A line segment with endpoints at #(5, 1)# and #(7, 2)# is rotated clockwise by #pi/2#. What are the new endpoints of the line segment?

Question

A line segment with endpoints at #(5, 1)# and #(7, 2)# is rotated clockwise by #pi/2#.  What are the new endpoints of the line segment?

Allison Allen · Accepted Answer

#(1,-5)" and " (2,-7)# #"I am assuming the rotation is centred at the origin"# #"under a clockwise rotation about the origin of " pi/2# #"a point " (x,y)to(y,-x)# #rArr(5,1)to(1,-5)# #rArr(7,2)to(2,-7)#

Parker Allen · Answer

undefined To rotate a point $(x, y)$ clockwise about the origin by an angle $ \theta $, the new coordinates $(x', y')$ are given by:

$$ x' = x \cos(\theta) + y \sin(\theta) $$
$$ y' = -x \sin(\theta) + y \cos(\theta) $$

Given the endpoints $(5, 1)$ and $(7, 2)$, and rotating by $ \frac{\pi}{2} $ clockwise:

For the point $(5, 1)$:
$$ x' = 5 \cdot \cos\left(\frac{\pi}{2}\right) + 1 \cdot \sin\left(\frac{\pi}{2}\right) = 5 \cdot 0 + 1 \cdot 1 = 1 $$
$$ y' = -5 \cdot \sin\left(\frac{\pi}{2}\right) + 1 \cdot \cos\left(\frac{\pi}{2}\right) = -5 \cdot 1 + 1 \cdot 0 = -5 $$

For the point $(7, 2)$:
$$ x' = 7 \cdot \cos\left(\frac{\pi}{2}\right) + 2 \cdot \sin\left(\frac{\pi}{2}\right) = 7 \cdot 0 + 2 \cdot 1 = 2 $$
$$ y' = -7 \cdot \sin\left(\frac{\pi}{2}\right) + 2 \cdot \cos\left(\frac{\pi}{2}\right) = -7 \cdot 1 + 2 \cdot 0 = -7 $$

Therefore, the new endpoints of the line segment after rotating clockwise by $ \frac{\pi}{2} $ are $(1, -5)$ and $(2, -7)$.