# Class 10 SELINA Solutions Maths Chapter 12 - Reflection (In x-axis, y-axis, x=a, y=a and the origin ; Invariant Points)

## Reflection (In x-axis, y-axis, x=a, y=a and the origin ; Invariant Points) Exercise TEST YOURSELF

### Solution 1(a)

Correct option: (iv) y-axis

When (5, 6) is reflected in a line to get the point (–5, 6), only x-coordinate changes.

Therefore, the line of reflection is y-axis.

### Solution 1(b)

Correct option: (i) Point A

Let A ≡ (a, b)

When A is reflected in y-axis, we get B ≡ (–a, b).

When B is reflected in origin, we get C ≡ (a, –b).

When C is reflected in y = 0 i.e. x-axis, we get P ≡ (a, b).

Therefore, point A coincides with point P.

### Solution 1(c)

Correct option: (ii) y’ = x’

The point P(x, y) is reflected in the line y = x to point P’(x’, y’), then y’ = x’.

### Solution 1(d)

Correct option: (i) (10, 0)

As (5, 7) is reflected in x-axis, so the coordinates of P’ are (5, –7).

O’ is the image of origin O in the line PP’.

So, the coordinates of O’ are (10, 0).

### Solution 1(e)

Correct option: (iv) no-one

Let P ≡ (a, b)

When P is reflected in x = 0 i.e. y-axis, we get P’ ≡ (–a, b).

When P’(–a, b) is reflected in y = 0 i.e. x-axis, we get P” ≡ (–a, –b).

Thus, no two points are invariant.

### Solution 1(f)

Correct option: (iv) Δ ABC ≠ Δ A”B”C”

As ABC is reflected in y-axis, so only x-coordinate of each vertex will change by a negative sign.

As A’B’C’ is reflected in the line y = 0 i.e. x-axis, so only y-coordinate of each vertex will change by a negative sign.

The lengths of the sides and the angles remains the same.

Thus, Δ A’B’C’ ∼Δ A”B”C”, Δ A’B’C’≅Δ A”B”C” and Δ ABC ≅Δ A”B”C”.

Also, Δ ABC = Δ A”B”C”.

### Solution 2

Reflection in y-axis is given by M_{y} (x, y) = (-x, y)

A' = Reflection of A(4, -1) in y-axis = (-4, -1)

Reflection in x-axis is given by M_{x} (x, y) = (x, -y)

B' = Reflection of B in x-axis = (-2, 5)

Thus, B = (-2, -5)

### Solution 3

(a) We know that reflection in the line x = 0 is the reflection in the y-axis.

It is given that:

Point (-5, 0) on reflection in a line is mapped as (5, 0).

Point (-2, -6) on reflection in the same line is mapped as (2, -6).

Hence, the line of reflection is x = 0, that is y-axis.

(b) It is known that M_{y} (x, y) = (-x, y)

Co-ordinates of the image of (5, -8) in y-axis are (-5, -8).

### Solution 4

(i) Co-ordinates of P' and O' are (3, -4) and (6, 0) respectively.

(ii) PP' = 8 units and OO' = 6 units.

(iii) From the graph it is clear that all sides of the quadrilateral POP'O' are equal.

In right PQO',

PO' =

So, perimeter of quadrilateral POP'O' = 4 PO' = 4 5 units = 20 units

(iv) Quadrilateral POP'O' is a rhombus.

### Solution 5

Quadrilateral ABCD is an isosceles trapezium.

Co-ordinates of A', B', C' and D' are A'(-1, -1), B'(-5, -1), C'(-4, -2) and D'(-2, -2) respectively.

It is clear from the graph that D, A, A' and D' are collinear.

### Solution 6

(a) Any point that remains unaltered under a given transformation is called an invariant.

It is given that P (0, 5) is invariant when reflected in an axis. Clearly, when P is reflected in the y-axis then it will remain invariant. Thus, the required axis is the y-axis.

(b) The co-ordinates of the image of Q (-2, 4) when reflected in y-axis is (2, 4).

(c) (0, k) on reflection in the origin is invariant. We know the reflection of origin in origin is invariant. Thus, k = 0.

