Point A is at $(8 ,1 )$ and point B is at $(2 ,-3 )$ . 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 · 2017-12-24T14:24:41

The new coordinates are $=(-1,8)$ and the distance has changed by $=4.19u$

The point $A=(8,1)$

The point $B=(2,-3)$

The distance

$AB=sqrt((2-8)^2+(-3-(1))^2)=sqrt(36+16)=sqrt52$

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

$A'B=sqrt((2-(-1))^2+(-3-8)^2)=sqrt(9+121)=sqrt130$

The change in the distance is

$A'B-AB=sqrt(130)-sqrt52=4.19u$

Point A is at (8 ,1 )(8 ,1 ) and point B is at (2 ,-3 )(2 ,-3 ). 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?