Reflection

(i) We know that,

Rule to find reflection of a point in x-axis :

1. Retain the abscissa i.e. x-coordinate.
2. Change the sign of ordinate i.e. y-coordinate.

∴ Point P'(2, 5) is the image on reflection.

Hence, (2, 5) is the image of the point (2, -5) on reflection in x-axis.

(ii) We know that,

Rule to find reflection of a point in x-axis :

1. Retain the abscissa i.e. x-coordinate.
2. Change the sign of ordinate i.e. y-coordinate.

∴ Point P'$\Big(-\frac{3}{2}, \frac{1}{2}\Big)$ is the image on reflection.

Hence, $\Big(-\frac{3}{2}, \frac{1}{2}\Big)$ is the image of the point $\Big(-\frac{3}{2}, -\frac{1}{2}\Big)$ on reflection in x-axis.

(iii) We know that,

Rule to find reflection of a point in x-axis :

1. Retain the abscissa i.e. x-coordinate.
2. Change the sign of ordinate i.e. y-coordinate.

∴ Point P'(-7, 0) is the image on reflection.

Hence, (-7, 0) is the image of the point (-7, 0) on reflection in x-axis.

(i) We know that,

Rule to find reflection of a point in y-axis :

1. Change the sign of abscissa i.e. x-coordinate.
2. Retain the ordinate i.e. y-coordinate.

∴ Point P'(-2, -5) is the image on reflection.

Hence, (-2, -5) is the image of the point (2, -5) on reflection in y-axis.

(ii) We know that,

Rule to find reflection of a point in y-axis :

1. Change the sign of abscissa i.e. x-coordinate.
2. Retain the ordinate i.e. y-coordinate.

∴ Point P'$(\frac{3}{2}, \frac{1}{2})$ is the image on reflection.

Hence, $(\frac{3}{2}, \frac{1}{2})$ is the image of the point $(-\frac{3}{2}, -\frac{1}{2})$ on reflection in y-axis.

(iii) We know that,

Rule to find reflection of a point in y-axis :

1. Change the sign of abscissa i.e. x-coordinate.
2. Retain the ordinate i.e. y-coordinate.

∴ Point P'(0, -7) is the image on reflection.

Hence, (0, -7) is the image of the point (0, -7) on reflection in y-axis.

(i) We know that,

Rules to find the reflection of a point in the origin :

1. Change the sign of abscissa i.e. x-coordinate.
2. Change the sign of ordinate i.e. y-coordinate.

∴ Point P'(-2, 5) is the image on reflection.

Hence, (-2, 5) is the image of the point (2, -5) on reflection in origin.

(ii) We know that,

Rules to find the reflection of a point in the origin :

1. Change the sign of abscissa i.e. x-coordinate.
2. Change the sign of ordinate i.e. y-coordinate.

∴ Point P'$(\frac{3}{2}, \frac{1}{2})$ is the image on reflection.

Hence, $(\frac{3}{2}, \frac{1}{2})$ is the image of the point $(-\frac{3}{2}, -\frac{1}{2})$ on reflection in origin.

(iii) We know that,

Rules to find the reflection of a point in the origin :

1. Change the sign of abscissa i.e. x-coordinate.
2. Change the sign of ordinate i.e. y-coordinate.

∴ Point P'(0, 0) is the image on reflection.

Hence, (0, 0) is the image of the point (0, 0) on reflection in origin.

Since, (5, -2) is the image of P under reflection in x-axis, from graph we get,

The coordinates of P are (5, 2).

(i) Since, (8, -6) is the image of P under reflection in x-axis, let this point be P' so from graph we get,

The coordinates of P are (8, 6).

(ii) We know that,

Rule to find reflection of a point in y-axis :

1. Change the sign of abscissa i.e. x-coordinate.
2. Retain the ordinate i.e. y-coordinate.

∴ Point P''(-8, 6) is the image on reflection.

Hence, (-8, 6) is the image of the point (8, 6) on reflection in y-axis.

(i) Since, (2, -5) is the image of P under reflection in origin, let this point be P' so from graph we get,

The coordinates of P are (-2, 5).

(ii) We know that,

Rule to find reflection of a point in x-axis :

1. Retain the abscissa i.e. x-coordinate.
2. Change the sign of ordinate i.e. y-coordinate.

∴ Point P''(-2, -5) is the image on reflection.

