Reflection — Exercise 12

A point P is its own image in a line l, then point P is :

1. on the line l

2. below the line l

3. above the line l

4. none of the above

Answer

Since, point P is its own image in a line l, so it is an invariant point, which lies on line itself.

Hence, Option 1 is the correct option.

A point P(-7, 8) is first reflected in the origin and then in x-axis to get the point Q. The co-ordinates of point Q are :

1. (-7, 8)

2. (7, 8)

3. (7, -8)

4. (-7, -8)

Answer

We know that,

On reflecting in origin the sign of both x and y co-ordinate changes.

Let on reflection in origin P(-7, 8) becomes P'.

P(-7, 8) = P'(7, -8)

We know that,

On reflecting in x-axis the sign of y co-ordinate changes.

According to question,

On reflection in x-axis P'(7, 8) becomes Q.

P'(7, -8) = Q(7, 8).

Hence, Option 2 is the correct option.

A point P is reflected first in the y-axis and then in the origin to get the point (6, 6). The co-ordinates of the point P are :

1. (6, -6)

2. (-6, -6)

3. (-6, 6)

4. (6, 6)

Answer

Let co-ordinates of point P are (x, y) and P" be (6, 6).

We know that,

On reflecting in y-axis the sign of x co-ordinate changes.

Let on reflection in y-axis P(x, y) becomes P'.

⇒ P' = (-x, y)

We know that,

On reflecting in origin the sign of both x and y co-ordinate changes.

Let on reflection in origin P'(-x, y) becomes P".

⇒ P" = (x, -y)

⇒ (6, 6) = (x, -y)

⇒ x = 6 and -y = 6

⇒ x = 6 and y = -6.

P = (x, y) = (6, -6).

Hence, Option 1 is the correct option.

The point P(4, -8) is reflected in the line x = 0 to get the point R. The co-ordinates of point R are :

1. (4, -8)

2. (-4, 8)

3. (4, 8)

4. (-4, -8)

Answer

Equation of y-axis is x = 0.

We know that,

On reflecting in y-axis the sign of x co-ordinate changes.

Given,

On reflecting in line x = 0 (or y-axis) point P becomes R.

∴ P(4, -8) = R(-4, -8).

Hence, Option 4 is the correct option.

The point P(5, 5) is first reflected in line y = 0 and then in x-axis to get the point Q. The co-ordinates of point Q are :

1. (5, 5)

2. (-5, 5)

3. (5, -5)

4. (-5, -5)

Answer

Given,

The point P(5, 5) is first reflected in line y = 0 and then in x-axis to get the point Q.

We know that,

Equation of x-axis is y = 0.

Thus, we can say that,

Point P is first reflected in x-axis and then again in x-axis to get point Q.

We know that,

On reflecting in x-axis the sign of y co-ordinate changes.

Let on first reflection in x-axis point P becomes P'.

P(5, 5) = P'(5, -5)

On second reflection in x-axis point P' becomes Q.

P'(5, -5) = Q(5, 5)

Hence, Option 1 is the correct option.

State the co-ordinates of the following points under reflection in the line x = 0 :

(i) (-6, 4)

(ii) (0, 5)

(iii) (3, -4)

Answer

On y-axis, x = 0.

Hence, reflection in the line x = 0 means reflection in y-axis.

Reflection in y-axis is given by,

M_y(x, y) = (-x, y) ..........(1)

(i) Substituting (-6, 4) in equation 1 we get,

M_y(-6, 4) = (6, 4).

Hence, co-ordinate of (-6, 4) under reflection in the line x = 0 is (6, 4).

(ii) Substituting (0, 5) in equation 1 we get,

M_y(0, 5) = (0, 5).

Hence, co-ordinate of (0, 5) under reflection in the line x = 0 is (0, 5).

(iii) Substituting (3, -4) in equation 1 we get,

M_y(3, -4) = (-3, -4).

Hence, co-ordinate of (3, -4) under reflection in the line x = 0 is (-3, -4).

State the co-ordinates of the following points under reflection in the line y = 0 :

(i) (-3, 0)

(ii) (8, -5)

(iii) (-1, -3)

Answer

On x-axis, y = 0.

Hence, reflection in the line y = 0 means reflection in x-axis.

Reflection in x-axis is given by,

M_x(x, y) = (x, -y) ..........(1)

(i) Substituting (-3, 0) in equation 1 we get,

M_x(-3, 0) = (-3, 0).

Hence, co-ordinate of (-3, 0) under reflection in the line y = 0 is (-3, 0).

