# ML Aggarwal ICSE Solutions Class 10 Math Tenth Chapter Reflection

### 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 XoXI and YoYI 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 PI so that QP1 = QP.

Thus, the point PI is the reflection of p in the line x = 4. And hence, the co-ordinates of pI are (6, 3)

(2) (i) First we drawn axes XoXI and YoYI 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 pI so that pq = qpI.

Thus, the point pI is the image or reflection of point p in the line x = -1.

Hence, the co-ordinates of points PI are (-3, -2)  (Q8) (i) The point P (2,4) on reflection in the line y = 1 is mapped onto PI find the co-ordinates of PI.

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

Solution:

(i) Initially we drawn axes XOXI and YOYI 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 PI so that QPI = PQ

Thus, PI is the reflection of point ‘p’ whose co-ordinates are found to be PI (2, -2) as shown below. (ii) Initially, we will draw axes XOXI and YOYI 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 PI so that PQI = QP

Hence, the image P is found to be PI (-3, 1) (Q9) The point P (-4, -5) on reflection in Y-axis is mapped on PI. The  point PI on reflection in the origin is mapped on PII. Find the co-ordinate of PI and PII. Write down a single transformation that maps P onto PII.

Solution:

Given that, PI is the image of point P (-4, -5) in Y-axis.

Thus, Co-ordinates of PI will be PI (4, -5).

Again, given that PII is the image P of point PI under reflection in origin will be (-4, 5)

Thus, the single transformation that maps P onto PII 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 XOXI and YOYI 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 P1 so that QPI = PQ and here PI is found to be image of the point ‘P’ in x-axis.
• Thus, co-ordinates of point PI (3, -1)
• Now, draw a perpendicular from PI which meets at point R on the line x = 1. (Q12) If PI (-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 P1 (-4, -3) is the image of a point P under reflection in the origin.

• As the reflection of point P is found to be PI (-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 PII so that QP11 = PQ.
• Thus, the PII is the image of point P in the line y = -2

Hence, the co-ordinates of point PII are PII (+4, -7) (Q13) A point P (a, b) is reflected in the x-axis to PI (2, -3), write down the values of a and b. PII is the image of P, when reflected in the y-axis. Write down the co-ordinates of PII. Find the co-ordinates of PII 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 PI (2, -3). And PII is the image of P, when reflected in the y-axis. Thus co-ordinates of P1 are found to be PI (a, -b) but here PI (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 PII are found to be PII (-2, 3).

• Now, we will draw a line x = 4 which is parallel to y-axis.
• Here PIII is the image o point P when it is reflected in the line x = 4
• Thus, co-ordinates of point PIII are found to be PIII (6, 3) (Q14) (i) Point P (a, b) is reflected in the x-axis to PI (5, -2).

Write down the values of a and b.

(ii) PII is the image of P when reflected in the y-axis. Write down the co-ordinates of PII.

(iii) Name a single transformation that maps PI to PII

Solution:

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

(ii) Here, PII is the image of P when reflected in the y-axis

Hence, co-ordinates of point PII are found to be PII (-5, -2)

(iii) A single transformation that maps PI to PII is the origin.

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

(i) The image AI of A under reflection in the x-axis.

(ii) The image BI of B under reflection in the line AAI.

Solution:

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

(i) AI is the image of A reflected in the x-axis having co-ordinates AI (2, -5)

(ii) BI is the image of B reflected in the line AAI whose co-ordinates are found to be BI (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
• AI (-6, 2), BI (-3, -1) and CI (2, 4) respectively as shown in figure below.
• Now, by joining the points AI, BI and CI we found a triangle △AIBICI 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 AIBICI. The triangle AIBICI is then reflected in the origin to triangle AIIBIICII. Write down the co-ordinates of AII, BII, CII. Write down a single transformation that maps ABC to AII BII CII.

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 AI, BI and CI respectively.
• Hence, the co-ordinates of points AI (1, -2), BI (4, -8) and CI (6, -8) respectively.
• Again points AI, BI and CI are reflected in origin to form a △AII BII CII
• Thus, the co-ordinates of point AII, BII and CII are found to be AII (-1, 2), BII (-4, 8) and CI (-6, 8) respectively.
• And the single transformation which maps ABC onto AII BII CII y-axis only.

(Q19) The image of a point P on reflection in a line l is point PI. 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 BI, CI, DI respectively.

(iii) Write down the co-ordinates of BI, CI and DI

(iv) Join the points A, B, C, DI, CI, BI, AI in order and give a name to the closed figure ABCDDICIB

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 BI, CI and DI also

(iii) The co-ordinates are found to be BI (-3, 0), CI (-1, 0) and DI (-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

= √62 + 82 = √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√32 + 42

= 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 Q11 (2, -4 ) in x-axis is (2, -4) which is the point Q’. Updated: February 15, 2023 — 1:43 pm