Subject: Optional Mathematics

In reflection, the line joining the object and the image is perpendicular to the mirror line. It means the mirror line is perpendicular bisector of the line segment joining object and image. The mirror line is also called the axis of reflection.

A reflection is a transformation that flips a figure across a line. The line work as a plane mirror. In reflection, the line joining the object and the image is perpendicular to the mirror line. It means the mirror line is perpendicular bisector of the line segment joining object and image. The mirror line is also called the axis of reflection.

When geometrical figures are reflected in the axis of reflection, the following properties are found.

a. Coordinates can be used for finding images of geometrical figures after the reflection in the lines like X- axis, Y- axis, a line parallel to X- axis, a line parallel to Y- axis, the line y = x, the line y = -x, etc.The distance of the object from the axis of reflection is equal to the distance of reflection is equal to the distance of the image from the axis is a reflection.

OP = OP' as shown in fig 1.

b. The shape of objects and images are laterally inverted. It means top remains at the top, bottom remains at the bottom but left side goes to the right side and right side goes to the left side as shown in fig 2.

c. The lines joining the same ends of the object and image are perpendicular to reflecting axis.

Axis of reflection is the perpendicular bisector of the line segment joining same ends of object and image.

XX'is perpendicular bisector of AA', BB' and CC' as in fig 3.

d. The points on the axis of reflection are invariant points.

e. Use of co-ordinates in Reflection.

f. In reflection, the object figure and its image figure are congruent to each other.

**Reflection in the X- axis**Equation of X- axis is y = 0. So, reflection in X- axis means reflection in the line y = 0. Let P(x, y) be any point in the plane. Draw a perpendicular PL from the point P to the X- axis and produce it to the point P' such that PL = LP'. Then P' is the image of P after reflection in X- axis.

Here, co-ordinates of L are (x, 0). Let the co- ordinates of P' be (x', y'). Since L is the mid- point of line segment PP', then by mid- point formula,

x = \(\frac {x+x'}2\) and 0 = \(\frac {y+y'}2\) or, x + x' = 2x and y + y' = 0 or, x' = x and y' = -y

∴ Image of point P(x, y) after reflection in X- axis is P'(x, -y).

Hence, if R_{x}denotes the reflection in X- axis, then:

Rx: P (x, y)→ P' (x, -y)**Reflection in the Y- axis**Equation of Y- axis is x = 0. So, reflection in Y- axis means reflection in the line x = 0.

Let P (x, y) be any point in the plane. Draw a perpendicular PM from the point P to the Y- axis and produce it to the point P' such that PM = MP'. Then P' is the image of P after reflection in Y- axis.

Here, co-ordinate of M is (0, y). Let the co- ordinates of P' be (x', y'). Since M is the mid-point of line segment PP', then by mid- point formula,

0 = \(\frac{x + x'}2\) and y = \(\frac {y + y'}2\) or, x + x' = 0 and y + y' = 2y or, x' = -x and y' = y

∴ Image of point P(x, y) after reflection in Y- axis P' (-x, y).

Hence, if R_{y}denotes the reflection in Y- axis, then:**Ry: P (x, y)→ P' (-x, y)**

**Reflection in the line parallel to X- axis**The equation of a line parallel to X- axis is given by y = k where k is Y- intercept of the line. So, reflection in the line parallel to X- axis means reflection in the line y = k.

Let P (x, y) be any point in the plane. Draw a perpendicular PM from P to the line y = k and produce it to the point P' such that PM = PM'. Then P' is the image of P after reflection in

Here, co- ordinates of M are (x, k). Let thye co- ordinates of P' be (x', y'). Since, M is the mid-point of the line segment PP', then by mid- point formula,x = \(\frac{x + x'}2\) and k = \(\frac{y + y'}2\) or, x + x' = 2x and y + y' = 2k or, x' = x and y' = 2k - y

∴ Image of point P (x, y) after reflection in the line y = k is P' (x, 2k - y).

Hence, if R denotes the reflection in the line y = k, then:

R: P (x, y)→ P' (x, 2k - y)**Reflection in the line parallel to Y- axis**The equation of a line parallel to Y- axis is given by x = k where k is X - intercept of the line. So, the reflection in a line parallel to Y- axis means reflection in the line x = k.

Let P (x, y) be any point in the plane. Draw a perpendicular PM from P to the line x = k and produce it to the point P' such that PM = MP'. Then P' is the image of P after reflection in the line x = k.

Let the co- ordinates of P' be (x', y').

