The $x ^ { \prime } y ^ { \prime }$ -coordinate system has been rotated $\theta$ degrees from the $x y$ -coordinate system.The coordinates of a point in the $x y$ -coordinate system are given.Find the coordinates of the point in the rotated coordinate system. $\theta = 45 ^ { \circ }$ , $( 9,2 )$
A) $\left( \frac { 11 \sqrt { 2 } } { 2 } , \frac { \sqrt { 2 } } { 2 } \right)$
B) $\left( - \frac { 11 \sqrt { 2 } } { 2 } , - \frac { 7 \sqrt { 2 } } { 2 } \right)$
C) $\left( \frac { 11 } { 2 } , - \frac { 7 \sqrt { 2 } } { 2 } \right)$
D) $\left( \frac { 11 \sqrt { 2 } } { 2 } , \frac { 7 \sqrt { 2 } } { 2 } \right)$
E) $\left( \frac { 11 \sqrt { 2 } } { 2 } , - \frac { 7 \sqrt { 2 } } { 2 } \right)$

Question

The $x ^ { \prime } y ^ { \prime }$ -coordinate system has been rotated $\theta$ degrees from the $x y$ -coordinate system.The coordinates of a point in the $x y$ -coordinate system are given.Find the coordinates of the point in the rotated coordinate system.  $\theta = 45 ^ { \circ }$ , $( 9,2 )$ 
A) $\left( \frac { 11 \sqrt { 2 } } { 2 } , \frac { \sqrt { 2 } } { 2 } \right)$
B) $\left( - \frac { 11 \sqrt { 2 } } { 2 } , - \frac { 7 \sqrt { 2 } } { 2 } \right)$
C) $\left( \frac { 11 } { 2 } , - \frac { 7 \sqrt { 2 } } { 2 } \right)$
D) $\left( \frac { 11 \sqrt { 2 } } { 2 } , \frac { 7 \sqrt { 2 } } { 2 } \right)$
E) $\left( \frac { 11 \sqrt { 2 } } { 2 } , - \frac { 7 \sqrt { 2 } } { 2 } \right)$

Quizplus · Accepted Answer

To find the coordinates in the rotated system, use the rotation formulas: $x' = x \cos \theta + y \sin \theta$ and $y' = -x \sin \theta + y \cos \theta$. For $\theta = 45^\circ$, $\cos 45^\circ = \sin 45^\circ = \frac{\sqrt{2}}{2}$. Substituting $x = 9$ and $y = 2$, we get $x' = 9 \cdot \frac{\sqrt{2}}{2} + 2 \cdot \frac{\sqrt{2}}{2} = \frac{11\sqrt{2}}{2}$ and $y' = -9 \cdot \frac{\sqrt{2}}{2} + 2 \cdot \frac{\sqrt{2}}{2} = -\frac{7\sqrt{2}}{2}$. Thus, the coordinates are $\left( \frac{11\sqrt{2}}{2}, -\frac{7\sqrt{2}}{2} \right)$.

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