Point A is at #(9 ,3 )# and point B is at #(1 ,-3 )#. 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 #(9   ,3   )# and point B is at #(1   ,-3   )#.  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?

Ella Armstrong · Accepted Answer

#color(green)("Change in distance " = 2.65 " units")#

#A (9, 3), B (1, -3), "Point A rotated about origin clockwise by " (pi/2)#

To find change in distance between A & B due to this rotation.

Using distance formula,

#vec(AB) = sqrt ((9-1)^2 + (3 + 3)^2) = 10#

#A((9),(3)) -> A'((-3),(9))#

#vec(A'B) = sqrt((-3-1)^2 + (9+3)^2) = 12.65#

#"Change in distance =" = vec(A'B) - vec(AB) = 12.65 - 10 = 2.65#

Noah Atkinson · Answer

undefined After rotating point A (9, 3) by $ \frac{\pi}{2} $ clockwise about the origin, its new coordinates can be obtained using the following rotation formulas:

For the new x-coordinate: $ x' = x \cdot \cos(\theta) - y \cdot \sin(\theta) $
For the new y-coordinate: $ y' = x \cdot \sin(\theta) + y \cdot \cos(\theta) $

Using $ \theta = \frac{\pi}{2} $ and the given coordinates of point A:
$ x = 9 $, $ y = 3 $

$ x' = 9 \cdot \cos\left(\frac{\pi}{2}\right) - 3 \cdot \sin\left(\frac{\pi}{2}\right) $
$ x' = 9 \cdot (0) - 3 \cdot (1) $
$ x' = -3 $

$ y' = 9 \cdot \sin\left(\frac{\pi}{2}\right) + 3 \cdot \cos\left(\frac{\pi}{2}\right) $
$ y' = 9 \cdot (1) + 3 \cdot (0) $
$ y' = 9 $

So, the new coordinates of point A after rotating it $ \frac{\pi}{2} $ clockwise about the origin are (-3, 9).

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

The distance formula between two points $ (x_1, y_1) $ and $ (x_2, y_2) $ is given by:
$ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} $

For the original coordinates:
$ x_1 = 9 $, $ y_1 = 3 $
$ x_2 = 1 $, $ y_2 = -3 $

$ d_{AB\_orig} = \sqrt{(1 - 9)^2 + (-3 - 3)^2} $
$ d_{AB\_orig} = \sqrt{(-8)^2 + (-6)^2} $
$ d_{AB\_orig} = \sqrt{64 + 36} $
$ d_{AB\_orig} = \sqrt{100} $
$ d_{AB\_orig} = 10 $

For the new coordinates:
$ x_1' = -3 $, $ y_1' = 9 $
$ x_2 = 1 $, $ y_2 = -3 $

$ d_{AB\_new} = \sqrt{(1 - (-3))^2 + (-3 - 9)^2} $
$ d_{AB\_new} = \sqrt{(1 + 3)^2 + (-12)^2} $
$ d_{AB\_new} = \sqrt{4^2 + (-12)^2} $
$ d_{AB\_new} = \sqrt{16 + 144} $
$ d_{AB\_new} = \sqrt{160} $
$ d_{AB\_new} = 4\sqrt{10} $

So, the distance between points A and B has changed from 10 units to $ 4\sqrt{10} $ units.