Question

Triangle ABC is reflected over the line y=2 to obtain triangle A'B'C'. Then triangle A'B'C' is translated by the directed line segment from (0,0) to (3,2) to obtain triangle A"B"C".

Answer 1

Answer: To obtain triangle A'B'C' from triangle ABC by reflection over the line y = 2, we reflect each point of the triangle over the line. This gives us:

A' = (A_x, 4 - A_y)

B' = (B_x, 4 - B_y)

C' = (C_x, 4 - C_y)

To obtain triangle A"B"C" from triangle A'B'C' by translation by the directed line segment from (0,0) to (3,2), we add the vector (3,2) to each point of the triangle. This gives us:

A" = (A'_x + 3, A'_y + 2) = (A_x + 3, 6 - A_y)

B" = (B'_x + 3, B'_y + 2) = (B_x + 3, 6 - B_y)

C" = (C'_x + 3, C'_y + 2) = (C_x + 3, 6 - C_y)

Therefore, the vertices of triangle A"B"C" are (A_x + 3, 6 - A_y), (B_x + 3, 6 - B_y), and (C_x + 3, 6 - C_y).

Explanation: