Point A is at #(1 ,3 )# and point B is at #(-7 ,-5 )#. Point A is rotated #pi/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 #(1 ,3  )# and point B is at #(-7  ,-5   )#.  Point A is rotated #pi/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

#color(blue)((3,-1)#

#color(blue)(0.54337889 \ \ units)# We can produce a rotation about the origin by using the transformation matrix: #((cos(theta),-sin(theta)),(sin(theta),cos(theta)))# This is for anticlockwise rotation, so for clockwise rotation we use the angle: #2pi-pi/2=(3pi)/2# So we have: #((cos((3pi)/2),-sin((3pi)/2)),(sin((3pi)/2),cos((3pi)/2)))=((0,1),(-1,0))# #A=(1,3)# #A'=((0,1),(-1,0))((1),(3))=((3),(-1))# Distance between A and B: #d=sqrt((1-(-7))^2+(3-(-5))^2)=sqrt(128)=8sqrt(2)# Distance between A' and B: #d=sqrt((3-(-7))^2+(-1-(-5))^2)=sqrt(116)=2sqrt(29)# Change in distance: #8sqrt(2)-2sqrt(29)=0.54337889#units

Autumn Armstrong · Answer

undefined To rotate point A, (1, 3), clockwise about the origin by π/2 radians, we can use the rotation matrix formula:

$$ \begin{pmatrix} x' \ y' \end{pmatrix} = \begin{pmatrix} \cos(\theta) & -\sin(\theta) \ \sin(\theta) & \cos(\theta) \end{pmatrix} \begin{pmatrix} x \ y \end{pmatrix} $$

Where $ \theta = \pi/2 $, and $ x = 1 $, $ y = 3 $.

$$ \begin{pmatrix} x' \ y' \end{pmatrix} = \begin{pmatrix} \cos(\pi/2) & -\sin(\pi/2) \ \sin(\pi/2) & \cos(\pi/2) \end{pmatrix} \begin{pmatrix} 1 \ 3 \end{pmatrix} $$

$$ \begin{pmatrix} x' \ y' \end{pmatrix} = \begin{pmatrix} 0 & -1 \ 1 & 0 \end{pmatrix} \begin{pmatrix} 1 \ 3 \end{pmatrix} $$

$$ \begin{pmatrix} x' \ y' \end{pmatrix} = \begin{pmatrix} 0 \ 1 \end{pmatrix} $$

So, the new coordinates of point A after rotating π/2 radians clockwise about the origin are (0, 1).

To find the change in distance between points A and B, we need to calculate the distance between the original point A and point B, and then between the new point A and point B.

Original distance between A and B:
$$ d_{AB} = \sqrt{(x_B - x_A)^2 + (y_B - y_A)^2} $$
$$ d_{AB} = \sqrt{(-7 - 1)^2 + (-5 - 3)^2} $$
$$ d_{AB} = \sqrt{(-8)^2 + (-8)^2} $$
$$ d_{AB} = \sqrt{128} $$
$$ d_{AB} = 8\sqrt{2} $$

New distance between A and B:
$$ d'_{AB} = \sqrt{(x_B - x'_A)^2 + (y_B - y'_A)^2} $$
$$ d'_{AB} = \sqrt{(-7 - 0)^2 + (-5 - 1)^2} $$
$$ d'_{AB} = \sqrt{(-7)^2 + (-6)^2} $$
$$ d'_{AB} = \sqrt{49 + 36} $$
$$ d'_{AB} = \sqrt{85} $$

The change in distance between points A and B is:
$$ \Delta d_{AB} = d'_{AB} - d_{AB} $$
$$ \Delta d_{AB} = \sqrt{85} - 8\sqrt{2} $$