**ML Aggarwal ICSE Solutions Class 10 Math 10th Chapter Reflection**

Class 10 Chapter 10 Reflection__Chapter – 10 ____Reflection__

**(1) Find the co-ordinates of the images of the following points under reflection in the x-axis. **

**(i) (2, -5) **

**Solution: **

The image of the point (2, -5) under reflection in the x-axis is (2, 5).

**(ii) (-3/2, -1/2) **

**Solution: **

The image of the point (-3/2, -1/2) under reflection in the x-axis is (-3/2, +1/2)

**(iii) (-7, 0) **

Solution:

The image of the point (-7/0) under reflection in the x-axis is (-7, 0)

**(Q2) Find the co-ordinates of the images of the following points under reflection in the y-axis is **

**(i) (2, -5) **

**Solution: **

The co-ordinates of the image of the point (2, -5) under reflection in the y-axis is (-2, 5)

**(ii) (-3/2, 1/2) **

**Solution: **

The co-ordinates of the image of the point (-3/2, 1/2) under reflection in the y-axis is (3/2, 1/2)

**(iii) (0, -7) **

**Solution: **

The co-ordinates of the image of the point (0, -7) under reflection in the y-axis is (0, -7)

**(Q3) Find the co-ordinates of the images of the following points under reflection in the origin: **

**(i) (2, -5) **

**Solution: **

**The co-ordinates of the images of the following point (2, -5) under reflection in the origin is (-2, 5). **

**(ii) (-3/2, -1/2) **

**Solution: **

The co-ordinates of the image of the point (-3/2, -1/2) under reflection in the origin is (3/2, 1/2)

**(iii) (0, 0) **

**Solution: **

The co-ordinates of the image of the point (0, 0) under reflection in the origin is (0, 0)

**(Q4) The image of a point ‘p’ under reflection in the x-axis is (5, -2). Write down the co-ordinates of point ‘p’. **

**Solution: **

As the image of a point ‘p’ under reflection in the x-axis is p. (5, -2) then the co-ordinates of the point p = (5, 2).

**(Q5) A point ‘p’ is reflected in the x-axis. Co-ordinates of its image are (8, -6). **

**(i) Find the co-ordinate of ‘p’ **

**(ii) Find the co-ordinates of the image p under reflection in y-axis. **

**Solution: **

Given that, the point ‘p’ is reflected in the x-axis. The coordinates of its image are (8, -6)

(i) Hence, the co-ordinates of point p = (8, -6)

(ii) Hence, the image of point ‘p’ under reflection in y-axis having co-ordinates is (-8, 6).

** **

**(Q6) A point ‘p’ is reflected in the origin. Coordinates of the image are (2, -5) and find **

**(i) The co-ordinates of ‘p’ **

**(ii) The co-ordinates of the image p in the x-axis. **

**Solution: **

Given that, a point ‘p’ is reflected in the origin. Co-ordinates of the image are (2, -5)

(i) Then the co-ordinates of point ‘p’ = (-2, 5)

(ii) And the co-ordinates of the image of point ‘p’ in the x-axis is (-2, -5)

** **

**(Q7) (i) The point p (2, 3) is reflected in the line x = 4 to the point P. **

**Find the coordinates of the point ‘p’ **

**(ii) Find the image of the point p (1, -2) in the line x = 1 **

**Solution: **

(1) Initially, we follow the following steps

(i) We will draw the axes XoX^{I} and YoY^{I} with origin ‘o’ and we will take 1cm = 1 unit

(ii) In XY – plane we will plot point P (2, 3) on it.

(iii) Then we drawn a line x = 4 which is parallel to y-axis

(iv) Now, we drawn a perpendicular from point p (43) on line x = 4 which intersects at point Q.

(v) Now, we will extend pq upto P^{I} so that QP^{1 }= QP.

Thus, the point P^{I} is the reflection of p in the line x = 4. And hence, the co-ordinates of p^{I} are (6, 3)

(2) (i) First we drawn axes XoX^{I} and YoY^{I} with 1cm = 1unit

(ii) We plot the point P (1, -2) on it as shown.

(iii) Now, we drawn a line x = – 1 which is parallel to Y-axis.

(iv) We drawn a perpendicular on the line x = -1, which meets points Q.

(v) Now, we extended pq upto p^{I} so that pq = qp^{I}.

Thus, the point p^{I} is the image or reflection of point p in the line x = -1.

