that x = 4. (1998)
Solution:
P’ (2, -3) is the reflection of P (a, b) in the x-axis
∴Co-ordinates of P’ will be P’ (a, – b) but P’ is (2, -3)
Comparing a = 2, b = 3 ∴Co-ordinates of P will be (2, 3)
P” is the image of P when reflected in y-axis
∴Co-ordinate of P” will be ( – 2, 3)
Draw a line x = 4, which is parallel to y-axis
and P’” is the image of P when it is reflected in the line x = 4,
then P’” is its reflection Co-ordinates of P”‘ will be (6, 3).

Q21. (i) Point P (a, b) is reflected in the x-axis to P’ (5, – 2). Write down the values of a and b.
(ii) P” is the image of P when reflected in the y-axis. Write down the co-ordinates of P”.
(iii) Name a single transformation that maps P’ to P”. (1997)
Solution:
(i) Image of P (a, b) reflected in the x-axis to P’ (5, -2)
∴ a = 5 and b = 2
(ii) P” is the image of P when reflected in the y-axis
∴ its co-ordinates will be (-5, -2).
(iii) The single transformation that maps P’ to P” is the origin.

Q22. Points A and B have co-ordinates (2, 5) and (0, 3). Find
(i) the image A’ of A under reflection in the x-axis.
(ii) the image B’ of B under reflection in the line AA’.
Solution:
Co-ordinates of A are (2, 5) and of B are (0, 3) (i) Co-ordinates of A’, the image of A reflected in the x-axis will be (2, -5)
(ii) Co-ordinates of B’, the image of B under reflection in the line AA’ will be (4, 3).

Q23. Plot the points A (2, – 3), B ( – 1, 2) and C (0, – 2) on the graph paper. Draw the triangle formed by reflecting these points in the x-axis. Are the two triangles congruent? Solution:
The points A (2, -3), B (-1, 2) and C(0, -2) has been plotted on the graph paper as shown and are joined to form a triangle ABC. The co-ordinates of the images of A, B and C reflected in x-axis will be A’ (2, 3), B’ (-1, -2), C’ (0, 2) respectively and are joined to from another ∆ A’B’C’
Yes, these two triangles are congruent.

Q23. The points (6, 2), (3, – 1) and ( – 2, 4) are the vertices of a right angled triangle. Check whether it remains a right angled triangle after reflection in the y-axis.
Solution:
Let A (6, 2), B (3, -1) and C (-2, 4) be the points of a right-angled triangle
then the co-ordinates of the images of A, B, C reflected in y-axis be
A’ (-6, 2), B’ (-3, -1) and C’ (2, 4). By joining these points, we find that ∆A’B’C’ is also a right angled triangle.

Q24. The triangle ABC where A (1, 2), B (4, 8), C (6, 8) is reflected in the x-axis to triangle A’ B’ C’. The triangle A’ B’ C’ is then reflected in.the origin to triangle A”B”C” Write down the co-ordinates of A”, B”, C”. Write down a single transformation that maps ABC onto A” B” C”.
Solution:
The co-ordinates of ∆ ABC are A (1, 2) B (4, 8), C (6, 8)
which are reflected in x- axis as A’, B’ and C’.
∴ The co-ordinates of A’ (1, -2), B (4, -8) and C (6, -8).
A’, B’ and C’ are again reflected in origins to form an ∆A”B”C”.
∴ The co-ordinates of A” will be (-1, 2), B” (-4, 8) and C” (-6, 8)
The single transformation that maps ABC onto A” B” C” is y-axis.

Q25. The image of a point P on reflection in a line l is point P’. Describe the location of the line l.
Solution:
The line will be the right bisector of the line segment joining P and P’.

Q26. Given two points P and Q, and that (1) the image of P on reflection in y-axis is the point Q and (2) the mid point of PQ is invariant on reflection in x-axis. Locate
(i) the x-axis
(ii) the y-axis and
(iii) the origin.
Solution:
Q is the image of P on reflection in y-axis
and mid point of PQ is invariant on reflection in x-axis (i) x-axis will be the line joining the points P and Q.
(ii) The line perpendicular bisector of line segment PQ is the y-axis.
(iii) The origin will be the mid point of line segment PQ.

Q27. The point ( – 3, 0) on reflection in a line is mapped as (3, 0) and the point (2, – 3) on reflection in the same line is mapped as ( – 2, – 3).
(i) Name the mirror line.
(ii) Write the co-ordinates of the image of ( – 3, – 4) in the mirror line.
Solution:
The point (-3,0) is the image of point (3, 0)
and point (2, -3) is image of point (-2, -3) reflected on the same line.
(i) It is clear that the mirror line will be y-axis.
(ii) The co-ordinates of the image of the point (-3, -4)
reflected in the same line i.e. y-axis will be (3, -4).

