Point A is at #(-3 ,4 )# and point B is at #(-8 ,1 )#. 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 #(-3   ,4  )# and point B is at #(-8  ,1   )#.  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

It can be seen from the diagram, that a rotation about the origin through an angle #theta# can be represented as:

#((1),(0))->((costheta),(sintheta))# and #((0),(1))->((-sintheta),(color(white)(8)costheta))#

So the transformation matrix will be:

#((costheta,-sintheta),(sintheta,color(white)(8)costheta))#

Matrix A:

#A=((-3),(color(white)(8)4))#

Transformation matrix will be:

#((cos((3pi)/2),-sin((3pi)/2)),(sin((3pi)/2),color(white)(8)cos((3pi)/2)))#
#:.#

#A'=((cos((3pi)/2),-sin((3pi)/2)),(sin((3pi)/2),color(white)(8)cos((3pi)/2)))((-3),(color(white)(8)4))=((4),(3))#

Coordinates:

#( 4 , 3 )#

Distance between A and B:

#d=sqrt((-3-(-8))^2+(4-1)^2)=sqrt(34)#

Distance between #A' and B#

#d=sqrt((4-(-8))^2+(3-1)^2)=4#

The distance has been reduced by a factor of #(2sqrt(34))/17#

#:.#

#(2sqrt(34))/17*sqrt(34)=4#

Sadie Adams · Answer

undefined To rotate a point $ A(x_1, y_1) $ clockwise about the origin by an angle $ \theta $, you use the following formulas:

$$ x' = x_1 \cos(\theta) - y_1 \sin(\theta) $$
$$ y' = x_1 \sin(\theta) + y_1 \cos(\theta) $$

Given that point A is at (-3, 4) and is rotated $ \frac{3\pi}{2} $ clockwise about the origin:

$$ x' = (-3) \cos\left(\frac{3\pi}{2}\right) - 4 \sin\left(\frac{3\pi}{2}\right) $$
$$ y' = (-3) \sin\left(\frac{3\pi}{2}\right) + 4 \cos\left(\frac{3\pi}{2}\right) $$

Solving these equations:

$$ x' = 4 $$
$$ y' = 3 $$

So the new coordinates of point A are (4, 3).

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

Before rotation, the distance between points A and B is $ d_{AB} = \sqrt{(-8 - (-3))^2 + (1 - 4)^2} = \sqrt{(-5)^2 + (-3)^2} = \sqrt{25 + 9} = \sqrt{34} $.

After rotation, the new coordinates of point A are (4, 3), and the coordinates of point B remain unchanged.

The distance between the new point A and point B is $ d_{A'B} = \sqrt{(4 - (-8))^2 + (3 - 1)^2} = \sqrt{(12)^2 + (2)^2} = \sqrt{144 + 4} = \sqrt{148} $.

The change in distance is $ \Delta d = d_{A'B} - d_{AB} = \sqrt{148} - \sqrt{34} $.