Hence, the co-ordinates of points P^{I} are (-3, -2)

**(Q8) (i) The point P (2,4) on reflection in the line y = 1 is mapped onto P ^{I} find the co-ordinates of P^{I}. **

**(ii) Find the image of the point P(-3, -5) in the line y = -2 **

Solution:

(i) Initially we drawn axes XOX^{I} and YOY^{I} so that 1cm = 1 unit.

We plot a point P (2, 4) in XY plane as shown

We drawn a line y = 1 which is parallel to x-axis

Now, we drawn a perpendicular from point P on y = 1

So that it meets at point Q as shown.

We extended PQ upto P^{I} so that QP^{I} = PQ

Thus, P^{I} is the reflection of point ‘p’ whose co-ordinates are found to be P^{I} (2, -2) as shown below.

(ii) Initially, we will draw axes XOX^{I} and YOY^{I} with origin ‘o’ as shown with 1cm = 1unit

We plot a point P (-3, -5) on it.

Then, we will draw a line y = -2 which is parallel to X-axis as shown.

Now, we drawn a perpendicular on y = -2 which meets at point Q.

Now, we extended PQ upto P^{I} so that PQ^{I} = QP

Hence, the image P is found to be P^{I} (-3, 1)

**(Q9) The point P (-4, -5) on reflection in Y-axis is mapped on P ^{I}. The point P^{I} on reflection in the origin is mapped on P^{II}. Find the co-ordinate of P^{I} and P^{II}. Write down a single transformation that maps P onto P^{II}. **

**Solution: **

Given that, P^{I} is the image of point P (-4, -5) in Y-axis.

Thus, Co-ordinates of P^{I} will be P^{I} (4, -5).

Again, given that P^{II} is the image P of point P^{I} under reflection in origin will be (-4, 5)

Thus, the single transformation that maps P onto P^{II} is th X-axis only.

** **

**(Q10) 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 the reflection in the y-axis. **

**(iv) Reflected in the origin **

**Solution: **

Given that, the co-ordinates of the image of the point are (3, -2)

(i) Hence, the co-ordinates of the image reflected in X-axis will be (3, 2)

(ii) The co-ordinates of the image reflected in y-axis found to be (-3, 2)

(iii) The co-ordinates of the point reflected in the x-axis followed by reflection in the y-axis found to be (-3, 2).

(iv) The co-ordinates of the point reflected in the origin are found to be (-3, 2)

** **

**(Q11) 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:**

- Initially, we will draw axes XOX
^{I}and YOY^{I}taking 1cm = 1unit. - We will plot a point P (3, 1) in this XY-plane.
- Then we will draw a line x = 1, which is parallel to y axis.
- Now, we will draw a perpendicular on x-axis meeting at Q as shown.
- We will extend P to P
^{1}so that QP^{I}= PQ and here P^{I}is found to be image of the point ‘P’ in x-axis. - Thus, co-ordinates of point P
^{I}(3, -1) - Now, draw a perpendicular from P
^{I}which meets at point R on the line x = 1.

**(Q12) If P ^{I} (-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: **

Given that, point P^{1} (-4, -3) is the image of a point P under reflection in the origin.

- As the reflection of point P is found to be P
^{I}(-4, -3) in the origin & hence the co-ordinates of point P found to be P (4, 3) - Now, we will draw a perpendicular from point P on the line y = -2 which intersects at point Q.
- Now, produce PQ upto P
^{II}so that QP^{11}= PQ. - Thus, the P
^{II}is the image of point P in the line y = -2

Hence, the co-ordinates of point P^{II} are P^{II} (+4, -7)

**(Q13) A point P (a, b) is reflected in the x-axis to P ^{I} (2, -3), write down the values of a and b. P^{II} is the image of P, when reflected in the y-axis. Write down the co-ordinates of P^{II}. Find the co-ordinates of P^{II} when P is reflected in the line parallel to y-axis so that x = 4. **

**Solution: **

Given that, a point P (a, b) is reflected in the x-axis to P^{I} (2, -3). And P^{II} is the image of P, when reflected in the y-axis. Thus co-ordinates of P^{1} are found to be P^{I} (a, -b) but here P^{I} (2, -3). Thus, on comparing => **a = 2, b = 3**

- Hence, Co-ordinate of point P are found to be (2, 3).

Also, the co-ordinates of point P^{II} are found to be P^{II} (-2, 3).

