Points A and B are at #(2 ,9 )# and #(7 ,5 )#, respectively. Point A is rotated counterclockwise about the origin by #(3pi)/2 # and dilated about point C by a factor of #1/2 #. If point A is now at point B, what are the coordinates of point C?

Answer 1

#C=(23,8)#

#"under a counterclockwise rotation about the origin of "(3pi)/2#
#• " a point "(x,y)to(-y,x)#
#rArrA(2,9)toA'(-9,2)" where A' is the image of A"#
#rArrvec(CB)=color(red)(1/2)vec(CA')#
#rArrulb-ulc=1/2(ula'-ulc)#
#rArrulb-ulc=1/2ula'-1/2ulc#
#rArr1/2ulc=ulb-1/2ula'#
#color(white)(rArr1/2ulc)=((7),(5))-1/2((-9),(2))#
#color(white)(rArr1/2ulc)=((7),(5))-((-9/2),(1))=((23/2),(4))#
#rArrulc=2((23/2),(4))=((23),(8))#
#rArrC=(23,8)#
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Answer 2

To find the coordinates of point C after the rotation and dilation, you can follow these steps:

  1. Rotate point A counterclockwise about the origin by ( \frac{3\pi}{2} ) radians. This rotation transforms point A into a new position.
  2. Dilate the rotated point A about point C by a factor of ( \frac{1}{2} ). This rescales the position of the point.

Let's first perform the rotation on point A:

Given point A: ( (2, 9) )

After rotating counterclockwise by ( \frac{3\pi}{2} ) radians, the new coordinates become:

( x' = x \cdot \cos(\frac{3\pi}{2}) - y \cdot \sin(\frac{3\pi}{2}) ) ( y' = x \cdot \sin(\frac{3\pi}{2}) + y \cdot \cos(\frac{3\pi}{2}) )

Substituting the coordinates of point A:

( x' = 2 \cdot \cos(\frac{3\pi}{2}) - 9 \cdot \sin(\frac{3\pi}{2}) = 9 ) ( y' = 2 \cdot \sin(\frac{3\pi}{2}) + 9 \cdot \cos(\frac{3\pi}{2}) = -2 )

Now, we have the coordinates after rotation: ( (9, -2) )

Next, let's dilate this point about an unknown point C by a factor of ( \frac{1}{2} ). Let the coordinates of point C be ( (x_C, y_C) ).

Using the dilation formula:

( x_C = \frac{x_A + x_B}{2} ) ( y_C = \frac{y_A + y_B}{2} )

Substituting the coordinates of point A (after rotation) and B:

( x_C = \frac{9 + 7}{2} = 8 ) ( y_C = \frac{-2 + 5}{2} = \frac{3}{2} )

Therefore, the coordinates of point C are ( (8, \frac{3}{2}) ).

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