Hence, P''(-2, -5) is the image of the point P(-2, 5) on reflection in x-axis.

We know that the reflection of the point (x, y) in the line x = a is the point (-x + 2a, y).

∴ The image of the point P(2, 3) under reflection in the line x = 4 is the point P'(-2 + 2 × 4, 3) i.e., the point P'(6, 3).

The coordinates of the point P' is (6, 3).

We know that the reflection of the point (x, y) in the line x = a is the point (-x + 2a, y).

∴ The image of the point P(1, -2) under reflection in the line x = -1 is the point P'(-1 + 2 × (-1), -2) i.e., the point P'(-3, -2).

The coordinates of the point P' is (-3, -2).

We know that the reflection of the point (x, y) in the line y = a is the point (x, -y + 2a).

∴ The image of the point P(2, 4) under reflection in the line y = 1 is the point P'(2, -4 + 2 × 1) i.e., the point P'(2, -2).

The coordinates of the point P' is (2, -2).

We know that the reflection of the point (x, y) in the line y = a is the point (x, -y + 2a).

∴ The image of the point P(-3, -5) under reflection in the line y = -2 is the point P'(-3, -(-5) + 2 × (-2)) i.e., the point P'(-3, 1).

Hence, the image of the point (-3, -5) in y = -2 is (-3, 1).

We know that,

Rule to find reflection of a point in y-axis :

1. Change the sign of abscissa i.e. x-coordinate.
2. Retain the ordinate i.e. y-coordinate.

∴ Point P(-4, -5) becomes P'(4, -5) on reflection in y-axis.

We know that,

Rules to find the reflection of a point in the origin :

1. Change the sign of abscissa i.e. x-coordinate.
2. Change the sign of ordinate i.e. y-coordinate.

∴ Point P'(4, -5) becomes P''(-4, 5) on reflection in origin.

On reflecting the point P(-4, -5) in x-axis, it becomes (-4, 5) which are the coordinates of P''.

Hence, the single transformation that maps P(-4, -5) onto P''(-4, 5) is reflection of point P in x-axis.

(i) We know that,

Rule to find reflection of a point in x-axis :

1. Retain the abscissa i.e. x-coordinate.
2. Change the sign of ordinate i.e. y-coordinate.

∴ Point P'(3, 2) is the image on reflection.

Hence, (3, 2) is the image of the point (3, -2) on reflection in x-axis.

(ii) We know that,

Rule to find reflection of a point in y-axis :

1. Change the sign of abscissa i.e. x-coordinate.
2. Retain the ordinate i.e. y-coordinate.

∴ Point P'(-3, -2) is the image on reflection.

Hence, (-3, -2) is the image of the point (3, -2) on reflection in y-axis.

(iii) On reflection in x-axis the point P(3, -2) becomes P'(3, 2) which on further reflection in y-axis becomes P''(-3, 2) as shown in the graph.

Hence, P''(-3, 2) is the final image after reflections.

(iv) We know that,

Rules to find the reflection of a point in the origin :

1. Change the sign of abscissa i.e. x-coordinate.
2. Change the sign of ordinate i.e. y-coordinate.

∴ Point P'(-3, 2) is the image on reflection.

Hence, (-3, 2) is the image of the point (3, -2) on reflection in origin.

We know that,

Rule to find reflection of a point in x-axis :

1. Retain the abscissa i.e. x-coordinate.
2. Change the sign of ordinate i.e. y-coordinate.

∴ Point P'(3, -1) is the image on reflection of (3, 1) in x-axis.

We know that the reflection of the point (x, y) in the line x = a is the point (-x + 2a, y).

∴ The image of the point P'(3, -1) under reflection in the line x = 1 is the point P'(-3 + 2 × 1, -1) i.e., the point P'(-1, -1).

Hence, the coordinates of the final image is P''(-1, -1).

(i) Since, P'(-4, -3) is the image of P under reflection in the origin, from graph P(4, 3).

Hence, the coordinates of P are (4, 3).

(ii) We know that the reflection of the point (x, y) in the line y = a is the point (x, -y + 2a).