Since, M is the mid- point of the line segment PP', then by mid- point formula,

k = \(\frac {x + x'}2\) and y = \(\frac {y + y'}2\) or, x + x' = 2k and y + y' = 2y or, x' = 2k - x and y' = y

∴ Image of point P (x, y) after reflection in the line x = k is P' (2k - x, y).

Hence, if R denotes the reflection in the line x = k, then:

R: P (x, y)→ P' (2k - x, y)**Reflection in the line y = x**y = x is the equation of the line which makes an angle of 45° with the positive direction of X- axis. Then, slope of the line, m_{1}= tan 45° = 1.

Let P (x, y) be any point in the plane. Draw perpendicular PM from the point P to the line y = x and produce it to the point P' such that PM = MP'. Then P' is the image of the point P under reflection about the line y = x.

Let (x', y') be the co- ordinates of the point P'.

Then, slope of the line segment PP', m_{2}= \(\frac {y - y'}{x - x'}\).

Since, the line segment PP' and the line y = x are perpendicular to each other. Then:

m_{1}× m_{2}= -1

or, 1× (\(\frac {y-y'}{x-x'}\)) = -1

or, y - y' = -x + x'

or, y + x = y' + x' .......................(i)

Again,

Co- ordinates of mid- point of the line segment PP' are: (\(\frac {x + x'}2\), \(\frac {y + y'}2\)).

This point lies in the line y = x.

So,

\(\frac {y + y'}2\) = \(\frac {x + x'}2\)

or, y + y' = x + x'

or, y - x = x' - y' .........................(ii)

Adding (i) and (ii) we get, 2y = 2x' or x' = y

Subtracting (ii) from (i) we get, 2x = 2y' or y' = x.

∴ Co- ordinates of the point P' are (y, x).

∴ Image of the point P (x, y) under the reflection about the line y = x is the point P (y, x).

Hence, if R denotes the reflection in the line y = x, then:

R: P (x, y)→ P' (y, x)**Reflection in the line y = -x**y = -x is an equation of the line which makes an angle of 135° with the positive direction of X- axis.

∴ Slope of the line y = -x is m_{1}= tan 135° = -1

Let P (x, y) be any point in the plane. Draw perpendicular PM from the point P to the line y = -x and produce it to the point P' such that PM = MP'.

Let the co- ordinates of the point P' be (x', y').

Slope of the line segment PP' is m_{2}= \(\frac {y - y'}{x - x'}\).

Since, the line segment PP' and the line y = -x are perpendicular to each other, then:

m _{1}× m_{2}= -1

or, (-1)× \(\frac {y - y'}{x - x'}\) = -1

or, y - y' = x - x' .........................(i)

Again,

Co- ordinates of the mid- point of line segment PP' are: (\(\frac {x + x'}2\), \(\frac {y + y'}2\)).

This point lies in the line y = -x.

So,

(\(\frac {y + y'}{2}\)) = - (\(\frac {x + x'}2\))

or, y + y' = -x' - x .......................(ii)

Adding (i) and (ii) we get, 2y = - 2x' or x' = -y

Subtracting (ii) from (i) we get, -2y' = 2x or y' = -x

∴ Co- ordinates of P' are (-y, -x)

∴ Image of the point P (x, y) after reflection in the line y = -x is P' (-y, -x).

Hence, if R denotes in the line y = -x, then:

R : P (x, y)→ P' (-y, -x)

- In reflection, the line joining the object and the image is perpendicular to the mirror line.
- The Axis of reflection is the perpendicular bisector of the line segment joining same ends of object and image.
- In reflection, the object figure and its image figure are congruent to each other.
- Reflection in X- axis means reflection in the line y = 0.
- The distance of the object from the axis of reflection is equal to the distance of reflection is equal to the distance of the image from the axis is a reflection.

- It includes every relationship which established among the people.
- There can be more than one community in a society. Community smaller than society.
- It is a network of social relationships which cannot see or touched.
- common interests and common objectives are not necessary for society.

In each of the following problems, m is the axis of reflection. Reflect each figure in m and draw the images.

a.

Soln:

Here, the perpendiculars AP, BQ and CR are drawn from the points A, B and C on the line m respectively. Again, AP = PA^{1}, BQ = QB^{1}and CR = CR^{1} are drawn by producing Ab, BQ and CR respectively. Then A', B' and C' are joined . So, that the imageΔA'B'C' of triangle ABC is formed.

In each of the following problems, m is the axis of reflection. Reflect each figure in m and draw the images.

b.

Soln:

As in the above figure, the perpendiculars AP, BQ, DR and CS are drawn in the mirror line 'm' from the points A, B, C and D respectively. Again AP = PA', BQ = QB', DR = RD' and CS = CS' are drawn by producing AP, BQ, QR and CS respectively. THen A',B',C' and D' are joined so that the image quadrilateral A'B'C'D' of the quadrilateral ABCD is formed.

