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?
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)
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 $\boxed{\textbf{(C)}\ S'(1, 4), T'(2, -3), \text{ and }U'(4, 0)}$.
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)
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 $\boxed{\textbf{(B)}}$.
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)
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 $\boxed{\textbf{(D)}\ (6,-2)}$.
To find the coordinates of the vertices of the triangle after the translation, you need to add the translation vector to each of the original coordinates.
The translation vector is (0, 3) since the triangle is translated up 3 units.
To get the coordinates of the image of vertex S, you add the translation vector to the original coordinates of S(1, 1):
S'(1 + 0, 1 + 3) = S'(1, 4)
To get the coordinates of the image of vertex T, you add the translation vector to the original coordinates of T(2, -3):
T'(2 + 0, -3 + 3) = T'(2, 0)
To get the coordinates of the image of vertex U, you add the translation vector to the original coordinates of U(4, 0):
U'(4 + 0, 0 + 3) = U'(4, 3)
Therefore, the coordinates of the vertices of the image triangle are S'(1, 4), T'(2, 0), and U'(4, 3).
The correct answer is B. S'(1, 4), T'(2, 0), and U'(4, 3).