∴ The image of the point P(4, 3) under reflection in the line y = -2 is the point P''(4, -(3) + 2 × (-2)) i.e., the point P''(4, -7).

Hence, the coordinates of P under reflection in the line y = -2 is (4, -7).

Since, P'(2, -3) is the image of P under reflection in the x-axis, from graph we get P(2, 3).

∴ The coordinates of P are (2,3), comparing with P(a, b) we get, a = 2 and b = 3.

We know that,

Rule to find reflection of a point in y-axis :

1. Change the sign of abscissa i.e. x-coordinate.
2. Retain the ordinate i.e. y-coordinate.

Since, P'' is image on reflection of P in y-axis.

∴ The coordinates of P'' are (-2,3).

We know that the reflection of the point (x, y) in the line x = a is the point (-x + 2a, y).

∴ The image of the point P(2, 3) under reflection in the line x = 4 is the point P'''(-2 + 2 × 4, 3) i.e the point P'''(6, 3).

Hence, the value of a = 2, b = 3 and the coordinates of P''(-2, 3), P'''(6, 3).

(i) Since, P'(5, -2) is the image of P under reflection in the x-axis, from graph P(5, 2).

∴ The coordinates of P are (5,2), comparing with P(a, b) we get, a = 5 and b = 2.

Hence, the value of a = 5 and b = 2.

(ii) We know that,

Rule to find reflection of a point in y-axis :

1. Change the sign of abscissa i.e. x-coordinate.
2. Retain the ordinate i.e. y-coordinate.

Since, P'' is image on reflection of P(5, 2) in y-axis.

Hence, the coordinates of P'' are (-5, 2).

(iii) The single transformation that maps P'(5, -2) into P''(-5, 2) is reflection in the origin.

The graph with the points A (2, 5), B (0, 3) and their images A' and B' is shown below:

(i) We know that,

Rule to find reflection of a point in x-axis :

1. Retain the abscissa i.e. x-coordinate.
2. Change the sign of ordinate i.e. y-coordinate.

∴ Point A'(2, -5) is the image on reflection of A(2, 5) in x-axis.

The coordinates of A' are (2, -5).

(ii) From graph we get,

The coordinates of B' are (4, 3).

The points A(2, -3), B(-1, 2) and C(0, -2) are plotted on the graph below:

From graph we get,

AB = A'B'
AC = A'C'
BC = B'C'

Hence, △ABC and △A'B'C' are congruent.

The right angled triangle with points (6, 2), (3, -1) and (-2, 4) as vertices is plotted below:

From graph we get,

AB = A'B'
AC = A'C'
BC = B'C'

∴ △ABC and △A'B'C' are congruent.

Since, △ABC is a right angled triangle hence, △A'B'C' is also a right angled triangle.

We know that,

Rule to find reflection of a point in x-axis :

1. Retain the abscissa i.e. x-coordinate.
2. Change the sign of ordinate i.e. y-coordinate.

∴ Coordinates of
⇒ A(1, 2) on reflection in x-axis becomes A'(1, -2).
⇒ B(4, 8) on reflection in x-axis becomes B'(4, -8).
⇒ C(6, 8) on reflection in x-axis becomes C'(6, -8).

We know that,

Rules to find the reflection of a point in the origin :

1. Change the sign of abscissa i.e. x-coordinate.
2. Change the sign of ordinate i.e. y-coordinate.

∴ Coordinates of
⇒ A'(1, -2) on reflection in origin becomes A''(-1, 2).
⇒ B'(4, -8) on reflection in origin becomes B''(-4, 8).
⇒ C'(6, -8) on reflection in origin becomes C''(-6, 8).

A single transformation that maps A ⇒ A'', B ⇒ B'' and C ⇒ C'' is reflection in y-axis.

Hence, the coordinates of A'', B'' and C'' are (-1, 2), (-4, 8) and (-6, 8) respectively, and a single transformation that maps ABC to A''B''C'' is reflection in y-axis.

Given, P on reflection in a line l is a point P'.

Hence, The line l is the right bisector of the line segment PP'.

Given, Q is the image of P on reflection in y-axis and midpoint of PQ is invariant on reflection in x-axis.

From graph we get,

(i) x-axis will be the line joining the points P and Q.

(ii) The line perpendicular bisector of the line segment PQ is the y-axis.

