A triangle has vertices at [tex]\(B (-3, 0)\)[/tex], [tex]\(C (2, -1)\)[/tex], and [tex]\(D (-1, 2)\)[/tex]. Which transformation would produce an image with vertices [tex]\(B' (1, -2)\)[/tex], [tex]\(C' (0, 3)\)[/tex], and [tex]\(D' (3, 0)\)[/tex]?

A. [tex]\((x, y) \rightarrow (x+1, y+1) \rightarrow (y, x)\)[/tex]

B. [tex]\((x, y) \rightarrow (x+1, y+1) \rightarrow (-x, y)\)[/tex]

C. [tex]\((x, y) \rightarrow (x, -y) \rightarrow (x+2, y+2)\)[/tex]

D. [tex]\((x, y) \rightarrow (-x, y) \rightarrow (x+2, y+2)\)[/tex]



Answer :

To find out which transformation maps the original triangle vertices [tex]\( B(-3,0), C(2,-1), D(-1,2) \)[/tex] to the new triangle vertices [tex]\( B '(1,-2), C '(0,3), D '(3,0) \)[/tex], we need to examine each proposed transformation step-by-step.

Let's analyze each of the given transformations to see which one correctly maps the original vertices to the transformed vertices.

### Transformation 1: [tex]\((x, y) \to (x+1, y+1) \to (y, x)\)[/tex]

1. Apply the first part of the transformation [tex]\((x + 1, y + 1)\)[/tex]:

- For [tex]\( B(-3,0) \)[/tex]: [tex]\((x+1, y+1) = (-3+1, 0+1) = (-2, 1)\)[/tex]
- For [tex]\( C(2,-1) \)[/tex]: [tex]\((x+1, y+1) = (2+1, -1+1) = (3, 0)\)[/tex]
- For [tex]\( D(-1,2) \)[/tex]: [tex]\((x+1, y+1) = (-1+1, 2+1) = (0, 3)\)[/tex]

2. Apply the second part of the transformation [tex]\((y, x)\)[/tex]:

- For [tex]\( (-2,1) \)[/tex]: [tex]\((y, x) = (1, -2)\)[/tex]
- For [tex]\( (3,0) \)[/tex]: [tex]\((y, x) = (0, 3)\)[/tex]
- For [tex]\( (0,3) \)[/tex]: [tex]\((y, x) = (3, 0)\)[/tex]

After both steps, the transformed points are: [tex]\( B'(1, -2), C'(0, 3), D'(3, 0) \)[/tex].

This matches the target points exactly: [tex]\( B'(1,-2), C'(0,3), D'(3,0) \)[/tex].

Thus, this transformation is correctly [tex]\((x, y)-(x+1, y+1) \to (y, x)\)[/tex].

Result: This is the correct transformation. No need to check the other transformations.