Question

14a. which rotation about the origin is (x, y) → (y, -x)? a. 270° ccw b. 180° c. 90° ccw d. reflection 14b. image of (1, 6) after 270° ccw rotation? a. (-6, 1) b. (6, -1) c. (-1, -6) d. (1, -6) 14c. image of (-2, -5) after 270° ccw rotation? a. (5, -2) b. (-5, 2) c. (2, 5) d. (-2, -5) 14d. image of (0, -8) after 270° ccw rotation? a. (8, 0) b. (-8, 0) c. (0, 8) d. (0, -8)

Explanation:

Step1: Recall rotation rules

The rule for a $270^{\circ}$ counter - clockwise (CCW) rotation about the origin is $(x,y)\to(y, - x)$.

Step2: Solve 14a

For the transformation $(x,y)\to(y, - x)$, the rotation is $270^{\circ}$ CCW. So the answer for 14a is A. $270^{\circ}$ CCW.

Step3: Solve 14b

Given the point $(1,6)$, using the $270^{\circ}$ CCW rotation rule $(x,y)\to(y, - x)$, we substitute $x = 1$ and $y = 6$. The image is $(6,-1)$. So the answer for 14b is B. $(6,-1)$.

Step4: Solve 14c

Given the point $(-2,-5)$, using the $270^{\circ}$ CCW rotation rule $(x,y)\to(y, - x)$, we substitute $x=-2$ and $y = - 5$. The image is $(-5,2)$. So the answer for 14c is B. $(-5,2)$.

Step5: Solve 14d

Given the point $(0,-8)$, using the $270^{\circ}$ CCW rotation rule $(x,y)\to(y, - x)$, we substitute $x = 0$ and $y=-8$. The image is $(-8,0)$. So the answer for 14d is B. $(-8,0)$.

Answer:

14a. A. $270^{\circ}$ CCW
14b. B. $(6,-1)$
14c. B. $(-5,2)$
14d. B. $(-8,0)$