(iii) The origin will be the mid point of the line segment PQ.

Given, the point (-3, 0) is the image of point (3, 0) and (-2, -3) is the image of point (2, -3) in the same line.

Considering graph,

(i) The mirror line is y-axis.

(ii) Since, the mirror line is y-axis. We know that,

Rule to find reflection of a point in y-axis :

1. Change the sign of abscissa i.e. x-coordinate.
2. Retain the ordinate i.e. y-coordinate.

Let C be the point (-3, -4).

∴ C' (3, -4) is the image on reflection of C (-3, -4) in y-axis.

Hence, the reflection of (-3, -4) in y-axis is (3, -4).

(i) The point A(0, 4), B(2, 2), C(5, 2) and D(4, 0), E(0, 0) are plotted on the graph below :

(ii) From graph, on reflecting B, C, D on y-axis we get,

B(2, 2) ⇒ B'(-2, 2)

C(5, 2) ⇒ C'(-5, 2)

D(4, 0) ⇒ D'(-4, 0).

(iii) Hence, the figure formed is a heptagon/septagon.

(i) The point A(-4, 0), B(-3, 2), C(0, 4), D(4, 1) and E(7, 3) are plotted on the graph below :

(ii) From graph, on reflecting B, C, D on x-axis we get,

B(-3, -2) ⇒ B'(-3, -2)

C(0, 4) ⇒ C'(0, -4)

D(4, 1) ⇒ D'(4, -1).

(iii) Points A, B, C, D, E, E', D', C', B' and A are joined in order.

(iv) The closed figure formed is a nonagon and resembles a fish.

(i) Reflected quadrilateral A'B'CD is shown in the graph below:

(ii) From graph we get,

The coordinates of A' and B' are (-2, 2) and (-2, -2) respectively.

(iii) From graph we get,

The two points which are invariant under the above reflection are C and D.

(iv) From graph we get,

The polygon(A'B'CD) formed is an isosceles trapezium.

Parts (i) and (ii) are solved in the graph below:

(iii) From graph we get,

The coordinates of B', C' and D' are (-3, 0), (-1, 0) and (-1, -5) respectively.

(iv) From graph we get,

The closed figure (ABCDD'C'B') formed is a heptagon/septagon (i.e., a polygon with seven sides).

The graph is given below:

(i) From graph we get,

The coordinate of Q is (-2, -4).

(ii) From graph we get,

The coordinate of R is (-2, 4).

(iii) From graph we get,

The image formed is right angled triangle.

(iv) We know that in the graph,

One block represents 1 unit.

Area of right angle triangle = $\frac{1}{2} \times$ base × height.

$$\begin{aligned}= \frac{1}{2} \times \text{PQ} \times \text{QR} \\[1em] = \frac{1}{2} \times 4 \text{ units } \times 8 \text{ units } \\[1em] = 16 \text{ sq. units}.\end{aligned}$$

Hence, the area of figure PQR is 16 sq. units.

(i) A' and B' are plotted on the graph below:

(ii) From graph we get,

The coordinates of A' and B' are (-6, -4) and (0, -4) respectively.

(iii) From graph we get,

ABA'B' formed is a parallelogram.

(iv) From graph we get,

By pythagoras theorem,

$$\begin{aligned}\text{AB}' = \sqrt{(\text{AB})^2 + (\text{BB}')^2} \\[1em] = \sqrt{6^2 + 8^2} \\[1em] = \sqrt{100} \\[1em] = 10 \text{ units }.\end{aligned}$$

Perimeter = Sum of all sides = AB' + B'A' + A'B + BA = 10 + 6 + 6 + 10 = 32 units.

Hence, the perimeter of the parallelogram is 32 units.

(i) Points A(4, 6) and B(1, 2) are plotted on the graph below:

(ii) From graph we get,

The coordinates of point A' = (4, -6).

(iii) From graph we get,

The coordinates of point B' = (7, 2).

(iii) From graph we get,

ABA'B' formed is a kite (or a Quadrilateral).

The graph is shown below:

(i) From graph we get,

The coordinates of A₁, B₁, C₁ are (-2, -3), (-4, -5) and (-7, -2) respectively.

