Question

question 1 (multiple choice worth 1 points) (02.01r mc) a triangle has vertices at b(-3,0), c(2,-1), d(-1,2). which transformation would produce an image with vertices b(1,-2), c(0,3), d(3,0)? (x,y)→(-x,y)→(x + 2,y + 2) (x,y)→(x,-y)→(x + 2,y + 2) (x,y)→(x + 1,y + 1)→(-x,y) (x,y)→(x + 1,y + 1)→(y,x)

Explanation:

Step1: Analyze transformation for B

Original B is (-3,0), new B' is (1,-2). Check each option. For option (x,y)→(x + 2,y + 2), when x=-3,y = 0, we get (-3+2,0 + 2)=(-1,2) which is wrong. For option (x,y)→(x - y), when x=-3,y = 0, we get (-3-0)=-3 (not in 2 - D form as required). For option (x,y)→(x+1,y + 1), when x=-3,y = 0, we get (-3 + 1,0+1)=(-2,1) which is wrong. Let's consider the general form of transformation (x,y)→(x + a,y + b). We want -3+a = 1 and 0 + b=-2, so a = 4 and b=-2.

Step2: Analyze for C

Original C is (2,-1), new C' is (0,3). For (x,y)→(x+1,y + 1), (2+1,-1 + 1)=(3,0) which is wrong. For (x,y)→(x - y), not in correct form. For (x,y)→(x + 2,y + 2), (2+2,-1+2)=(4,1) which is wrong.

Step3: Analyze for D

Original D is (-1,2), new D' is (3,0).
Let's check the transformation (x,y)→(x+4,y - 2). For B: (-3+4,0 - 2)=(1,-2). For C: (2+4,-1 - 2)=(6,-3) (wrong).
Let's check (x,y)→(x - y)→(x+2,y + 2). First for B: (-3,0)→(-3-0)=-3 (wrong form).
Let's check (x,y)→(x+1,y + 1)→(-x,y). For B: (-3,0)→(-3 + 1,0+1)=(-2,1)→(2,1) (wrong).
Let's check (x,y)→(x+1,y + 1)→(y,x). For B: (-3,0)→(-3 + 1,0+1)=(-2,1)→(1,-2). For C: (2,-1)→(2 + 1,-1+1)=(3,0)→(0,3). For D: (-1,2)→(-1 + 1,2+1)=(0,3)→(3,0).

Answer:

(x,y)→(x + 1,y + 1)→(y,x)