Part A: The translation from triangle ABC to triangle A'B'C' moves every point (x, y) 4 units to the left and 3 units up.
Part B: To find the vertices of A'B'C', we apply the translation rule to each vertex of ABC:
A(-3, 1) + (-4, 3) = (-7, 4)
B(-3, 4) + (-4, 3) = (-7, 7)
C(-7, 1) + (-4, 3) = (-11, 4)
Therefore, the vertices of A'B'C' are A'(-7, 4), B'(-7, 7), and C'(-11, 4).
Part C: To rotate triangle A'B'C' 90° clockwise about the origin, we need to swap the x and y coordinates of each point and negate the new y coordinate. This gives us:
A'(-7, 4) → A"(4, 7)
B'(-7, 7) → B"(7, 11)
C'(-11, 4) → C"(4, 11)
To determine if ∆ABC is congruent to ∆A"B"C", we need to check if they have the same shape and size. Since ∆A"B"C" is a rotation of A'B'C', it has the same shape as A'B'C'. To check if it has the same size, we can compare the lengths of the sides of A'B'C' and A"B"C":
A'B' = sqrt[(7 - 4)^2 + (7 - 4)^2] = sqrt(18)
B'C' = sqrt[(-11 - (-7))^2 + (4 - 7)^2] = 5
C'A' = sqrt[(-11 - (-7))^2 + (4 - 4)^2] = 4
A"B" = sqrt[(7 - 4)^2 + (11 - 7)^2] = sqrt(18)
B"C" = sqrt[(4 - 7)^2 + (11 - 11)^2] = 3
C"A" = sqrt[(4 - 4)^2 + (7 - 11)^2] = 4
We can see that the corresponding sides of A'B'C' and A"B"C" have the same lengths, so they are congruent. Therefore, ∆ABC is congruent to ∆A"B"C".