Point A is at #(8 ,-1 )# and point B is at #(9 ,-7 )#. Point A is rotated #(3pi)/2 # clockwise about the origin. What are the new coordinates of point A and by how much has the distance between points A and B changed?

Question

Point A is at #(8   ,-1  )# and point B is at #(9  ,-7   )#.  Point A is rotated #(3pi)/2  # clockwise about the origin.  What are the new coordinates of point A and by how much has the distance between points A and B changed?

Noah Atkinson · Accepted Answer

The new coordinates are #=(1,8)# and the distace has changed by #=11.9u#

The point #A=(8,-1)#

The point #B=(9,-7)#

The distance

#AB=sqrt((9-8)^2+(-7-(-1))^2)=sqrt(1+25)=sqrt26#

The matrix for a rotation of angle #theta# is

#r(theta)=((costheta,-sintheta),(sintheta,costheta))#

And when #theta=-3/2pi#

#r(-3/2pi)=((cos(-3/2pi),-sin(-3/2pi)),(sin(-3/2pi),cos(-3/2pi)))#

#=((0,-1),(1,0))#

Therefore,

The coordinates of the point #A'# after the rotation of the point #A# clockwise by #3/2pi# is

#((x),(y))=((0,-1),(1,0))*((8),(-1))=((1),(8))#

The distance

#A'B=sqrt((9-1)^2+(-7-8)^2)=sqrt(64+225)=sqrt289=17#

The change in the distance is

#AB-A'B=17-sqrt26=11.9u#

Parker Allen · Answer

undefined The new coordinates of point A after rotating clockwise about the origin by $ \frac{3\pi}{2} $ radians are:

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

Plugging in the values $ (x, y) = (8, -1) $ and $ \theta = \frac{3\pi}{2} $, we get:

$$ x' = 8 \cos\left(\frac{3\pi}{2}\right) - (-1) \sin\left(\frac{3\pi}{2}\right) $$
$$ y' = 8 \sin\left(\frac{3\pi}{2}\right) + (-1) \cos\left(\frac{3\pi}{2}\right) $$

Solving these equations:

$$ x' = 8 \cdot 0 - (-1) \cdot (-1) = 0 - 1 = -1 $$
$$ y' = 8 \cdot (-1) + (-1) \cdot 0 = -8 + 0 = -8 $$

So, the new coordinates of point A after the rotation are $ (-1, -8) $.

To find the change in distance between points A and B, we first calculate the distance between the original points A and B, and then the distance between the new point A and B. The difference between these distances gives us the change in distance.

Original distance between A and B:
$$ d_{AB} = \sqrt{(9-8)^2 + (-7-(-1))^2} = \sqrt{1^2 + (-6)^2} = \sqrt{1 + 36} = \sqrt{37} $$

New distance between A and B:
$$ d'_{AB} = \sqrt{(-1-9)^2 + (-8-(-7))^2} = \sqrt{(-10)^2 + (-1)^2} = \sqrt{100 + 1} = \sqrt{101} $$

Change in distance:
$$ \Delta d_{AB} = d'_{AB} - d_{AB} = \sqrt{101} - \sqrt{37} $$