In the xy-plane, a circle with its center at the origin contains the point P(6, -2\sqrt{3}… — SAT

In the $xy$ -plane, a circle with its center at the origin contains the point $P(6, -2\sqrt{3})$ . Point $P$ is rotated counterclockwise about the origin by an angle of $\frac{5\pi}{6}$ radians to a new position $Q$ . What are the coordinates of point $Q$ ?

A
$(0, -4\sqrt{3})$
B
$(-4\sqrt{3}, 0)$
$(-2\sqrt{3}, 6)$ Answer
D
$(-6, 2\sqrt{3})$

Answer

The coordinates of point

Q

are

(-2\sqrt{3}, 6)

The correct coordinates are found by calculating the circle's radius

R = 4\sqrt{3}

and identifying that the initial point

P(6, -2\sqrt{3})

has an angle of

-\frac{\pi}{6}

radians. Rotating counterclockwise by

\frac{5\pi}{6}

radians results in a new angle of

\frac{2\pi}{3}

radians (

120^\circ

). Using circular trigonometry,

x = R\cos(120^\circ) = -2\sqrt{3}

and

y = R\sin(120^\circ) = 6

, giving the point

(-2\sqrt{3}, 6)

Step-by-Step Solution

Find the radius

R

of the circle using the coordinates of point

P(6, -2\sqrt{3})

R = \sqrt{6^2 + (-2\sqrt{3})^2} = \sqrt{36 + 12} = \sqrt{48} = 4\sqrt{3}

The radius of the circle is needed to determine the coordinates of the rotated point

Q

on the same circle.

Determine the initial angle

\theta_P

of point

P

in standard position.

Since

\cos(\theta_P) = \frac{6}{4\sqrt{3}} = \frac{\sqrt{3}}{2}

and

\sin(\theta_P) = \frac{-2\sqrt{3}}{4\sqrt{3}} = -\frac{1}{2}

, the point lies in the fourth quadrant with

\theta_P = -\frac{\pi}{6}

radians (or

330^\circ

Finding the initial position in standard angle form allows us to apply the rotation.

Calculate the new angle

\theta_Q

after the counterclockwise rotation of

\frac{5\pi}{6}

radians.

\theta_Q = -\frac{\pi}{6} + \frac{5\pi}{6} = \frac{4\pi}{6} = \frac{2\pi}{3}

radians.

Adding the counterclockwise rotation angle to the initial angle gives the new standard position angle of the terminal ray.

Convert the new angle to degrees and compute the coordinates of point

Q

Converting the angle gives

\frac{2\pi}{3} \times \frac{180^\circ}{\pi} = 120^\circ

. The coordinates of

Q

are

x_Q = 4\sqrt{3} \cos(120^\circ) = -2\sqrt{3}

and

y_Q = 4\sqrt{3} \sin(120^\circ) = 6

Converting to degrees allows the use of standard trigonometric values to find the exact coordinates.

Key Concept

Radian-to-degree conversion and application of rotations on a coordinate plane using circular trigonometry.

Alternative Method

Instead of working entirely in radians, convert the coordinates of

P

to degrees first. Since

\tan(\theta_P) = -\frac{\sqrt{3}}{3}

in Quadrant IV,

\theta_P = -30^\circ

. The rotation of

\frac{5\pi}{6}

radians is converted to degrees:

\frac{5\pi}{6} \times \frac{180^\circ}{\pi} = 150^\circ

. Adding these gives the new angle

\theta_Q = -30^\circ + 150^\circ = 120^\circ

. Finally, compute the coordinates using

x = R\cos(120^\circ)

and

y = R\sin(120^\circ)

Estimated Time:3m 0s

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Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Alternative Method

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