(ii) Substituting (8, -5) in equation 1 we get,

M_x(8, -5) = (8, 5).

Hence, co-ordinate of (8, -5) under reflection in the line y = 0 is (8, 5).

(iii) Substituting (-1, -3) in equation 1 we get,

M_x(-1, -3) = (-1, 3).

Hence, co-ordinate of (-1, -3) under reflection in the line y = 0 is (-1, 3).

A point P is reflected in the x-axis. Co-ordinates of its image are (-4, 5).

(i) Find the co-ordinates of P.

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

Answer

(i) Reflection in x-axis is given by,

M_x(x, y) = (x, -y) .........(1)

Given, image of point P(x, y) on reflection in x-axis is (-4, 5)

Comparing with equation 1 we get,

(x, -y) = (-4, 5)

⇒ x = -4 and -y = 5

⇒ x = -4 and y = -5.

P = (x, y) = (-4, -5).

Hence, co-ordinates of P = (-4, -5).

(ii) Reflection in y-axis is given by,

M_y(x, y) = (-x, y)

Substituting (-4, -5) in above equation we get,

M_y(-4, -5) = (4, -5).

Hence, co-ordinates of image of P under reflection in y-axis = (4, -5).

A point P is reflected in the origin. Co-ordinates of its image are (-2, 7).

(i) Find the co-ordinates of P.

(ii) Find the co-ordinates of the image of P under reflection in the x-axis.

Answer

(i) Reflection in origin is given by,

M_o(x, y) = (-x, -y) .........(1)

Given, image of point P(x, y) on reflection in origin is (-2, 7)

Comparing with equation 1 we get,

(-x, -y) = (-2, 7)

⇒ -x = -2 and -y = 7

⇒ x = 2 and y = -7.

P = (x, y) = (2, -7).

Hence, co-ordinates of P = (2, -7).

(ii) Reflection in x-axis is given by,

M_x(x, y) = (x, -y)

Substituting (2, -7) in above equation we get,

M_x(2, -7) = (2, 7).

Hence, co-ordinates of image of P under reflection in x-axis = (2, 7).

The point P(a, b) is first reflected in the origin and then reflected in the y-axis to P'. If P' has co-ordinates (4, 6); evaluate a and b.

Answer

Let on reflection of P in origin,

M_o(a, b) = (-a, -b).

Now on reflection in y-axis,

M_y(-a, -b) = (a, -b)

Given, final co-ordinates after reflections = (4, 6).

Comparing with above equation we get,

⇒ (a, -b) = (4, 6)

⇒ a = 4 and -b = 6

⇒ a = 4 and b = -6.

Hence, a = 4 and b = -6.

The point A(-3, 2) is reflected in the x-axis to the point A'. Point A' is then reflected in the origin to point A".

(i) Write down the co-ordinates of A".

(ii) Write down a single transformation that maps A onto A".

Answer

(i) Reflection in x-axis is given by,

M_x(x, y) = (x, -y)

∴ Image on reflection of A(-3, 2) in x-axis = A'(-3, -2)

Reflection in origin is given by,

M_o(x, y) = (-x, -y)

∴ Image on reflection of A'(-3, -2) in origin = A"(3, 2)

Hence, co-ordinates of A" = (3, 2).

(ii) Transformation from A to A" is,

A(-3, 2) = A"(3, 2).

Reflection in y-axis is given by,

M_y(x, y) = (-x, y)

∴ Image on reflection of A(-3, 2) in y-axis = A''(3, 2)

Hence, the single transformation that maps A onto A" is reflection in y-axis.

The triangle ABC, where A is (2, 6), B is (-3, 5) and C is (4, 7), is reflected in the y-axis to triangle A'B'C'. Triangle A'B'C' is then reflected in the origin to triangle A"B"C".

(i) Write down the co-ordinates of A", B" and C".

(ii) Write down a single transformation that maps triangle ABC onto triangle A"B"C".

Answer

(i) On reflection of point A in y-axis,

A(2, 6) = A'(-2, 6)

On reflection of point A' in origin,

A'(-2, 6) = A"(2, -6).

On reflection of point B in y-axis,

B(-3, 5) = B'(3, 5)

On reflection of point B' in origin,

B'(3, 5) = B"(-3, -5).

On reflection of point C in y-axis,

C(4, 7) = C'(-4, 7)

On reflection of point C' in origin,

C'(-4, 7) = C"(4, -7).

Hence, co-ordinates of A" = (2, -6), B" = (-3, -5), C" = (4, -7).

