FROM THE GIVEN POINTS OF REFLECTION FIND THE RULE OF REFLECTION

Problem 1 :

L(0, 1), K(0, 2), J(3, 3), I(5, 1)

to

L'(0, -1), K'(0, -2), J'(3, -3), I'(5, -1)

Solution :

By comparing the corresponding coordinates

L (0, 1) ==> L' (0, -1)

K (0, 2) ==> K' (0, -2)

J (3, 3) ==> J' (3, -3)

I (5, 1) ==>  I' (5, -1)

There is no change in x-coordinate.

(x, y) ==> (x, -y)

So,

Reflection across x-axis.

Problem 2 :

H (-3, -5), I (-5, -2), J (-1, -1), K (0, -4)

to

H' (-3, 5), I' (-5, 2), J' (-1, 1), K' (0, 4)

Solution :

By comparing the corresponding coordinates

H (-3, -5) ==> H' (-3, 5)

 I (-5, -2) ==> I' (-5, 2)

J (-1, -1) ==> J' (-1, 1)

K (0, -4) ==>  K' (0, 4)

There is no change in x-coordinate. (x, y) ==> (x, -y)

So,

Reflection across x-axis.

Problem 3 :

P (-4, -3), Q (-1, 1), R (0, -4)

to

P'(-4, 3), Q'(-1, -1), R'(0, 4)

Solution :

By comparing the corresponding coordinates

P (-4, -3) ==> P' (-4, 3)

Q (-1, 1) ==> Q' (-1, -1)

R (0, -4) ==> R' (0, 4)

There is no change in x-coordinate. (x, y) ==> (x, -y)

So,

Reflection across x-axis.

Problem 4 :

E (3, 1), F (3, 4), G (5, 1)

to

E' (-1, -3), F' (-4, -3), G' (-1, -5)

Solution :

By comparing the corresponding coordinates

E (3, 1) ==> E' (-1, -3)

F (3, 4) ==> F' (-4, -3)

G (5, 1) ==> G' (-1, -5)

There is no change in x-coordinate. (x, y) ==> (-y, -x)

So,

Reflection across y = -x

Problem 5 :

The reflection of a point P in the y-axis is P' (– 4, –2). The coordinates of point P are:

(a) (– 4, 2)     (b) (4, 2)      (c) (4, –2)      (d) (–2, 4) 

Solution :

Given that the point is P' (– 4, –2)

Reflection across y-axis, (x, y) ==> (-x, y)

P'(-4, -2) ==> (4, -2)

So, option c is correct.

Problem 6 :

Point (0, –2) is invariant under reflection in:

(a) x-axis      (b) y-axis     (c) origin     (d) none

Solution :

Given that the point is (0, -2)

Reflection across x-axis, (x, y) ==> (x, -y)

(0, -2) ==> (0, 2)

Reflection across y-axis, (x, y) ==> (-x, y)

(0, -2) ==> (0, -2)

There is no change after reflection across y-axis, then option b is correct.

Problem 7 :

Point (5, 0) is invariant under reflection in :

(a) x-axis     (b) y-axis     (c) origin     (d) none

Solution :

Given that the point is (5, 0)

Reflection across x-axis, (x, y) ==> (x, -y)

(5, 0) ==> (0, -5)

Reflection across y-axis, (x, y) ==> (-x, y)

(5, 0) ==> (-5, 0)

Reflection across y-axis, (x, y) ==> (-x, -y)

(5, 0) ==> (-5, 0)

So, option d is correct.

Problem 8 :

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.