Point A is at #(-2 ,8 )# and point B is at #(-1 ,3 )#. 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?

Answer 1

The change in distance is:

#Deltad = d_2 - d_1#

#Deltad = sqrt82 - sqrt26#

The first distance from B to A is:

#d_1 = sqrt((-2 - -1)^2 + (8 - 3)^2)#
#d_1 = sqrt((-1)^2 + (5)^2)#
#d_1 = sqrt(1 + 25)#
#d_1 = sqrt26#

Compute the polar coordinates of point A:

#r = sqrt((-2)^2 + (8)^2)#
#r = sqrt(4 + 64)#
#r = sqrt68#
#theta_1 = tan^-1(8/-2) + pi#
Note: We add #pi#, because the inverse tangent function returns either first quadrant angles or negative rotations. In this case, it will return a negative angle.
To rotate the point we add #(3pi)/2# to the original angle:
#theta_2 = tan^-1(8/-2) + (5pi)/2#

Compute the x and y coordinates for new point A:

#(sqrt68cos(tan^-1(8/-2) + (5pi)/2), sqrt68sin(tan^-1(8/-2) + (5pi)/2))#
#(8, 2)#

Compute the new distance:

#d_2 = sqrt((8 - -1)^2 + (2 - 3)^2)#
#d_2 = sqrt((9)^2 + (-1)^2)#
#d_2 = sqrt82#

The change in distance is:

#Deltad = d_2 - d_1#
#Deltad = sqrt82 - sqrt26#
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Answer 2

The new coordinates of point A after rotating ((3\pi)/2) clockwise about the origin are ((8, 2)).

To find the new coordinates, we use the following rotation formulas:

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

For point A ((-2, 8)), substituting (\theta = (3\pi)/2), we get:

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

So, the new coordinates of point A are ((8, 2)).

The original distance between points A and B can be calculated using the distance formula:

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

For points A ((-2, 8)) and B ((-1, 3)), the original distance is:

[ d_{\text{original}} = \sqrt{(-1 - (-2))^2 + (3 - 8)^2} = \sqrt{1^2 + 5^2} = \sqrt{26} ]

After the rotation, the new coordinates of point A are ((8, 2)). So, the distance between the new point A and B ((-1, 3)) is:

[ d_{\text{new}} = \sqrt{(8 - (-1))^2 + (2 - 3)^2} = \sqrt{9^2 + (-1)^2} = \sqrt{82} ]

The change in distance is given by:

[ \Delta d = d_{\text{new}} - d_{\text{original}} = \sqrt{82} - \sqrt{26} ]

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