Point A is at $(6, - 2)$ $(6 ,-2 )$ and point B is at $(- 3, 5)$ $(-3 ,5 )$ . Point A is rotated $\frac{π}{2}$ $pi/2$ 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 $(6, - 2)$ $(6 ,-2 )$ and point B is at $(- 3, 5)$ $(-3 ,5 )$ . Point A is rotated $\frac{π}{2}$ $pi/2$ 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?

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Answer 1 · 2016-09-28T01:52:16

Use $d = \sqrt{{(x_{A} - x_{B})}_{A}^{2} + {(y_{A} - y_{B})}_{A}^{2}}$ $d = sqrt((x_A - x_B)^2 + (y_A - y_B)^2)$ to compute the distance, convert A to polar form and rotate $- \frac{π}{2}$ $-pi/2$ , convert A back to Cartesian form, compute the new distance, and the difference.

Explanation:

Compute the distance from B to A:

$d = \sqrt{{(6 - - 3)}^{2} + {(- 2 - 5)}^{2}}$ $d = sqrt((6 - -3)^2 + (-2 - 5)^2)$

$d = \sqrt{130}$ $d = sqrt(130)$

Convert A to polar form:

$r = \sqrt{6^{2} + {(- 2)}^{2}}$ $r = sqrt(6^2 + (-2)^2)$
$r = \sqrt{40}$ $r = sqrt(40)$
$θ_{1} = {tan}^{- 1} (- \frac{2}{6})$ $theta_1 = tan^-1(-2/6)$

Rotate:

$r = \sqrt{6^{2} + {(- 2)}^{2}}$ $r = sqrt(6^2 + (-2)^2)$
$r = \sqrt{40}$ $r = sqrt(40)$
$θ_{2} = {tan}^{- 1} (- \frac{2}{6}) - \frac{π}{2}$ $theta_2 = tan^-1(-2/6) - pi/2$

Convert back to Cartesian form:

$(r cos (θ_{2}), r sin (θ_{2}) = (- 2, - 6)$ $(rcos(theta_2), rsin(theta_2) = (-2, -6)$

Compute the distance from B to the new A:

$d_{2} = \sqrt{{(- 2 - - 3)}^{2} + {(- 6 - 5)}^{2}}$ $d_2 = sqrt((-2 - -3)^2 + (-6 - 5)^2)$

$d_{2} = \sqrt{1 + {(- 11)}^{2}}$ $d_2 = sqrt(1 + (-11)^2$

$Deltad = sqrt(130) - sqrt(122)$

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Point A is at $(6, - 2)$ $(6 ,-2 )$ and point B is at $(- 3, 5)$ $(-3 ,5 )$ . Point A is rotated $\frac{π}{2}$ $pi/2$ 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?

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Explanation:

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Point A is at (6 ,-2 )(6,−2)(6 ,-2 ) and point B is at (-3 ,5 )(−3,5)(-3 ,5 ). Point A is rotated pi/2 π2pi/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

Explanation:

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Point A is at $(6, - 2)$ $(6 ,-2 )$ and point B is at $(- 3, 5)$ $(-3 ,5 )$ . Point A is rotated $\frac{π}{2}$ $pi/2$ 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?