(ii) From graph we get,

The coordinates of A₂, B₂, C₂ are (2, -3), (4, -5) and (7, -2) respectively.

(iii) From graph we get,

BCC₂B₂ formed is an isosceles trapezium.

We know that,

Area of trapezium = $\frac{1}{2} \times$ sum of parallel sides $\times$ distance between them

$$\begin{aligned}= \frac{1}{2} \times (\text{BB}_2 + \text{CC}_2) \times 3 \\[1em] = \frac{1}{2} \times (10 + 4) \times 3 \\[1em] = \frac{1}{2} \times 42 \\[1em] = 21 \text{ sq. units}\end{aligned}$$

The graph is shown below:

(i) From graph we get,

The coordinates of P' and O' are (3, -4) and (6, 0) respectively.

(ii) From graph we get,

Length of PP' = 8 units and OO' = 6 units.

(iii) OP = PO' = O'P' = P'O, as all are hypotenuse for same base and perpendicular.

∴ Perimeter of POP'O' is = 4 × OP

$$\begin{aligned}= 4 \times \sqrt{(OQ)^2 + (PQ)^2} \\[1em] = 4 \times \sqrt{(3)^2 + (4)^2} \\[1em] = 4 \times \sqrt{9 + 16} \\[1em] = 4 \times \sqrt{25} \\[1em] = 4 \times 5 \\[1em] = 20 \text{ units }.\end{aligned}$$

The perimeter of the quadrilateral POP'O' is 20 units.

The graph for the question is shown below:

(i) The point P(0, 5) is invariant in y-axis.

(ii) From graph we get,

The image of Q(-2, 4) on reflection in y-axis is Q'(2, 4).

(iii) Given, (0, k) on reflection in the origin is invariant.

From graph we get, (0, 0) on reflection in the origin is invariant. Comparing (0, k) with (0, 0) we get k = 0.

Hence, the value of k = 0.

(iv) From graph we get,

The coordinates of image of Q after reflection in origin and then in x-axis is (2, 4).

We know that,

On reflection in x-axis :

The sign of y-coordinate changes and x-coordinate remains the same.

So, on reflection in x-axis coordinates of P(-3, 5) becomes (-3, -5).

Hence, option 2 is the correct option.

We know that,

Rule to find reflection of a point in y-axis :

1. Change the sign of abscissa i.e. x-coordinate.
2. Retain the ordinate i.e. y-coordinate.

∴ Coordinates of point P(-2, 3) on reflection in y-axis is (2, 3).

Hence, Option 1 is the correct option.

Since, (-3, 2) is the image of the point P on reflection in x-axis so from graph P(-3, -2).

Hence, Option 2 is the correct option.

We know that the reflection of the point (x, y) in the line y = a is the point (x, -y + 2a).

∴ The image of the point P(1, -2) under reflection in the line y = -1 is the point (1, -(-2) + 2 × (-1)) i.e the point (1, 0).

Hence, Option 4 is the correct option.

We know that the reflection of the point (x, y) in the line x = a is the point (-x + 2a, y).

∴ The image of the point A(4, -1) under reflection in the line x = 2 is the point (-4 + 2 × 2, -1) i.e., the point P'(0, -1).

Hence, Option 1 is the correct option.

We know that,

Rules to find the reflection of a point in the origin :

1. Change the sign of abscissa i.e. x-coordinate.
2. Change the sign of ordinate i.e. y-coordinate.

∴ Point (3, 0) is the image on reflection.

Hence, Option 3 is the correct option.

Since, in (3, -2), y = -2, it means this point lies on the line. Hence, it a invariant point.

Hence, Option 2 is the correct option.

Reflection of a point (x, y) in the origin gives P'(-x, -y).

Thus, reflection of a P(2, 4) in the origin gives P'(-2, -4).

P'(-2, -4) lies in the third quadrant, where both x and y are negative.

So, assertion (A) is false.

Reflection in the origin preserves distance from the origin.

Both P and P' are at the same distance from the origin:

∴ PO = P'O

So, reason (R) is true.

Thus, Assertion (A) is false, but Reason (R) is true.

Hence, option 2 is the correct option.

A point is invariant with respect to a line if lies on the line.

So, if a point is invariant under reflection over y = 2, it must lie on the line y = 2.

So, reason (R) is true.

