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Math  /  Geometry

QuestionThe x′y′x^{\prime} y^{\prime}-coordinate system is rotated θ\theta degrees from the xyx y-coordinate system. The coordinates of a point in the xyx y-coordinate system are given. Find tt coordinates of the point in the rotated coordinate system. θ=60∘,(3,1)\theta=60^{\circ},(3,1)

Studdy Solution

STEP 1

1. The original coordinates of the point are given as (3,1)(3, 1) in the xyxy-coordinate system.
2. The rotation angle θ\theta is 60∘60^\circ.

STEP 2

1. Recall the rotation transformation formulas.
2. Substitute the given values into the formulas.
3. Calculate the new coordinates in the rotated system.

STEP 3

Recall the rotation transformation formulas for rotating a point (x,y)(x, y) by an angle θ\theta to get the new coordinates (x′,y′)(x', y'):
x′=xcos⁡θ+ysin⁡θx' = x \cos \theta + y \sin \theta y′=−xsin⁡θ+ycos⁡θy' = -x \sin \theta + y \cos \theta

STEP 4

Substitute the given values (x,y)=(3,1)(x, y) = (3, 1) and θ=60∘\theta = 60^\circ into the formulas. Recall that cos⁡60∘=12\cos 60^\circ = \frac{1}{2} and sin⁡60∘=32\sin 60^\circ = \frac{\sqrt{3}}{2}:
x′=3⋅12+1⋅32x' = 3 \cdot \frac{1}{2} + 1 \cdot \frac{\sqrt{3}}{2} y′=−3⋅32+1⋅12y' = -3 \cdot \frac{\sqrt{3}}{2} + 1 \cdot \frac{1}{2}

STEP 5

Calculate the new coordinates:
x′=32+32=3+32x' = \frac{3}{2} + \frac{\sqrt{3}}{2} = \frac{3 + \sqrt{3}}{2} y′=−332+12=−33+12y' = -\frac{3\sqrt{3}}{2} + \frac{1}{2} = \frac{-3\sqrt{3} + 1}{2}
The coordinates of the point in the rotated x′y′x'y'-coordinate system are:
(3+32,−33+12)\boxed{\left( \frac{3 + \sqrt{3}}{2}, \frac{-3\sqrt{3} + 1}{2} \right)}

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