Point A is at $(8 ,-1 )$ and point B is at $(9 ,-7 )$ . 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 · 2017-12-23T07:53:27

The new coordinates are $=(1,8)$ and the distace has changed by $=11.9u$

The point $A=(8,-1)$

The point $B=(9,-7)$

The distance

$AB=sqrt((9-8)^2+(-7-(-1))^2)=sqrt(1+25)=sqrt26$

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

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

The distance

$A'B=sqrt((9-1)^2+(-7-8)^2)=sqrt(64+225)=sqrt289=17$

The change in the distance is

$AB-A'B=17-sqrt26=11.9u$

Point A is at (8 ,-1 )(8 ,-1 ) and point B is at (9 ,-7 )(9 ,-7 ). 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?