- Now, we will draw a line x = 4 which is parallel to y-axis.
- Here P
^{III}is the image o point P when it is reflected in the line x = 4 - Thus, co-ordinates of point P
^{III}are found to be P^{III}(6, 3)

**(Q14) (i) Point P (a, b) is reflected in the x-axis to P ^{I} (5, -2). **

**Write down the values of a and b. **

**(ii) P ^{II} is the image of P when reflected in the y-axis. Write down the co-ordinates of P^{II}. **

**(iii) Name a single transformation that maps P ^{I} to P^{II} **

**Solution: **

(i) Here, given that point P (a, b) is the reflection in the x-axis to P^{I} (5, -2). Hence, a = 5 and b = 2

(ii) Here, P^{II} is the image of P when reflected in the y-axis

Hence, co-ordinates of point P^{II} are found to be P^{II} (-5, -2)

(iii) A single transformation that maps P^{I} to P^{II} is the origin.

**(Q15) Points A and B have co-ordinates (2, 5) and (0, 3). Find **

**(i) The image A ^{I} of A under reflection in the x-axis. **

**(ii) The image B ^{I} of B under reflection in the line AA^{I}. **

**Solution: **

Given that, co-ordinates of points A and B are (2, 5) and (0, 3) respectively.

(i) A^{I} is the image of A reflected in the x-axis having co-ordinates A^{I} (2, -5)

(ii) B^{I} is the image of B reflected in the line AA^{I} whose co-ordinates are found to be B^{I} (4, 3)

** **

**(Q17) The points (6, 2), (3, -1) and (-2, 4) are the vertices of a right angle triangle. Check whether it remains a right angled triangle after reflection in the y-axis. **

**Solution: **

- Given that, the points (6, 2), (3, -1) and (-2, 4) are the vertices of a right angled triangle.
- Then, the co-ordinates of the images of points A, B, C which are reflected I y-axis are found to be
- A
^{I}(-6, 2), B^{I}(-3, -1) and C^{I}(2, 4) respectively as shown in figure below. - Now, by joining the points A
^{I}, B^{I}and C^{I}we found a triangle △A^{I}B^{I}C^{I}which is also a right angled triangle.

** **

**(Q18) The triangle ABC where A (1, 2), B (4, 8), C (6, 8) is reflected in the x-axis to triangle A ^{I}B^{I}C^{I}. The triangle A^{I}B^{I}C^{I }is then reflected in the origin to triangle A^{II}B^{II}C^{II}. Write down the co-ordinates of A^{II}, B^{II}, C^{II}. Write down a single transformation that maps ABC to A^{II} B^{II} C^{II}. **

**Solution: **

Here, given that

- The co-ordinates of vertices of a △ABC are found to be A (1, 2), B (4, 8) an C (6, 8) which are reflected in the X-axis to the points A
^{I}, B^{I}and C^{I}respectively. - Hence, the co-ordinates of points A
^{I}(1, -2), B^{I}(4, -8) and C^{I}(6, -8) respectively. - Again points A
^{I}, B^{I}and C^{I}are reflected in origin to form a △A^{II}B^{II}C^{II} - Thus, the co-ordinates of point A
^{II}, B^{II}and C^{II}are found to be A^{II}(-1, 2), B^{II}(-4, 8) and C^{I}(-6, 8) respectively. - And the single transformation which maps ABC onto A
^{II}B^{II}C^{II}y-axis only.

** **

**(Q19) The image of a point P on reflection in a line l is point P ^{I}. Describe the location of the line l. **

**Solution: **

Given that, the image of a point P on reflection in a line l is point P!

Hence, the line is the right bisector of the line segment joining points P and P!

** **

**(Q20) Given two points P and Q and (1) is the image of P on reflection y-axis is the point Q and (2) the midpoint of PQ is invariant on reflection in x-axis. Locate **

**(i) The x-axis **

**(ii) The y-axis and (iii) The origin **

**Solution: **

Here, Given that, the two points P and Q, (1) is the image P on reflection in Y – axis and (2) is the midpoint of PQ is invariant on reflection in x-axis.

(i) Here, x-axis is the line joining the points P and Q as shown.

(ii) Y – axis is the line perpendicular bisector of line segment PQ as shown.

(iii) The origin is the midpoint of line segment PQ

**(21) 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 coordinates of the image of (-3, -4) in the mirror line. **

**Solution: **

Given that, the point (-3,0) on reflection in a line mapped as (3, 0) and the point (2, -3) on reflection in the same line is mapped as (-2, -3).

