Question

Which series of transformations will not map figure H onto itself

Answer 1

Answer:

D

Step-by-step explanation:

Given a square with vertices at points (2,1), (1,2), (2,3) and (3,2).

Consider option A.

1st transformation $(x+0,y-2)$ will map vertices of the square into points

• $(2,1)\rightarrow (2,-1);$
• $(1,2)\rightarrow (1,0);$
• $(2,3)\rightarrow (2,1);$
• $(3,2)\rightarrow (3,0).$

2nd transformation = reflection over y = 1 has the rule (x,2-y). So,

• $(2,-1)\rightarrow (2,3);$
• $(1,0)\rightarrow (1,2);$
• $(2,1)\rightarrow (2,1);$
• $(3,0)\rightarrow (3,2)$

These points are exactly the vertices of the initial square.

Consider option B.

1st transformation $(x+2,y-0)$ will map vertices of the square into points

• $(2,1)\rightarrow (4,1);$
• $(1,2)\rightarrow (3,2);$
• $(2,3)\rightarrow (4,3);$
• $(3,2)\rightarrow (5,2).$

2nd transformation = reflection over x = 3 has the rule (6-x,y). So,

• $(4,1)\rightarrow (2,1);$
• $(3,2)\rightarrow (3,2);$
• $(4,3)\rightarrow (2,3);$
• $(5,2)\rightarrow (1,2)$

These points are exactly the vertices of the initial square.

Consider option C.

1st transformation $(x+3,y+3)$ will map vertices of the square into points

• $(2,1)\rightarrow (5,4);$
• $(1,2)\rightarrow (4,5);$
• $(2,3)\rightarrow (5,6);$
• $(3,2)\rightarrow (6,5).$

2nd transformation = reflection over y = -x + 7 will map vertices into points

• $(5,4)\rightarrow (3,2);$
• $(4,5)\rightarrow (2,3);$
• $(5,6)\rightarrow (1,2);$
• $(6,5)\rightarrow (2,1)$

These points are exactly the vertices of the initial square.

Consider option D.

1st transformation $(x-3,y-3)$ will map vertices of the square into points

• $(2,1)\rightarrow (-1,-2);$
• $(1,2)\rightarrow (-2,-1);$
• $(2,3)\rightarrow (-1,0);$
• $(3,2)\rightarrow (0,-1).$

2nd transformation = reflection over y = -x + 2 will map vertices into points

• $(-1,-2)\rightarrow (4,3);$
• $(-2,-1)\rightarrow (3,4);$
• $(-1,0)\rightarrow (2,3);$
• $(0,-1)\rightarrow (3,2)$

These points are not the vertices of the initial square.