Soln:

From the given points P and Q, perpendiculars PA and QB are drawn on the mirror line 'm'. Also, PA and QB are produced into P' and Q' so that PA = AP' and QB = BQ'. So the line P'Q' joining P' and Q' is the image of the line PQ.

Soln:

From the given figure, the mirror line 'm' is the perpendicular bisector of the line PQ. So the image P' and Q and image Q' of Q is at P. So, the image of PQ is Q'P' which is the same line PQ.

Soln:

Here, the mirror line 'm' is perpendicular to both the lines AD and BC. Let 'm' cuts AD at P and BC at Q. Now DP is produced up to D' so that DP = PD'. Similarly, APis produced up to A' so that AP = PA' BQ up to B' so that BQ = QB' and CQ up to C' so that CQ = QC'. Now the image A'B'C'D' of ABCD is formed.

Write down the coordinates of the image of each of these points A(3, 0), B(4, -3), C(0, -7), D(-2, 6), E(-3, -3), F(-3, -9), G(6, 6) and H(7, -7)when they are reflected in the following line.

A. the x-axis

Soln:

We have, reflection of a point on x-axis remains the x-coordinate same but the sign of y-coordinate changes. i.e. P(a, b)→ P'(a, -b).

So the images of the above points under the reflection on x - axis are:

A(3, 0)→A'(3, 0), B(4, -3)→ B'(4, 3), C(0, -7)→ C'(0, 7), D(-2, 6)→ D'(-2, -6), E(-3, -3)→ E'(-3, 3), F(-3, -9)→ F'(-3, 9), G(6, 6)→ G'(6, -6), H(7, -7)→ H'(7, 7). Ans

Write down the coordinates of the image of each of these points A(3, 0), B(4, -3), C(0, -7), D(-2, 6), E(-3, -3), F(-3, -9), G(6, 6) and H(7, -7)when they are reflected in the following line.

the y-axis

Soln:

We have the reflection of the point P(a, b) on the y-axis is P'(-a, b). i.e. P(a, b)→ P'(-a, b). So the images of the given points are:

A(3, 0)→ A'(-3, 0), B(4, -3)→ B'(-4, -3), C(0, -7)→ C'(0, -7), D(-2, 6)→ D'(2, 6), E(-3, -3)→ E'(3, -3), F(-3, -9)→ F'(3, -9), G(6, 6)→ G'(-6, 6) and H(7, -7)→ H'(-7, -7). Ans.

Write down the coordinates of the image of each of these points A(3, 0), B(4, -3), C(0, -7), D(-2, 6), E(-3, -3), F(-3, -9), G(6, 6) and H(7, -7)when they are reflected in the following line.

x = y

Soln:

We have, a reflection of a point P(a, b) on the line x = y is P'(b, a). i. e. P(a, b) → P'(b, a). So the images of the given points are:

A(3, 0)→ A'(0, 3), B(4, -3)→ B'(-3, 4), C(0, -7)→ C'(-7, 0), D(-2, 6)→ D'(6, -2), E(-3, -3)→ E'(-3, -3), F(-3, -9)→ F'(-9, -3), G(6, 6)→ G'(6, 6), H(7, -7)→ H'(-7, 7). Ans.

If P(1, 1), Q(3, 1) and R(3, -1) are the coordinates of a triangle PQR.

Show in graph the figure of triangle PQR, after reflecting about x-axis.

Soln:

We have, a reflection on x-axis is P(a, b)→ P'(a, -b). So reflection of triangle PQR on x-axis given the images coordinates as:

P(1, 1) →P'(1, -1)

Q(3, 1)→ Q'(3, -1)

R(3, -1)→ R'(3, 1)

Here, the shaded portion triangle P'Q'R' is the image of triangle PQR.

If P(1, 1), Q(3, 1) and R(3, -1) are the coordinates of triangle PQR.

Show in graph the figure of triangle PQR, after reflecting about line y = -x.

Soln:

We have, reflection on the line y = -x is P(a, b)→ P'(-b, -a). So, we have,

P(1, 1) →P(-1, -1)

Q(3, 1)→ Q'(-1, -3)

R(3, -1)→ R'(1, -3)

The graph is given below:

The shaded portion triangle P'Q'R' is the image of triangle PQR.

If P(1, 1), Q(3, 1) and R(3, -1) are the coordinates of triangle PQR.

Show in graph the figure of triangle PQR after reflecting about the line y = 3.

