Reflect triangle ABC across the line x = 2. Then the reflect (image) A' B' C' across the line x = -3.

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)

yes.. it's proven to be 36.