Question #36cc4

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Answer 1 · 2017-06-22T19:22:49

$B'= (-12,6)$

Assuming that the rotation is about the origin, then the rotation matrix is:

$R(theta)= [ (cos(theta),-sin(theta)), (sin(theta), cos(theta)) ]$

Evaluating this matrix at $theta = 90^@$ :

$R(90^@)= [ (0,-1), (1, 0) ]$

Hence, we have the transformation:

$[ (x'), (y') ] = [ (0,-1), (1, 0) ][ (x), (y) ]$

This gives us the two equations:

$x' = -y$
$y'= x$

Using the point $B=(6,12)$

$x' = -12$
$y' = 6$

$B'= (-12,6)$

Answer 2 · 2017-06-22T19:29:47

$B'(-12,6)$

$"I am assuming a rotation about the origin"$

$"under a rotation about the origin of "90^@$

$"a point " (x,y)to(-y,x)$

$rArrB(6,12)toB'(-12,6)$