Soln:

We have, a reflection on the line y = k is P(a, b)→ P'(a, 2k - b) where k = 3. So,

P(1, 1)→ P'(1, 2×3 - 1) = P'(1, 5)

Q(3, 1)→ Q'(3, 2× 3 -1) = Q'(3, 5)

R(3, -1)→ R[3, 2× 3 - (-1)] = R'(3, 7)

The graph is as shown below:

The shaded portion triangle P'Q'R' is teh image of triangle PQR.

Find the axis of reflection of the following cases:

Point A(3, 4) reflected by R to A'(3, -4)

Soln:

Here, the sign of y-coordinate of the point A(3, 4) is changed in the image A'(3, -4) under the reflection R. So the axis of reflection is x-axis.

Find the axis of reflection of the following cases:

Point A(3, 4) reflected by Y to A'(-1, 4)

Soln:

Here, Y reflects the point of A(3, 4) to A'(-1, 4), is which y-coordinate is same but x-coordinate changes from 3 to -1. So the line AA' is parallel to x-axis. Here, the mirror line is the perpendicular bisector of AA' or the mirror line is parallel to y-axis which passes through the mid-point of AA' i. e. \(\frac{3-1}{2}\), \(\frac{4+4}{2}\) = (1, 4). So, the line parallel to y-axis (1, 4) is x = 1.

∴ Axis of reflection is x = 1. Ans

Find the axis of reflection of the following cases:

Point A(4, 5) reflected by R to A'(-5,-4)

Soln:

Here, reflection R reflects the point A(4, 5) to A'(-5, -4). So, x and y coordinates are interchanged and the signs are also changed. i. e. P(a, b)→ P'(-b, -a). So the axis of reflection is the line y = -x. Ans.

Find the axis of reflection of the following cases:

Point A(-2, 3) reflected by P to A'(6, 3)

Soln:

Here, reflection P reflects the point A (-2, 3) to A'(6, 3) is which y-coordinate is same but x-coordinate changes from -2 to 6. In this case as in the reflection of the line x = h, P(a, b)→ P'(2h-a, b)

A(-2, 3)→ A'(6, 3). So,

2h - a = 6

or, 2h -(-2) = 6

or, 2h + y = 6

or, 2h + y = 6

or, 2h = 6 - 2 = 4

∴ h = 2

∴ The axis of reflection is h = 2 which is parallel to y-axis.

Find the axis of reflection of the following cases:

Point A(3, 4) reflected by Q to A'(3, 2)

Soln:

Here, reflection Q reflects the point A(3, 4) to A'(3, 2) is which x-coordinate is same but y-coordinate changes from 4 to 2. In this axis the reflection of the line y = k, P(a, b)→ P'(a, 2k - b),

A(3, 4)→ A'(3, 2).

so, 2k - b = 2

or, 2k - 4 = 2

or, 2k = 2 + 4 = 6

∴ k = 3

∴ The axis of reflection is the line k = 3 which is parallel to x-axis. Ans.

A line passes through the points (3, 0) and (5, 6). Draw the image of this line when it is reflected in (a) the axis and y-axis.

Soln:

X-axis

The reflection on x-axis is P(a, b)→ P'(a, -b). So,

A(3, 0)→ A'(3, 0)

B(5, 6)→ B'(5, -6)

The image A'B' of the line AB is as shown in the graph below:

Y-axis

The reflection of the line AB joining A(3, 0) and B(5, 6) as

P(a, b)→ P'(-a, b), we get,

A(3, 0) → A'(-3, 0)

B(5, 6)→ B'(-5, 6)

The graph line AB and its image A'B' is as shown below:

Points p(2, 1), Q(1, -2), R(-3, -2) and S(-5, 1) are the coordinates of a trapezium. Write down the coordinates of the image and draw the image of PQRS when reflected under the following conditions:

i)axis of reflection is the x-axis

ii)axis of reflection is the y-axis

Soln:

i)Reflection of the given trapezium PQRS under, x-axis is as

P(a, b) → P'(a, -b) we get,

P(2, 1)→ P'(2, -1)

Q(1, -2)→ Q'(1, 2)

R(-3, -2)→ R'(-3, 2) and

S(-5, 1)→ S'(-5, -1)

The graph of trapezium PQRS and itsn image P'Q'R'S' is as shown below:

ii)Reflection of trapezium on y-axis as P(a, b)→ P'(-a, b) we get,

P(2, 1)→ P'(-2, 1)

Q(1, -2)→ Q'(-1, -2)

R(-3, -2)→ R'(3, -2) and S(-5, 1)→ S'(5, 1)

The graph of trapezium PQRS and its image P'Q'R'S' is shown below:

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