(d) Co-ordinates of image of Q (-2, 4) when reflected in origin = (2, -4)

Co-ordinates of image of (2, -4) when reflected in x-axis = (2, 4)

Thus, the co-ordinates of the point are (2, 4).

### Solution 7

(a) P (2, -4) is reflected in (x = 0) y-axis to get Q.

P(2, -4) Q (-2, -4)

(b) Q (-2, -4) is reflected in (y = 0) x-axis to get R.

Q (-2, -4) R (-2, 4)

(c) The figure PQR is right angled triangle.

(d) Area of

### Solution 8

(a)

### Solution 9

i. A' = (4, 4) AND B' = (3, 0)

ii. The figure is an arrow head.

### Solution 10

(i)Plotting A(0,4), B(2,3), C(1,1) and D(2,0).

(ii) Reflected points B'(-2,3), C'(-1,1) and D'(-2,0).

(iii) The figure is symmetrical about x = 0

## Reflection (In x-axis, y-axis, x=a, y=a and the origin ; Invariant Points) Exercise Ex. 12

### Solution 2

(i) (-6, 4)

The co-ordinates of the given point under reflection in the line x = 0 are (6, 4).

(ii) (0, 5)

The co-ordinates of the given point under reflection in the line x = 0 are (0, 5).

(iii) (3, -4)

The co-ordinates of the given point under reflection in the line x = 0 are (-3, -4).

### Solution 3

(i) (-3, 0)

The co-ordinates of the given point under reflection in the line y = 0 are (-3, 0).

(ii) (8, -5)

The co-ordinates of the given point under reflection in the line y = 0 are (8, 5).

(iii) (-1, -3)

The co-ordinates of the given point under reflection in the line y = 0 are (-1, 3).

### Solution 4

(i) Since, M_{x} (-4, -5) = (-4, 5)

So, the co-ordinates of P are (-4, -5).

(ii) Co-ordinates of the image of P under reflection in the y-axis are (4, -5).

### Solution 5

(i) Since, M_{O} (2, -7) = (-2, 7)

So, the co-ordinates of P are (2, -7).

(ii) Co-ordinates of the image of P under reflection in the x-axis (2, 7).

### Solution 6

M_{O} (a, b) = (-a, -b)

M_{y} (-a, -b) = (a, -b)

Thus, we get the co-ordinates of the point P' as (a, -b). It is given that the co-ordinates of P' are (4, 6).

On comparing the two points, we get,

a = 4 and b = -6

### Solution 7

(i) The reflection in x-axis is given by M_{x} (x, y) = (x, -y).

A' = reflection of A (-3, 2) in the x- axis = (-3, -2).

The reflection in origin is given by M_{O} (x, y) = (-x, -y).

A'' = reflection of A' (-3, -2) in the origin = (3, 2)

(ii) The reflection in y-axis is given by M_{y} (x, y) = (-x, y).

The reflection of A (-3, 2) in y-axis is (3, 2).

Thus, the required single transformation is the reflection of A in the y-axis to the point A''.

### Solution 8

(i) Reflection in y-axis is given by M_{y} (x, y) = (-x, y)

A' = Reflection of A (2, 6) in y-axis = (-2, 6)

Similarly, B' = (3, 5) and C' = (-4, 7)

Reflection in origin is given by M_{O} (x, y) = (-x, -y)

A'' = Reflection of A' (-2, 6) in origin = (2, -6)

Similarly, B'' = (-3, -5) and C'' = (4, -7)

(ii) A single transformation which maps triangle ABC to triangle A''B''C'' is reflection in x-axis.

### Solution 9

(c)

(i) From graph, it is clear that ABB'A' is an isosceles trapezium.

(ii) The measure of angle ABB' is 45°.

(iii) A'' = (-3, -2)

(iv) Single transformation that maps A' to A" is the reflection in y-axis.

### Solution 10

(i) We know that every point in a line is invariant under the reflection in the same line.

Since points (3, 0) and (-1, 0) lie on the x-axis.

So, (3, 0) and (-1, 0) are invariant under reflection in x-axis.

Hence, the equation of line L_{1} is y = 0.

Similarly, (0, -3) and (0, 1) are invariant under reflection in y-axis.

Hence, the equation of line L_{2} is x = 0.

