Point A is at #(4 ,1 )# and point B is at #(-6 ,-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 #(4   ,1  )# and point B is at #(-6  ,-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?

Ella Armstrong · Accepted Answer

The new point A is at #(-1, 4)#
The changed in distance between points A and B
#=2sqrt41-sqrt(146)=0.723203# Rotating point A (4, 1) about the origin by #(3pi)/2# clockwise will place it exactly at (-1, 4) at the 2nd quadrant. The new distance between A and B is #=sqrt146#. The distance between point B(-6, -7) and the original location of point A(4, 1) is #=sqrt(164)=2sqrt(41)# difference in distance#=sqrt(164)-sqrt(146)=0.723203#

Aaliyah Anderson · Answer

undefined The new coordinates of point A after rotating clockwise about the origin by $ \frac{3\pi}{2} $ radians are (-1, 4). The distance between points A and B remains unchanged after the rotation. Therefore, the distance between the two points is the same as before, which can be calculated using the distance formula:

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

Substituting the coordinates of points A and B into the formula:

$$ d = \sqrt{(-6 - (-1))^2 + (-7 - 4)^2} = \sqrt{(-5)^2 + (-11)^2} = \sqrt{25 + 121} = \sqrt{146} $$

So, the distance between points A and B remains $ \sqrt{146} $ units.