Point A is at $(8 ,-4 )$ and point B is at $(-9 ,6 )$ . 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?

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Answer 1 · 2017-12-21T14:16:57

The new coordinates are $=(4,8)$ and the distance has changed by $=6.32$

The point $A=(8,-4)$

The point $B=(-9,6)$

The distance

$AB=sqrt((-9-8)^2+(6-(-4))^2)=sqrt(289+100)=sqrt325$

The matrix for a rotation of angle $theta$ is

$r(theta)=((costheta,-sintheta),(sintheta,costheta))$

And when $theta=-3/2pi$

$r(-3/2pi)=((cos(-3/2pi),-sin(-3/2pi)),(sin(-3/2pi),cos(-3/2pi)))$

$=((0,-1),(1,0))$

Therefore,

The coordinates of the point $A'$ after the rotation of the point $A$ clockwise by $3/2pi$ is

$((x),(y))=((0,-1),(1,0))*((8),(-4))=((4),(8))$

The distance

$A'B=sqrt((-2-9)^2+(8-4)^2)=sqrt(121+16)=sqrt137$

The change in the distance is

$AB-A'B=sqrt325-sqrt137=6.32u$

Point A is at (8 ,-4 )(8 ,-4 ) and point B is at (-9 ,6 )(-9 ,6 ). Point A is rotated (3pi)/2 (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?