(ii) P' = Image of P (3, 4) in L_{1} = (3, -4)

Q' = Image of Q (-5, -2) in L_{1} = (-5, 2)

(iii) P'' = Image of P (3, 4) in L_{2} = (-3, 4)

Q'' = Image of Q (-5, -2) in L_{2} = (5, -2)

(iv) Single transformation that maps P' onto P" is reflection in origin.

### Solution 11

(i) We know reflection of a point (x, y) in y-axis is (-x, y).

Hence, the point (-2, 0) when reflected in y-axis is mapped to (2, 0).

And, the point (5, -6) when reflected in y-axis is mapped to (-5, -6).

Thus, the mirror line is the y-axis and its equation is x = 0.

(ii) Co-ordinates of the image of (-8, -5) in the mirror line (i.e., y-axis) are (8, -5).

### Solution 12

The line y = 3 is a line CD parallel to x-axis and at a distance of 3 units from it.

Mark points P(4, 1) and Q(-2, 4).

Mark P' at the same distance above CD as P is below it. Since, P is 2 units below CD, its image P' will be 2 units above CD.

Hence, the co-ordinates of P' are (4, 5).

Similarly, mark Q' at the same distance below CD as Q is above it. Since, Q is 1 unit above CD, its image Q' will be 1 unit below CD.

Hence, the co-ordinates of Q' are (-2, 2).

### Solution 13

The line x = 2 is a line CD parallel to y-axis and at a distance of 2 units from it.

Mark point P(-2, 3).

Mark P' at the same distance to the right of CD as P is to the left of it.

Hence, the co-ordinates of P' are (6, 3).

### Solution 14

A point P (a, b) is reflected in the x-axis to P' (2, -3).

We know M_{x} (x, y) = (x, -y)

Thus, co-ordinates of P are (2, 3). Hence, a = 2 and b = 3.

P" = Image of P reflected in the y-axis = (-2, 3)

P''' = Reflection of P in the line (x = 4) = (6, 3)

### Solution 15

(a) A' = Image of A under reflection in the x-axis = (3, -4)

(b) B' = Image of B under reflection in the line AA' = (6, 2)

(c) A" = Image of A under reflection in the y-axis = (-3, 4)

(d) B" = Image of B under reflection in the line AA" = (0, 6)

### Solution 16

(i) The points A (3, 5) and B (-2, -4) can be plotted on a graph as shown.

(ii) A' = Image of A when reflected in the x-axis = (3, -5)

(iii) C = Image of B when reflected in the y-axis = (2, -4)

B' = Image when C is reflected in the origin = (-2, 4)

(iv) Isosceles trapezium

(v) Any point that remains unaltered under a given transformation is called an invariant.

Thus, the required two points are (3, 0) and (-2, 0).

### Solution 17

(a) Co-ordinates of P' = (-5, -3)

(b) Co-ordinates of M = (5, 0)

(c) Co-ordinates of N = (-5, 0)

(d) PMP'N is a parallelogram.

(e) Area of PMP'N = 2 (Area of D PMN)

### Solution 1(a)

Correct option: (i) on the line *l*

A point P is its own image in a line *l*, then point P is on the line *l*.

### Solution 1(b)

Correct option: (ii) (7, 8)

When point P(–7, 8) is reflected in the origin, its coordinates become (7, –8).

It is then reflected in the x-axis to get the point Q.

Therefore, Q ≡ (7, 8).

### Solution 1(c)

Correct option: (i) (6, –6)

Let P be (a, b).

When reflected in the y-axis, its coordinates are (–a, b).

Then it is reflected in the origin. So, the coordinates are (a, –b).

But (a, –b) = (6, 6)

i.e. a = 6 & b = –6

The coordinates of P are (6, –6).

### Solution 1(d)

Correct option: (iv) (–4, –8)

When x = 0, we get y-axis.

The point P(4, –8) is reflected in the y-axis.

So, the co-ordinates of point R are (–4, –8).

### Solution 1(e)

Correct option: (i) (5, 5)

When y = 0, we get x-axis.

Point P(5, 5) is first reflected in the x-axis and so the co-ordinates are (5, –5).

It is then reflected to x-axis to get the point Q.

Therefore, Q ≡ (5, 5).