Point A is at $(-2 ,-8 )$ and point B is at $(-5 ,3 )$ . Point A is rotated $pi/2$ 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

A line segment with endpoints at $(5, 1)$ and $(7, 2)$ is rotated clockwise by $pi/2$ . What are...
A line segment with endpoints at $(5, 5)$ and $(1, 2)$ is rotated clockwise by $pi/2$ . What are...
A line segment with endpoints at $(1, 5)$ and $(1, -2)$ is rotated clockwise by $pi/2$ . What...
A line segment with endpoints at $(1, 3)$ and $(8, -2)$ is rotated clockwise by $pi/2$ . What...
A line segment with endpoints at $(-1, 1)$ and $(3, -5)$ is rotated clockwise by $pi/2$ . What...
A line segment with endpoints at $(-1 , -9)$ and $(3, -5 )$ is rotated clockwise by $pi/2$ . ...
A line segment with endpoints at $(1 , -2 )$ and $(1, -4 )$ is rotated clockwise by $pi/2$ ....
A line segment with endpoints at $(1 , -2 )$ and $(6, -7 )$ is rotated clockwise by $pi/2$ ....
A line segment with endpoints at $(5 , -2 )$ and $(2, -7 )$ is rotated clockwise by $pi/2$ ....
A line segment with endpoints at $(5 , -9 )$ and $(2, -7 )$ is rotated clockwise by $pi/2$ ....

1415 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-02-22T06:30:06

Increase in distance due to rotation of point A 1.6816

enter image source here

$A (-2, -8), B (-5, 3)$

Point A rotated clockwise about origin by $pi/2$

$A ((-2), (-8)) -> A’ ((8),(2))$ from III to I quadrant.

$vec(AB) = sqrt((-2+5)^2 + (-8-3)^2) = sqrt130 = 11.4018$

$vec (A’B) = sqrt ((8+5)^2 + (2-3)^2) = sqrt170 = 13.0834$

Change in distance due to rotation of point A

$d = 13.0834 - 11.4018 = color(purple)(1.6816$

Point A is at (-2 ,-8 )(-2 ,-8 ) and point B is at (-5 ,3 )(-5 ,3 ). Point A is rotated pi/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?