Q28. A ( – 2, 4) and B ( – 4, 2) are reflected in the y-axis. If A’ and B’ are images of A and B respectively, find (i) the co-ordinates of A’ and B’.
(ii) Assign special name to quad. AA’B’B.
(iii) State whether AB’ = BA’.
Solution:
A (-2, 4) and B (-4, 2) are reflected in y- axis as A’ and B’. (i) Co-ordinates of A’ are (2, 4) and of B are (4, 2).
(ii) The quadrilateral AA’ B’ B is an isosceles trapezium.
(iii) yes, AB’ = BA’

Q29. Use 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 figure PQR. (2007)
Solution:
(i) Since the point Q is the reflection of the point P (2, -4) in the line x = 0,
the co-ordinates of Q are (2, 4).
(ii) Since R is the reflection of Q (2, 4) about the line y = 0,
the co-ordinates of R are ( – 2, 4).
(iii) Figure PQR is the right angled triangle PQR. (iv) Area of ∆ PQR = 12 x QR x PQ
= 12 x 4 x 8
= 16 sq. units.

Q30. Use graph paper for this question. The point P (5, 3) was reflected in the origin to get the image P’.
(i) Write down the co-ordinates of P’.
(ii) If M is the foot of perpendicular from P to the x-axis, find the co-ordinates of M.
(iii) If N is the foot of the perpendicular from P’ to the x-axis, find the co-ordinates of N.
(iv) Name the figure PMP’N.
(v) Find the area of the figure PMP’N. (2001)
Solution:
P’ is the image of point P (5, 3) reflected in the origin. (i) Co-ordinates of P’ will be (-5, -3).
(ii) M is the foot of the perpendicular from P to the x-axis.
Co-ordinates of M will be (5, 0)
(iii) N is the foot of the perpendicular from P’ to x-axis.
Co-ordinates of N will be (-5, 0).
(iv) By joining the points, the figure PMP’N is a parallelogram.
(v) Area of the parallelogram = 2 x area of ∆ MPN
= 2 x 12 x MN x PM = MN x PM
= 10 x 3 = 30 sq. units. Ans.

Q31. 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:
(i) A (6, 4), B (0, 4) Perimeter = Sum of all sides = 6 + 10 + 6 + 10 = 32 units

Q32. Use graph paper to answer this question
(i) Plot the points A (4, 6) and B (1, 2).
(ii) If A’ is the image of A when reflected in x-axis, write the co-ordinates of A’.
(iii) If B’ is the image of B when B is reflected in the line AA’, write the co-ordinates of B’.
(iv) Give the geometrical name for the figure ABA’B’. (2009)
Solution:
(i) Plotting the points A (4, 6) and B (1, 2) on the given graph.
(ii) A’ = (4, -6)
(iii) B’ = (7, 2)
(iv) In the quadrilateral ABA’B’, we have AB = AB’ and A’B = A’B’
Hence, ABA’B’ is a kite.

Q33. The points A (2, 3), B (4, 5) and C (7, 2) are the vertices’s of ∆ABC. (2006)
(i) Write down the co-ordinates of A1, B1, C1 if ∆ A1B1C1 is the image of ∆ ABC when reflected in the origin.
(ii) Write down the co-ordinates of A2, B2, C2 if ∆ A2B2C2 is the image of ∆ ABC when reflected in the x-axis.
(iii) Assign the special name to the quadrilateral BCC2B2 and find its area.  Solution:
Points A (2, 3), B (4, 5) and C (7, 2) are the vertices’s of ∆ ABC.
A1, B1 and C1 are the images of A, B and C reflected in the origin.
(i) Co-ordinates of A1 = (-2, -3) of B1 (-4, -5) and of C1 (-7, -2).
(ii) Co-ordinates of A2, B2 and C2 the images of A, B and C
when reflected in x-axis are A2 (2, – 3), B2 (4, – 5), C2 (7, – 2)
(iii) The quadrilateral formed by joining the points,
BCC2B2 is an isosceles trapezium and its area  Q34. The point P (3, 4) is reflected to P’ in the x-axis and O’ is the image of O (origin) in the line PP’. Find :
(i) the co-ordinates of P’ and O’,
(ii) the length of segments PP’ and OO’.
(iii) the perimeter of the quadrilateral POP’O’.
Solution:
P’ is the image of P (3, 4) reflected in x- axis
and O’ is the image of O the origin in the line P’P.
(i) Co-ordinates of P’ are (3, -4)
and co-ordinates of O’ reflected in PP’ are (6, 0)
(ii) Length of PP’ = 8 units and OO’ = 6 units
(iii) Perimeter of POP’O’ is Q35. Use a graph paper for this question. (Take 10 small divisions = 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 x-axis. (2005)
Solution: (i) Two points P (0, 5) and Q (-2, 4) are given As the abscissa of P is 0.
It is invariant when is reflected in y-axis.
(ii) Let Q’ be the image of Q on reflection in y-axis.
Co-ordinate of Q’ will be (2, 4)
(iii) (0, k) on reflection in the origin is invariant.
co-ordinates of image will be (0, 0). k = 0
(iv) The reflection of Q in the origin is the point Q”
and its co-ordinates will be (2, – 4)
and reflection of Q” (2, – 4) in x-axis is (2, 4) which is the point Q’ 