Point A is at #(2 ,-1 )# and point B is at #(3 ,-4 )#. Point A is rotated #pi # 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 #(2  ,-1   )# and point B is at #(3  ,-4   )#.  Point A is rotated #pi  # 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?

Sadie Allen · Accepted Answer

#|bar(AB) - bar(A'B)| = sqrt(26) - 5sqrt(2)|# Given : Two Points #A(2, -1)# and #B(3, -4)#, Rotate #A# by #pi# Required: New coordinates of #A(2, -1) =>_(R(pi)) A'(x, y)# and distance #|bar(AB) - bar(A'B)|# Solution strategy : a) Rotate A b) Use distance formula to find #AB and A'B# c) Calculate #|bar(AB) - bar(A'B)|# #color(red)a)# #A' = R(pi)A# where #R(pi)# is the #2xx2# rotation matrix #R(pi)=[(costheta, -sintheta ), (sintheta, costheta)] _(theta=pi)# #R(pi)=[(1, 0), (0, 1)] # so then, #A' = [(-1, 0), (0, -1)][(2), (-1)] = [(-2), (1)]# that is #A'(-2,1)# You can also get this geometrically or by inspection... #color(red)b)# #bar(AB)= sqrt((2-3)^2 + (-1-4)^2) = sqrt(26)# #bar(A'B) = sqrt((-2-3)^2 + (1-(-4))^2) = 5sqrt(2)# #|bar(AB) - bar(A'B)| = sqrt(26) - 5sqrt(2)|#

James Atkins · Answer

undefined To rotate a point $ (x, y) $ about the origin by an angle $ \theta $ clockwise, we use the following formulas:

New $ x $ coordinate: $ x' = x \cdot \cos(\theta) - y \cdot \sin(\theta) $

New $ y $ coordinate: $ y' = x \cdot \sin(\theta) + y \cdot \cos(\theta) $

For point A $ (2, -1) $ rotated by $ \pi $ radians clockwise, we have:

New $ x $ coordinate: $ x' = 2 \cdot \cos(\pi) - (-1) \cdot \sin(\pi) = 2 \cdot (-1) - (-1) \cdot 0 = -2 $

New $ y $ coordinate: $ y' = 2 \cdot \sin(\pi) + (-1) \cdot \cos(\pi) = 2 \cdot 0 + (-1) \cdot (-1) = -1 $

So, the new coordinates of point A are $ (-2, -1) $.

To find the change in distance between points A and B, we calculate the distances before and after the rotation:

Before rotation: $ d_{AB} = \sqrt{(3 - 2)^2 + (-4 - (-1))^2} = \sqrt{1^2 + (-3)^2} = \sqrt{1 + 9} = \sqrt{10} $

After rotation, the coordinates of point B remain unchanged. The new distance between points A and B is the distance between $ (-2, -1) $ and $ (3, -4) $:

After rotation: $ d'_{AB} = \sqrt{(3 - (-2))^2 + (-4 - (-1))^2} = \sqrt{5^2 + (-3)^2} = \sqrt{25 + 9} = \sqrt{34} $

The change in distance between points A and B is $ \sqrt{34} - \sqrt{10} $.