REFLECTION OVER VERTICAL AND HORIZONTAL LINES

Problem 1 :

Reflection across y = -1

Solution :

By observing the figure given above,

W (-4, -1), X (-5, 3), Y (-3, 5) and Z (-1, 1)

How to get W’?

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

W and W’ are the same.

How to get X’?

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

X (-5, 3) ==> X’ (-5, -5)

How to get Y’?

To reach Y from y = -1, we move up 6 unit. So, to reach the reflection of Y, that is Y’ we have to move down 6 unit.

Y (-3, 5) ==> Y’ (-3, -7)

How to get Z’?

To reach Z from y = -1, we move up 2 unit. So, to reach the reflection of Z, that is Z’ we have to move down 2 unit.

Z (-1, 1) ==> Z’ (-1, -3)

Problem 2 :

Reflection across y = -1

C (2, 0), D (5, 3), E (4, -1)

Solution :

Plotting the points and creating the sides.

How to get C’?

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

C (2, 0) ==> C’ (2, -2)

How to get D’?

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

D (5, 3) ==> D’ (5, -5)

How to get E’?

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

E and E’ are the same.

Problem 3 :

Reflection across y = -1

Q (-3, -4), R (-4, 0), S (-2, 0)

Solution :

Plotting the points and creating the sides.

How to get Q’?

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

Q (-3, -4) ==> Q’ (-3, 2)

How to get R’?

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

R (-4, 0) ==> R’ (-4, -2)

How to get S’?

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

S (-2, 0) ==> S’ (-2, -2)

Problem 4 :

Reflection across x = -2

Q (-4, 1), R (-3, 5), S (0, 3), T (-3,-1)

Solution :

Plotting the points and creating the sides.

How to get Q’?

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

Q (-4, 1) ==> Q’ (0, 1)

How to get R’?

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

R (-3, 5) ==> R’ (-1, 5)

How to get S’?

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

S (0, 3) ==> S’ (-4, 3)

How to get T’?

To reach T from x = -2, we move left 1 unit. So, to reach the reflection of T, that is T’ we have to move right 1 unit.

T (-3, -1) ==> T’ (-1, -1)

Problem 5 :

(a) Point P (a, b) is reflected in x-axis to P′ (4, – 3). Write down the values of a and b.

(b) P′′ is the image of P when reflected in y-axis. Write down the co-ordinates of P′′.

(c) Name a single transformation that maps P′ to P′′.

Solution :

(a) Rx (a, b) = (a, –b) ⇒ (a, –b) = (4, –3) ⇒ a = 4 and b = 3

(b) Ry (4, 3) = (–4, 3) Hence, P'' has coordinates (– 4, 3)

(c) P' → P" means (4, –3) → (– 4, 3)

Also, Ro(4, –3), = (–4, 3)

Problem 6 :

The point A (–6, 4) on reflection in y-axis is mapped as A′. Point A′ on reflection in the origin is mapped as A′′.

(a) Find the co-ordinates of A′.

(b) Find the co-ordinates of A′′.

(c) Write down a single transformation that maps A to A′′.

Solution :

(a) Ry (–6, 4) = (6, 4)

(b) Ro (6, 4) = (– 6, – 4)

(c) A → A" means (– 6, 4) → (– 6, – 4)

Also, Rx (– 6, 4) = (– 6, – 4).

So, reflection in the x-axis is the required single transformation.