Answer:
1. ΔA'B'C' = A' = (1, 1), B'(-1, 1), and C'(1, 4)
2. ΔA''B''C'' = A''(-7 1), B''(-15 1), and C''(-7, 4)
Step-by-step explanation:
1. The coordinates of the triangle ΔABC are; A(3, 1), B(5, 1), and C(3, 4)
The reflection of a line across the line x = 2, is given as follows
The distance between the x-coordinate of the preimage and the line of reflection which is parallel to the x-axis = The distance of the x-coordinate of the preimage from the line of reflection but opposite in sign
The distance from A(3, 1) from x = 2 is 3 - 2 = 1, therefore, the x-coordinate of the image, A' = 2 - 1 = 1, therefore, we have;
The coordinate of A' = (1, 1)
Similarly, we have; B(5, 1) (reflection across x = 2) → B'(-1, 1)
C(3, 4) (reflection across x = 2) → C'(1, 4)
Therefore when we reflect ABC across the line x = 2, we get ΔA'B'C', with A' = (1, 1), B'(-1, 1), and C'(1, 4)
2. Reflection of A'B'C' across the line x = -3, gives;
A'(1, 1) (reflection across x = -3) → A''(-7 1)
B'(-1, 1) (reflection across x = -3) → B''(-15 1)
C'(1, 4) (reflection across x = -3) → C''(-7, 4)
The coordinates of the reflection of ΔA'B'C' across the line x = -3 is ΔA''B''C'' = A''(-7 1), B''(-15 1), and C''(-7, 4)