(ii) Transformation,

A(2, 6) = A"(2, -6), B(-3, 5) = B"(-3, -5) and C(4, 7) = C"(4, -7)

The single transformation that maps above transformation is reflection in x-axis.

Hence, reflection in x-axis maps triangle ABC onto triangle A"B"C".

Attempt this question on graph paper.

(a) Plot A (3, 2) and B (5, 4) on graph paper. Take 2 cm = 1 unit on both the axes.

(b) Reflect A and B in the x-axis to A' and B' respectively. Plot these points also on the same graph paper.

(c) Write down :

(i) the geometrical name of the figure ABB'A';

(ii) the measure of angle ABB';

(iii) the image A" of A, when A is reflected in the origin.

(iv) the single transformation that maps A' to A".

Answer

The graph is shown below:

(c) (i) From figure,

ABB'A' is an isosceles trapezium.

(ii) On measuring,

∠ABB' = 45°.

Hence, ∠ABB' = 45°.

(iii) From figure,

When A is reflected in origin,

A(3, 2) = A"(-3, -2).

Hence, co-ordinates of A" = (-3, -2).

(iv) From figure,

A' = A" (On reflection in y-axis)

Hence, reflection of A' in y-axis maps A' to A".

Points (3, 0) and (-1, 0) are invariant points under reflection in the line L₁; points (0, -3) and (0, 1) are invariant points on reflection in line L₂.

(i) Name and write equations for the lines L₁ and L₂.

(ii) Write down the images of points P(3, 4) and Q(-5, -2) on reflection in L₁. Name the images as P' and Q' respectively.

(iii) Write down the images of P and Q on reflection in L₂. Name the images as P" and Q" respectively.

(iv) State or describe a single transformation that maps P' onto P".

Answer

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

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

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

So, L₁ = x axis.

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

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

So, L₂ = y axis.

Hence, L₁ = x-axis whose equation is y = 0 and L₂ = y-axis whose equation is x = 0.

(ii) Line L₁ is x axis.

Reflection in x-axis is given by,

M_x(x, y) = (x, -y)

∴ Image on reflection of P(3, 4) in L₁ (x-axis) = P'(3, -4)

Similarly, image on reflection of Q(-5, -2) in L₁ (x-axis) = Q'(-5, 2)

Hence, co-ordinates of P' = (3, -4) and Q' = (-5, 2).

(iii) Line L₂ is y axis.

Reflection in y-axis is given by,

M_y(x, y) = (-x, y)

∴ Image on reflection of P(3, 4) in L₂ (y-axis) = P''(-3, 4)

Similarly, image on reflection of Q(-5, -2) in L₂ (y-axis) = Q''(5, -2)

Hence, co-ordinates of P" = (-3, 4) and Q" = (5, -2).

(iv) P' = (3, -4) and P" = (-3, 4)

P'(3, -4) ⇒ P"(-3, 4)

Since sign of both abscissa and ordinate is changed, this transformation is possible on reflection in origin.

Hence, reflection in origin maps P' onto P".

The point (-2, 0) on reflection in a line is mapped to (2, 0) and the point (5, -6) on reflection in the same line is mapped to (-5, -6).

(i) State the name of mirror line and write its equation.

(ii) State the co-ordinates of the image of (-8, -5) in the mirror line.

Answer

(i) Given,

(-2, 0) ⇒ (2, 0) and (5, -6) ⇒ (-5, -6)

In both above transformation the sign of abscissa changes, which is possible after reflection in y-axis.

Hence, mirror line = y-axis, whose equation is x = 0.

(ii) Reflection in y-axis is given by,

M_y(x, y) = (-x, y)

∴ Image on reflection of (-8, -5) in mirror line (y-axis) = (8, -5)

Hence, co-ordinates of the image of (-8, -5) in the mirror line = (8, -5).

The points P(4, 1) and Q(-2, 4) are reflected in line y = 3. Find the co-ordinates of P', the image of P and Q', the image of Q.

Answer

Since, y = 3 is a straight line parallel to x-axis and at a distance of 3 units from it, therefore in the figure, AB represents y = 3.

Steps of construction :

1. Mark P(4, 1) and Q(-2, 4) on the graph.
2. From point P draw a straight line Perpendicular to AB and produce.
3. On this line mark a point P' which is at same distance behind AB as P(4, 1) before it.
4. From point Q draw a straight line Perpendicular to AB and produce.
5. On this line mark a point Q' which is at same distance behind AB as Q(-2, 4) before it.

