QUESTION IMAGE
Question
the point v(4, -1) is rotated 90° clockwise around the origin. what are the coordinates of its image v? 〇 v(1, 4) 〇 v(-1, 4) 〇 v(-1, -4) 〇 v(4, 1)
Step1: Recall 90° clockwise rotation rule
For a point $(x,y)$, 90° clockwise rotation around the origin transforms it to $(y, -x)$.
Step2: Substitute $x=4, y=-1$
Substitute into the rule: $(y, -x) = (-1, -4)$? No, correct substitution: $(y, -x) = (-1, -4)$ is wrong, correct: $(y, -x) = (-1, - (4))$? No, correct rule: 90° clockwise rotation is $(x,y) \to (y, -x)$. So $x=4$, $y=-1$: new $x$ is $y=-1$, new $y$ is $-x=-4$? No, wait, correct rule: 90° clockwise around origin: $(x,y) \to (y, -x)$. So $(4,-1) \to (-1, -4)$? No, wait no, let's correct: 90° clockwise rotation formula is $(x,y) \mapsto (y, -x)$. So $x=4$, $y=-1$: new coordinate is $(-1, -4)$? No, wait no, another way: 90° clockwise is equivalent to 270° counterclockwise, formula $(x,y) \to (y, -x)$. So $(4,-1)$: $y=-1$, $-x=-4$, so $(-1, -4)$? Wait no, let's plot: original point (4,-1) is in 4th quadrant. Rotate 90° clockwise around origin, it moves to 3rd quadrant? No, wait 4th quadrant, rotate 90° clockwise: down to 3rd? No, 4th quadrant: x positive, y negative. Rotate 90° clockwise: the x becomes original y, y becomes -original x. So $(4,-1) \to (-1, -4)$. Wait but let's check options: V'(-1,-4) is an option. Wait no, wait maybe I mixed up the rule. Wait 90° counterclockwise is $(x,y) \to (-y,x)$, 90° clockwise is $(x,y) \to (y,-x)$. Yes, so (4,-1): y is -1, -x is -4, so (-1, -4). Wait but let's verify with another method: the vector from origin to (4,-1) is $\vec{v} =
$. The rotation matrix for 90° clockwise is
? No, wait rotation matrix for $\theta$ clockwise is
. $\theta=90°$, $\cos90=0$, $\sin90=1$. So matrix is
. Multiply by
: $0*4 +1*(-1) = -1$, $-1*4 +0*(-1) = -4$. So result is
.
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V'(-1, -4)