A line segment with endpoints at #(1 , -4 )# and #(2, 7 )# is rotated clockwise by #(3 pi)/2#. What are the new endpoints of the line segment?

Question

A line segment with endpoints at #(1 , -4   )# and #(2, 7    )# is rotated clockwise by #(3 pi)/2#.  What are the new endpoints of the line segment?

Autumn Armstrong · Accepted Answer

#(4,1)" and " (-7,2)# #"assuming rotation about the origin"# #"under a clockwise rotation about the origin of " (3pi)/2# #• " a point " (x,y)to(-y,x)# #rArr(1,-4)to(4,1)# #rArr(2,7)to(-7,2)#

Sadie Adams · Answer

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

1. Find the midpoint of the original line segment using the formula:
   $$ \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) $$

2. Rotate the midpoint using the rotation formula for a point $ (x, y) $ about the origin by angle $ \theta $:
   $$ \left( x' = x \cos(\theta) - y \sin(\theta), \quad y' = x \sin(\theta) + y \cos(\theta) \right) $$

3. Add or subtract the distances between the original endpoints and the midpoint to find the new endpoints.

Using these steps:

- Midpoint: 
  $$ \left( \frac{1 + 2}{2}, \frac{-4 + 7}{2} \right) = \left( \frac{3}{2}, \frac{3}{2} \right) $$

- Rotating the midpoint by $ \frac{3\pi}{2} $:
  $$ \left( x' = \frac{3}{2} \cos\left(\frac{3\pi}{2}\right) - \frac{3}{2} \sin\left(\frac{3\pi}{2}\right), \quad y' = \frac{3}{2} \sin\left(\frac{3\pi}{2}\right) + \frac{3}{2} \cos\left(\frac{3\pi}{2}\right) \right) $$

- Calculating:
  $$ x' = \frac{3}{2} \times 0 - \frac{3}{2} \times (-1) = \frac{3}{2} $$
  $$ y' = \frac{3}{2} \times (-1) + \frac{3}{2} \times 0 = -\frac{3}{2} $$

- The new midpoint is: $ \left( \frac{3}{2}, -\frac{3}{2} \right) $

- Using the new midpoint to find the new endpoints:
  $$ \text{Endpoint 1}: \left( 1 - \frac{3}{2}, -4 - \left(-\frac{3}{2}\right) \right) = \left( -\frac{1}{2}, -\frac{5}{2} \right) $$
  $$ \text{Endpoint 2}: \left( 2 - \frac{3}{2}, 7 - \left(-\frac{3}{2}\right) \right) = \left( \frac{1}{2}, \frac{17}{2} \right) $$

So, the new endpoints of the line segment after rotating it clockwise by $ \frac{3\pi}{2} $ are $ \left( -\frac{1}{2}, -\frac{5}{2} \right) $ and $ \left( \frac{1}{2}, \frac{17}{2} \right) $.