Question

10
rotate
270°
clockwise
about the
origin
i (-4, -3)
purple
i (3, -4)
dark blue
i (4, -3)
pink

Explanation:

Step1: Recall rotation rule

270° clockwise about origin: $(x,y)→(y,-x)$

Step2: Identify point I's coordinates

From graph, point I is at $(-3,4)$

Step3: Apply rotation rule

Substitute $x=-3,y=4$: $(4,-(-3))=(4,3)$? Wait, correction: 270° clockwise = 90° counterclockwise? No, correct rule: 270° clockwise: $(x,y)→(y,-x)$. If I was $(-3,4)$: $(4, 3)$? Wait, maybe I misread. Wait, if original I is $(-3,4)$, 270 clockwise: $(4, 3)$ no—wait, standard rule: 90° clockwise: (x,y)→(y,-x); 180°: (-x,-y); 270° clockwise: (-y,x). Oh right! 270° clockwise about origin: $(x,y)→(-y,x)$. Let’s confirm: if I is $(-3,4)$, then $-y=-4$, $x=-3$? No, wait let's take a sample: (1,0) rotated 270 clockwise becomes (0,-1)? No, (1,0) 90 clockwise is (0,-1); 180 is (-1,0); 270 clockwise is (0,1). Using rule $(x,y)→(-y,x)$: (1,0)→(0,1) which is correct. So if original I is, say, (-3,4): $(-y,x)=(-4,-3)$? Wait the purple option is (-4,-3). Wait maybe original I is (3,-4)? No, the options: purple is (-4,-3), dark blue (3,-4), pink (4,-3). Wait the correct rule for 270° clockwise is $(x,y)→(-y,x)$. Suppose original I is (3,4): $(-4,3)$ no. Wait maybe the original point I is (-3,-4): $(-(-4),-3)=(4,-3)$ which is pink. But wait the most common mistake: 270° clockwise is same as 90° counterclockwise, rule $(x,y)→(-y,x)$. If the original point I is (3,-4): $-y=4$, $x=3$→(4,3) no. Wait the pink option is (4,-3). Let’s think again: maybe the original point I is (-3,4): $(-y,x)=(-4,-3)$ (purple). If the graph shows I at (-3,4), then purple is correct. But wait the user's graph—since the options include (-4,-3) as purple, which fits the rule if original I is (-3,4). So the correct answer is purple.

Answer:

A. I' (-4,-3) PURPLE