Point A is at #(8 ,-2 )# and point B is at #(5 ,-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   ,-2  )# and point B is at #(5  ,-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?

Violet Aldrich · Accepted Answer

#color(purple)("Increase in distance due to rotation " = 10.72 " units"#

#"Coordinates of " A(8,-2), B (5, -7)#

#vec (AB) = sqrt((8-5)^2 + (-2+7)^2) = 5.83#

Point A rotated about the origin by #(3pi)/2# clockwise.

#A (8, -2) -> A' (-2, 8)#

#vec(A'B) = sqrt((-2-5)^2 + (8+7)^2) = 16.55#

#color(purple)("Increase in distance " = 16.55 - 5.83 = 10.72#

Sadie Allen · Answer

undefined The new coordinates of point A after rotating it $ \frac{3\pi}{2} $ clockwise about the origin can be found using the following rotation formulas:

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

where $ (x, y) $ are the original coordinates of point A, and $ (x', y') $ are the new coordinates after rotation.

For point A at (8, -2), the rotation angle is $ \frac{3\pi}{2} $. Substituting the values into the rotation formulas:

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

Solving these equations will give us the new coordinates of point A after rotation.

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

$$ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} $$

where $ (x_1, y_1) $ and $ (x_2, y_2) $ are the coordinates of points A and B, respectively. Subtracting the initial distance from the distance after rotation will give us the change in distance.