If P(-5, m) is invariant under reflection over y = 2, so it lies on the line y = 2, so m = 2.

So, assertion (A) is false.

Thus, Assertion (A) is false, but Reason (R) is true.

Hence, option 2 is the correct option.

We know that,

The y-coordinate of any point on x-axis equals to 0.

Substituting y = 0 into the equation 3x - 5y = 3, we get :

⇒ 3x - 5.0 = 3

⇒ 3x = 3

⇒ x = $\frac{3}{3}$ = 1

So the point of intersection is P(1, 0).

So, reason (R) is false.

A point is invariant with respect to a line if it lies on that line.

Since, P is the point of intersection of line 3x - 5y = 3 and x-axis, so it lies on line 3x - 5y = 3.

So, assertion (A) is true.

Thus, Assertion (A) is true, but Reason (R) is false.

Hence, option 1 is the correct option.

A point (x, y) reflected in the y-axis becomes (−x, y).

This changes the sign of the x-coordinate but preserves the y-coordinate.

As a result, distances and angles are preserved, and so is area.

Reflection over an axis is a rigid transformation (i.e., it preserves size and shape)

Thus, the area remains unchanged.

So, assertion (A) is true.

Since all corresponding sides and angles remain equal under reflection.

Thus, the two triangles are congruent.

So, reason (R) is true and reason (R) is the correct explanation of assertion (A).

Thus, both Assertion (A) and Reason (R) are correct, and Reason (R) is the correct reason for Assertion (A).

Hence, option 3 is the correct option.

We know that,

Point invariant to both x-axis and y-axis is their point of intersection i.e. the origin.

So, assertion (A) is true.

A point is invariant to itself.

So, reason (R) is true.

Both Assertion (A) and Reason (R) are correct, and Reason (R) is the correct reason for Assertion (A).

Hence, Option 3 is the correct option.

We know that,

Rule to find reflection of a point in x-axis :

1. Retain the abscissa i.e. x-coordinate.
2. Change the sign of ordinate i.e. y-coordinate.

∴ Coordinates of point P(4, -7) on reflection in x-axis is P'(4, 7).

We know that,

Rule to find reflection of a point in y-axis :

1. Change the sign of abscissa i.e. x-coordinate.
2. Retain the ordinate i.e. y-coordinate.

∴ Coordinates of point P'(4, 7) on reflection in y-axis is P''(-4, 7).

The single transformation that maps P(4, -7) onto P''(-4, 7) is reflection in the origin.

We know that,

Rules to find the reflection of a point in the origin :

1. Change the sign of abscissa i.e. x-coordinate.
2. Change the sign of ordinate i.e. y-coordinate.

∴ Coordinates of point P(a, b) on reflection in origin is (-a, -b).

We know that,

Rule to find reflection of a point in y-axis :

1. Change the sign of abscissa i.e. x-coordinate.
2. Retain the ordinate i.e. y-coordinate.

∴ Coordinates of point (-a, -b) on reflection in y-axis is (a, -b).

According to question after reflection in y-axis the point is P'(3, -4).

∴ (3, -4) = (a, -b) or, a = 3 and b = 4.

Hence, the value of a = 3, b = 4.

We know that,

Rule to find reflection of a point in x-axis :

1. Retain the abscissa i.e. x-coordinate.
2. Change the sign of ordinate i.e. y-coordinate.

∴ Coordinates of point P(a, b) on reflection in x-axis is P'(a, -b).

According to question after reflection in x-axis P becomes (-2, c).

∴ (a, -b) = (-2, c) or,
⇒ a = -2
⇒ -b = c or b = -c.

We know that,

Rules to find the reflection of a point in the origin :

1. Change the sign of abscissa i.e. x-coordinate.
2. Change the sign of ordinate i.e. y-coordinate.

∴ Coordinates of point P(a, b) on reflection in origin is (-a, -b).

According to question after reflection in origin P becomes (d, 5).

∴ (-a, -b) = (d, 5) or,
⇒ -a = d or d = -(-2) = 2.
⇒ -b = 5 or b = -5.

Since, b = -c = -5, so, c = 5.

Hence, the value of a = -2, b = -5, c = 5 and d = 2.

The graph for this problem is shown below:

First the triangle is reflected in y-axis.

