PROBLEMS ON REFLECTION OVER HORIZONTAL AND VERTICAL LINE

Problem 1 :

Reflection across y = -1

R (-3, -5), N (-4, 0), V (-2, -1) E (0, -4)

Solution :

Plotting the points and creating the sides.

How to get R’?

To reach R from y = -1, we move down 4 unit. So, to reach the reflection of R, that is R’ we have to move up 4 unit.

R (-3, -5) ==> R’ (-3, 3)

How to get N’?

To reach N from y = -1, we move up 1 unit. So, to reach the reflection of N, that is N’ we have to move down 1 unit.

N (-4, 0) ==> N’ (-4, -2)

How to get V’?

Distance between y = -1 and V is 0 unit. So, the same point is V’.

V and V’ are the same.

How to get E’?

To reach E from y = -1, we move down 3 unit. So, to reach the reflection of E, that is E’ we have to move up 3 unit.

E (0, -4) ==> E’ (0, 2)

Problem 2 :

Reflection across x = 3

F (2, 2), W (2, 5), K (3, 2)

Solution :

Plotting the points and creating the sides.

How to get F’?

To reach F from x = 3, we move left 1 unit. So, to reach the reflection of F, that is F’ we have to move right 1 unit.

F (2, 2) ==> F’ (4, 2)

How to get K’?

Distance between x = 3 a nd K is 0 unit. So, the same point is K’.

K and K’ are the same.

How to get W’?

To reach W from x = 3, we move left 1 unit. So, to reach the reflection of W, that is W’ we have to move right 1 unit.

W (2, 5) ==> W’ (4, 5)

Problem 3 :

Reflection across x = -1

V (-3, -1), Z (-3, 2), G (-1, 3), M (1, 1)

Solution :

Plotting the points and creating the sides.

How to get V’?

To reach V from x = -1, we move left 2 unit. So, to reach the reflection of V, that is V’ we have to move right 2 unit.

V (-3, -1) ==> V’ (1, -1)

How to get Z’?

To reach Z from x = -1, we move left 2 unit. So, to reach the reflection of Z, that is Z’ we have to move right 2 unit.

Z (-3, 2) ==> Z’ (1, 2)

How to get G’?

Distance between x = -1 and G is 0 unit. So, the same point is G’.

G and G’ are the same.

How to get M’?

To reach M from x = -1, we move right 2 unit. So, to reach the reflection of M, that is M’ we have to move left 2 unit.

M (1, 1) ==> M’ (-3, 1)

Problem 4 :

Reflection across x = 3

Solution :

By observing the triangle given above,

A (1, 1), B (4, 4) and C (3, 0)

How to get A’?

To reach A from x = 3, we move left 2 unit. So, to reach the reflection of A, that is A’ we have to move right 2 unit.

A (1, 1) ==> A’ (5, 1)

How to get B’?

To reach B from x = 3, we move right 1 unit. So, to reach the reflection of B, that is B’ we have to move left 1 unit.

B (4, 4) ==> B’ (2, 4)

How to get C’?

Distance between x = 3 and C is 0 unit. So, the same point is C’.

C and C’ are the same.

Problem 5 :

The points (3, 0) and (–1, 0) are invariant points under reflection in the line L1, while the points (0, –3) and (0, 1) are invariant points under reflection in the line L2.

(a) Name the lines L1 and L2

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

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

(d) State or describe a single transformation that maps P′ onto P′′.

Solution :

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

Rx     P(3, 4) = P′(3, – 4)

Rx      Q(–5, –2) = Q′ (–5, 2)

So, L1 represents x-axis, L2 represents y-axis.

(b) Rx    P(3, 4) = P′(3, –4)

Rx       Q(–5, –2) = Q′(–5, 2)

Images of the points P(3, 4) and Q(–5, –2) on reflection in L1 are P′ (3, – 4) and Q′ (–5, 2).

(c)

Ry P(3, 4) = P′′ (–3, 4)

Ry Q(–5, –2) = Q′′ (–5, –2)

Images of the points P(3, 4) and Q(–5, –2) on reflection in L2 are P′′ (– 3, 4) and Q′′ (5, – 2).

(d) P′ → P′′ means (3, – 4) → (–3, 4) Also,

Ro   (3, –4) = (–3, 4)

So, required transformation is Ro.