The graph is shown below:

From graph,

P' = (4, 5) and Q' = (-2, 2)

Hence, co-ordinates of P' = (4, 5) and Q' = (-2, 2).

A point P (-2, 3) is reflected in the line x = 2 to point P'. Find the co-ordinates of P'.

Answer

Since, x = 2 is a straight line parallel to y-axis and at a distance of 2 units from it, therefore in the figure, AB represents x = 2.

Steps of construction :

1. Mark P(-2, 3) on the graph.
2. From point P draw a straight line perpendicular to AB and produce.
3. On this line mark a point P' which is at same distance behind AB as P(-2, 3) before it.

The graph is shown below:

From graph,

P' = (6, 3).

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

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, 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.

Answer

When a point is reflected in x-axis, the sign of its ordinate changes.

Image of reflection of point P(a, b) in x-axis = P'(a, -b)

Comparing with P'(2, -3)

We get,

⇒ a = 2 and -b = -3

⇒ a = 2 and b = 3.

So, P(a, b) = (2, 3) and P'(a, -b) = (2, -3)

When a point is reflected in y-axis, the sign of its abscissa changes.

Image of reflection of point P(2, 3) in y-axis = P"(-2, 3)

Since, x = 4 is a straight line parallel to y-axis and at a distance of 4 units from it, therefore in the figure, AB represents x = 4.

Steps of construction :

1. Mark P(2, 3) on the graph.
2. From point P draw a straight line perpendicular to AB and produce.
3. On this line mark a point P''' which is at same distance behind AB as P(2, 3) before it.

From graph,

P''' = (6, 3).

Hence, a = 2, b = 3, P" = (-2, 3) and P''' = (6, 3).

Points A and B have co-ordinates (3, 4) and (0, 2) respectively. Find the image :

(a) A' of A under reflection in the x-axis.

(b) B' of B under reflection in the line AA'

(c) A" of A under reflection in the y-axis.

(d) B" of B under reflection in the line AA''.

Answer

The graph is shown below:

(i) From graph,

Co-ordinates of A' = (3, -4).

(ii) From graph,

Co-ordinates of B' = (6, 2).

(iii) From graph,

Co-ordinates of A" = (-3, 4).

(iv) From graph,

Co-ordinates of B" = (0, 6).

(i) Plot the points A(3, 5) and B(-2, -4). Use 1 cm = 1 unit on both the axes.

(ii) A' is the image of A when reflected in the x-axis. Write down the co-ordinates of A' and plot it on the graph paper.

(iii) B' is the image of B when reflected in the y-axis, followed by reflection in the origin. Write down the co-ordinates of B' and plot it on the graph paper.

(iv) Write down the geometrical name of the figure AA'BB'.

(v) Name two invariant points under reflection in the x-axis.

Answer

(i) The graph is shown below:

(ii) From graph,

Co-ordinates of A' = (3, -5).

(iii) Let B" be the point after reflection of B in y-axis.

From graph,

Co-ordinates of B' = (-2, 4).

(iv) From graph,

Figure AA'BB' is an isosceles trapezium.

(v) Invariant points to a line are points which lie on the line.

So, points invariant under reflection in x-axis lie on x-axis.

These can be any points with ordinate = 0.

Hence, (5, 0) and (-17, 0) are invariant points.

The point P(5, 3) was reflected in the origin to get the image P'.

(a) Write down the co-ordinates of P'.

(b) If M is the foot of the perpendicular from P to the x-axis, find the co-ordinates of M.

(c) If N is the foot of the perpendicular from P' to the x-axis, find the co-ordinates of N.

(d) Name the figure PMP'N.

(e) Find the area of the figure PMP'N.

Answer

The graph is shown below:

(a) From graph,

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

(b) From graph,

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

(c) From graph,

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

(d) From graph,

Figure PMP'N is a parallelogram.

(e) As diagonal divides parallelogram into two triangles of equal area.

From graph,

Diagonal NM divides parallelogram into two right angle triangle of equal area.

∴ area of △NMP = area of △NMP'

Since, 1 block = 1 unit

So, NM = 10 and NP = 3

area of △NMP = $\dfrac{1}{2}$ x NM x NP = $\dfrac{1}{2}$ x 10 x 3 = 15 sq. units.

∴ area of △NMP' = 15 sq. units

From graph,

area of || gm PMP'N = area of △NMP + area of △NMP' = 15 sq. units + 15 sq. units = 30 sq. units.

Hence, area of || gm PMP'N = 30 sq. units.