We know that,

Rule to find reflection of a point in y-axis :

1. Change the sign of abscissa i.e. x-coordinate.
2. Retain the ordinate i.e. y-coordinate.

∴ Coordinates of
⇒ A(4, -1) on reflection in y-axis becomes A'(-4, -1).
⇒ B(0, 7) on reflection in y-axis becomes B'(0, 7).
⇒ C(-2, 5) on reflection in y-axis becomes C'(2, 5).

Now the triangle is reflected in origin.

We know that,

Rules to find the reflection of a point in the origin :

1. Change the sign of abscissa i.e. x-coordinate.
2. Change the sign of ordinate i.e. y-coordinate.

∴ Coordinates of
⇒ A'(-4, -1) on reflection in y-axis becomes A''(4, 1).
⇒ B'(0, 7) on reflection in y-axis becomes B''(0, -7).
⇒ C'(2, 5) on reflection in y-axis becomes C''(-2, -5).

Hence, the coordinates of the final images of the vertices are (4, 1), (0, -7) and (-2, -5) respectively.

The graph for this problem is shown below:

First the parallelogram is reflected in y-axis.

We know that,

Rule to find reflection of a point in y-axis :

1. Change the sign of abscissa i.e. x-coordinate.
2. Retain the ordinate i.e. y-coordinate.

∴ Coordinates of
⇒ A(4, -11) on reflection in y-axis becomes A'(-4, -11).
⇒ B(5, 3) on reflection in y-axis becomes B'(-5, 3).
⇒ C(2, 15) on reflection in y-axis becomes C'(-2, 15).
⇒ D(1, 1) on reflection in y-axis becomes D'(-1, 1).

We know that,

Rules to find the reflection of a point in the origin :

1. Change the sign of abscissa i.e. x-coordinate.
2. Change the sign of ordinate i.e. y-coordinate.

∴ Coordinates of
⇒ A'(-4, -11) on reflection in origin becomes A''(4, 11).
⇒ B'(-5, 3) on reflection in origin becomes B''(5, -3).
⇒ C'(-2, 15) on reflection in origin becomes C''(2, -15).
⇒ D'(-1, 1) on reflection in origin becomes D''(1, -1)

From graph we can see that A"B"C"D" is also a parallelogram.

The single transformation that marks the following changes i.e.
A(4, -11) ⇒ A"(4, 11)
B(5, 3) ⇒ B"(5, -3)
C(2, 15) ⇒ C"(2, -15)
D(1, 1) ⇒ D"(1, -1), is reflection in x-axis.

Hence,

1. The coordinates of the final images of the vertices are (4, 11), (5, -3), (2, -15) and (1, -1) respectively.
2. These new images still form parallelogram.
3. The single transformation from ABCD ⇒ A"B"C"D" can be achieved by reflection in x-axis.

(i) The point A(0, 4), B(2, 3), C(1, 1) and D(2, 0) are plotted on the graph below:

(ii) From graph, on reflecting B, C, D on y-axis we get,

     B(2, 3) ⇒ B'(-2, 3)
     C(1, 1) ⇒ C'(-1, 1)
     D(2, 0) ⇒ D'(-2, 0).

(iii) From graph we see that the figure is divided into two symmetrical parts by y-axis. Hence, the equation of line of symmetry is x = 0.

The graph showing the given information is drawn below:

(i) Since, on reflection in the origin A becomes A'(-3, -4) and B becomes B'(0, -5) thus,

A = (3, 4) and B = (0, 5).

(iii) The quadrilateral ABA'B' represents a rectangle.

(iv) From graph we get,

The coordinates of A" is (3, -4).

(v) From graph we get,

The coordinates of B" is (0, 5).

(a) From graph,

Coordinates of A = (3, 3), B = (-2, 1), C = (3, -1) and D = (0, 1).

(b) From graph,

C is the image of point A in the line y = 1.

Hence, the line is BD and its equation is y = 1.

(c) Since, point D lies on y-axis, thus it is invariant under reflection in y-axis.

Hence, coordinates of point D on reflection in y-axis is (0, 1).

(d) A point whose image is the point itself is called invariant point.

(e) On joining the points A, B, C, D and A in order the figure formed is a concave quadrilateral or arrowhead.