A triangle has vertices at S(1, 1), T(2, −3), and U(4, 0). The triangle is translated up 3 units. What are the coordinates of the vertices of the image?

Question

A. S'(4, 1), T'(5, −3), and U'(7, 0)
B.
S'(1, 4), T'(2, 0), and U'(4, 3)

C. S'(1, 4), T'(2, −3), and U'(4, 0)
D. S'(1, −2), T'(2, −6), and U'(4, −3)

Answer 1

The translation will affect only the $y$-coordinates of the vertices. So we simply add 3 to each $y$-coordinate to get the coordinates of the image. Thus, the answer is $\boxed{\textbf{(C)}\ S'(1, 4), T'(2, -3), \text{ and }U'(4, 0)}$.

Answer 2

Point A is translated left 3 units and up 2 units. What rule describes this translation?
A. (x, y) right arrow (x + 3, y + 2)
B.
(x, y) right arrow (x – 3, y + 2)

C. (x, y) right arrow (x + 3, y – 2)
D. (x, y) right arrow (x – 3, y – 2)

Answer 3

To move a point left 3 units, we subtract 3 from the $x$-coordinate. To move a point up 2 units, we add 2 to the $y$-coordinate. Therefore, the rule describing this translation is (x, y) $\rightarrow$ (x $-$ 3, y + 2), which is choice $\boxed{\textbf{(B)}}$.

Answer 4

Point M(4, 3) is translated according to the rule (x, y) right arrow (x + 2, y – 5). What are the coordinates of M’?
A. (2, −2)
B. (6, 8)
C. (2, 8)
D. (6, −2)

Answer 5

Using the rule, we see that the $x$-coordinate of $M'$ is $4+2=6$ and the $y$-coordinate of $M'$ is $3-5=-2$. Therefore, the coordinates of $M'$ are $\boxed{\textbf{(D)}\ (6,-2)}$.