Point A is at #(-7 ,3 )# and point B is at #(5 ,4 )#. Point A is rotated #pi # 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

#A' = (7, -3)#
The distance has decreased by #sqrt(145) - sqrt(53) ~~ 4.76#

Given: #A(-7, 3), B(5, 4)# Point #A# is rotated #pi# clockwise about the origin.

A #pi# rotation clockwise is a #180^@# rotation CW.

The coordinate rule for a #180^@# rotation CW is: #(x, y) -> (-x, -y)#

Using the coordinate rule: #A' = (7, -3)#

The distance formula is #d = sqrt((x_2 - x_1)^2 + (y_2 - y_1)^2)#

#d_(AB) = sqrt((5 - -7)^2 + (4 - 3)^2) = sqrt(12^2 + 1^2) = sqrt(145)#

#d_(A'B) = sqrt((5-7)^2 + (4 - -3)^2) = sqrt(2^2 + 7^2) = sqrt(53)#

The distance between point #A# and #B# has not changed.

The distance between the rotated point #A# = #A'# and #B# is #sqrt(145) - sqrt(53) ~~4.76#

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Answer 2

After rotating point A (originally at (-7, 3)) by π radians clockwise about the origin, the new coordinates of point A will be (7, -3). The distance between the original point A and point B (at (5, 4)) can be calculated using the distance formula. After the rotation, the distance between the new point A and point B will be the same as the original distance, as rotation about the origin does not change the distance between two points.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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