Soln:

Here, the perpendiculars AP, BQ and CR are drawn from the points A, B and C on the line m respectively. Again, AP = PA1, BQ = QB1and CR = CR1 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.

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.

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

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.

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.

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.

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.

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.

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.

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

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.

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.

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.

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:

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: