ML Aggarwal Solution Class 10 Chapter 10 Reflection Exercise 10

 Exercise 10

Question 1

Find the co-ordinates of the images of the following points under reflection in the x- axis:

(i) (2, -5)

(ii) $-\frac{3}{2},-\frac{1}{2}$

(iii) ( – 7, 0)
Sol :
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 $-\frac{3}{2},-\frac{1}{2}$ will be $-\frac{3}{2}, \frac{1}{2}$

(iii) Image of ( -7, 0) will be (-7, 0)


Question 2

Find the co-ordinates of the images of the following points under reflection in the y-axis:

(i) (2, – 5)

(ii) $-\frac{3}{2}, \frac{1}{2}$

(iii) (0, – 7)

Sol :

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 $-\frac{3}{2}, \frac{1}{2}$ will be $\frac{3}{2}, \frac{1}{2}$

(iii) Image of (0, -7) will be (0, -7)


Question 3

Find the co-ordinates of the images of the following points under reflection in the origin:

(i) (2, – 5)

(ii) $\frac{-3}{2}, \frac{-1}{2}$

(iii) (0, 0)

Sol :

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 $\frac{-3}{2}, \frac{-1}{2}$ will be $\frac{3}{2}, \frac{1}{2}$

(iii) Image of (0, 0) will be (0, 0)


Question 4

The image of a point P under reflection in the x-axis is (5, – 2). Write down the co-ordinates of P.

Sol :

As the image of a point (5, -2) under x – axis is P

∴ Co-ordinates of P will be (5, 2)


Question 5

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.

Sol :

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)


Question 6

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.

Sol :

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)


Question 7

(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.

Sol :

(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)






















Question 8

(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.

Sol :

(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).





















Question 9

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”.

Sol :

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


Question 10

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)

Sol :

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)


Question 11

Find the co-ordinates of the image of (3, 1) under reflection in x-axis followed by a reflection in the line x = 1.

Sol :

(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)





















Question 12

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.

Sol :

(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)





















Question 13

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)

Sol :

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).


Question 14

(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)

Sol :

(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.


Question 15

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’.

Sol :

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).

Question 16

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?

















Sol :
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.

Question 17

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.
Sol :
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.

Question 18

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”.

Sol :

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.


Question 19

The image of a point P on reflection in a line l is point P’. Describe the location of the line l.

Sol :

The line will be the right bisector of the line segment joining P and P’.


Question 20

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.

Sol :

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.


Question 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 co-ordinates of the image of ( – 3, – 4) in the mirror line.

Sol :

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).


Question 22

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’.
Sol :
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’

Question 23

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)
Sol :
(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 $\Delta \mathrm{PQR}=\frac{1}{2} \times \mathrm{QR} \times \mathrm{PQ}$
$=\frac{1}{2} \times 4 \times 8$
=16 sq. units.

Question 24

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)

Sol :

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

$=2 \times \frac{1}{2} \times \mathrm{MN} \times \mathrm{PM}=\mathrm{MN} \times \mathrm{PM}$

= 10 x 3 = 30 sq. units. Ans.


Question 25

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.

Sol :

(i) A (6, 4), B (0, 4)

(ii) $\mathrm{A}(6,4) \xrightarrow{origin} \mathrm{A}^{\prime}(-6,-4)$

$\mathrm{B}(0,4) \xrightarrow{origin} \mathrm{B}^{\prime}(0,-4)$

(iii) ABA'B' is a parallelogram

(iv) $\mathrm{AB}^{\prime}=\sqrt{(\mathrm{AB})^{2}+\left(\mathrm{BB}^{\prime}\right)^{2}}=\sqrt{(6)^{2}+(8)^{2}}$

$=\sqrt{36+64}=\sqrt{100}=10$ units

Perimeter = Sum of all sides = 6 + 10 + 6 + 10 = 32 units


Question 26

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)

Sol :

(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.


Question 27

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.































Sol :

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
















(iii) The quadrilateral formed by joining the points, $\mathrm{BCC}_{2} \mathrm{~B}_{2}$ is an isosceles trapezium and its area

$=\frac{1}{2}\left(\mathrm{BB}_{2}+\mathrm{CC}_{2}\right) \times 3=\frac{1}{2}(10+4) \times 3$

$=\frac{1}{2} \times 14 \times 3=21$ sq. units


Question 28

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’.

Sol :

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

$4 \times \mathrm{OP}=4 \times \sqrt{(\mathrm{OQ})^{2}+(\mathrm{PQ})^{2}}=4 \sqrt{3^{2}+4^{2}}$











$=4 \sqrt{(9+16)}=4 \times \sqrt{25}=4 \times 5=20$ units


Question 29

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)

Sol :










(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’


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