(i) Thus, here the mirror line is found to be y – axis

(ii) The co-ordinates of the image of point (-3, -4) in the mirror line is found to be (3, -4)

** **

**(Q23) Use a graph for this question. **

**Take 1cm = 1 unit along both x and y axes. **

**(i) Plot the following points: A (0, 5), B (3, 0), C (1, 0) and D (1, -5) **

**(ii) Reflect the points B, C and on the y-axis and name them as B ^{I}, C^{I}, D^{I} respectively. **

**(iii) Write down the co-ordinates of B ^{I}, C^{I} and D^{I} **

**(iv) Join the points A, B, C, D ^{I}, C^{I}, B^{I}, A^{I} in order and give a name to the closed figure ABCDD^{I}C^{I}B **

**Solution: **

We drawn a graph by taking 1cm = 1unit along both x and Y axes

(i) We ploted the points A (0, 5), B (3, 0), C (1, 0) and D (1, -5) as shown

(ii) We located the points B^{I}, C^{I} and D^{I} also

(iii) The co-ordinates are found to be B^{I} (-3, 0), C^{I} (-1, 0) and D^{I} (-1, -5)

(iv) The closed figure formed is found to be arrow head.

**(Q24) 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 coordinates of Q. **

**(ii) Point Q is reflected about line y = 0 to get the image R. Fin the co-ordinate of R. **

**(iii) Name the figure PQR **

**(iv) Find the area of fig. PQR **

**Solution: **

Given that, the point P (2, -4) is reflected about the line x = 0 to get the image Q.

Hence, the co-ordinates of point Q are found to be (2, 4)

(ii) Given that, point Q is reflected about line y = 0 to get the image R.

Hence, the co-ordinates of point R are found to be (-2, 4).

(iii) The figure formed PQR is the right angle triangle as shown

(iv) Then A (△PQR) = 1/2 × QR × PQ

= 1/2 × 4 × 8

A (△PQR) = 16 sq. units

**(Q25) 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 image A’ and B’. **

**(ii) Write down the co-ordinate of A’ and B’ **

**(iii) State the geometrical name of the figure ABA’B’ **

**(iv) Find its perimeter . **

**Solution: **

Given points are A (6, 6) and B (0, 4), after reflection of points A and B in the origin we got the images A’ and B’ as shown below.

A (6, 4) à A’ (-6, -4) and B (0, 4) à B’ (0, -4)

(ii) Thus, the co-ordinates of points A’ and B’ are found to be A’ (-6, -4) and B’ (0, -4)

(iii) The geometrical figure ABA’B’ formed is the parallelogram

(iv) Here, AB’ = √(AB)^{2} + (BB’)^{2}

= √6^{2} + 8^{2} = √36+64 = √100 = 10 units

Thus, the perimeter of parallelogram is formed to be perimeter = 6 + 10 + 6 + 10 = 32 units

**(Q26) 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 AB’, write the co-ordinates of B’. **

**(iv) Give the geometrical name for the figure ABA’B’ **

**Solution: **

(i) We ploted the given points A (4, 6) and B (1, 2) on the graph as shown in fig.

(ii) Given that, A’ is the image of A when reflected in x-axis, then co-ordinates of A’ are formed found to be A’ (4, – 6).

(iii) Given that, B’ is the image of B when reflected in the line AA’ & hence the co-ordinates of B’ are found to be B’ (7, 2)

(iv) Here, ABA’B is the quadrilateral formed in which AB = AB’ and A’B = A’B’

Hence, the geometrical ABA’B’ formed is kite.

** **

**(Q28) 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: **

Given that, the point P (3, 4) is reflected to P’ in the x-axis and O’ is the image of origin in the line PP’

(i) Hence, the co-ordinates of P’ are found to be P’ (3, -4) and co-ordinates of O’ reflected in PP’ are found to be O’ (6, 0)

(ii) The length of segment PP’ = 8 units and the length of segment OO’ = 6 units

(iii) The perimeter of quadrilateral POP’O’ is found to be perimeter = 4 × OP

= 4 × √(OQ)^{2 }+ (PQ)^{2}

= 4√3^{2 }+ 4^{2}

= 4 √9+16

= 4√25 = 4×5

Perimeter = 20 units

**(Q29) Use a 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 (1) **

**(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 **

**Solution: **

Given points are P and Q having co-ordinates (0, 5) and (-2, 4).

(i) Given that, P is invariant when reflected in an axis.

The points P (0, 5) and Q (-2, 4) are given where abscissa of P is 0. Hence, the required axis is y-axis.

(ii) Let us consider the point Q’ is the image of Q on reflection in Y-axis hence, the co-ordinates of Q are found to be Q’ (2, 4)

(iii) (0, K) on reflection in the origin is invariant. And hence the co-ordinates of the image are found to be (0, 0) where K = 0

(iv) Given that, the reflection of Q in the origin is the point Q’’ and its co-ordinates are found to be (2, -4)

Hence, reflection of Q^{11} (2, -4 ) in x-axis is (2, -4) which is the point Q’.