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?

1 Answer
Douglas K.
Sep 30, 2016

The change in distance is:

Deltad = d_2 - d_1

Deltad = sqrt82 - sqrt26

Explanation:

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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