Question

Triangle ABC is located at A (-2, 2), B (-2, 4), and C (0, 0). The triangle is then transformed using the rule (x+3, y-2) to form the image A'B'C'. What are the new coordinates of A', B', and C'? Describe what characteristics you would find if the corresponding vertices were connected with line segments

Answer 1

$A'=(-2+3,2-2)=(1,0)\\ B'=(-2+3,4-2)=(1,2)\\ C'=(0+3,0-2)=(3,-2)$

Answer 2

In this problem, we have three vertices, namely:

$A(-2,2) \\ \\ B(-2,4) \\ \\ C(0,0)$

Our transformation rule is $(x+3,y-2)$ and this rule forms the image A'B'C'. So:

1. What are the new coordinates of A', B', and C'?

To find these new coordinates we have:

$A' \rightarrow (-2+3,2-2) \rightarrow \boxed{A'(1,0)} \\ \\ B' \rightarrow (-2+3,4-2) \rightarrow \boxed{B'(1,2)} \\ \\ C' \rightarrow (0+3,0-2) \rightarrow \boxed{C'(3,-2)}$

2. Describe what characteristics you would find if the corresponding vertices were connected with line segments.

Taking a look in the First Figure, we can see the two triangles. The red one is the triangle ABC and the blue one is A'B'C'. In the Second Figure, the corresponding vertices have been connected with line segments, as indicated the green lines. The characteristics of those three lines are two, namely:

• They are parallel.
• They have the same length.