Point A is at $(- 6, 1)$ $(-6 ,1 )$ and point B is at $(3, 8)$ $(3 ,8 )$ . Point A is rotated $\frac{3 π}{2}$ $(3pi)/2$ 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?

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Answer 1 · 2017-06-24T07:20:29

The new coordinates are $= (- 1, - 6)$ $=(-1,-6)$ and the distance has changed by $= 3.2$ $=3.2$

The matrix of a rotation clockwise by $\frac{3}{2} π$ $3/2pi$ about the origin is

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

Therefore, the transformation of point $A$ is

$A'= ((0,-1),(1,0)) ((-6),(1))=((-1),(-6))$

The distance $AB$ is

$=sqrt((3-(-6))^2+(8-1)^2)$

$=sqrt(81+49)$

$=sqrt130$

The distance $A'B$ is

$=sqrt((3-(-1))^2+(8-(-6))^2)$

$=sqrt(16+196)$

$=sqrt212$

The distance has changed by

$=sqrt212-sqrt130$

$=3.2$

Point A is at (-6 ,1 )(−6,1)(-6 ,1 ) and point B is at (3 ,8 )(3,8)(3 ,8 ). Point A is rotated (3pi)/2 3π2(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?