A line segment with endpoints at #(3 , -2 )# and #(5, 8 )# is rotated clockwise by #(3 pi)/2#. What are the new endpoints of the line segment?

Question

A line segment with endpoints at #(3 , -2   )# and #(5, 8    )# is rotated clockwise by #(3 pi)/2#.  What are the new endpoints of the line segment?

Aaliyah Anderson · Accepted Answer

( 3,-2) would change to (2,3)
and (5,8) would change to (-8,5)

To find the new coordinates, the process of clockwise rotation by #(3pi)/2#can be split in two parts: I. First rotate by #pi/2# in which (x,y) #-># (y,-x) II Next rotate by #pi# in which (x,y) #-># (-x,-y) Thus any point (x,y) on rotation by #(3pi)/2# would become (-y,x) Thus ( 3,-2) would change to (2,3) and (5,8) would change to (-8,5)

Violet Aldrich · Answer

undefined To find the new endpoints of the line segment after rotating it clockwise by $ \frac{3\pi}{2} $ radians, you can use the following rotation matrix:

$$ \begin{pmatrix} \cos(\theta) & -\sin(\theta) \ \sin(\theta) & \cos(\theta) \end{pmatrix} $$

In this case, $ \theta = \frac{3\pi}{2} $.

Applying the rotation matrix to each endpoint of the line segment:

For the point (3, -2):
$$ \begin{pmatrix} \cos(\frac{3\pi}{2}) & -\sin(\frac{3\pi}{2}) \ \sin(\frac{3\pi}{2}) & \cos(\frac{3\pi}{2}) \end{pmatrix} \begin{pmatrix} 3 \ -2 \end{pmatrix} = \begin{pmatrix} 0 \ 3 \end{pmatrix} $$

For the point (5, 8):
$$ \begin{pmatrix} \cos(\frac{3\pi}{2}) & -\sin(\frac{3\pi}{2}) \ \sin(\frac{3\pi}{2}) & \cos(\frac{3\pi}{2}) \end{pmatrix} \begin{pmatrix} 5 \ 8 \end{pmatrix} = \begin{pmatrix} -8 \ 5 \end{pmatrix} $$

Therefore, the new endpoints of the line segment are (0, 3) and (-8, 5).