Points A and B are at #(4 ,6 )# and #(7 ,5 )#, respectively. Point A is rotated counterclockwise about the origin by #pi/2 # and dilated about point C by a factor of #5 #. If point A is now at point B, what are the coordinates of point C?

2 Answers

#(-37/4,15/4)#

Explanation:

Rotating Point A counterclockwise by #pi/2# will give you a point at A' #(-6,4)#

The distance between #A':(-6,4) and B:(7,5)# is #sqrt(13^2+1^2)=sqrt(170)#

The distance between A' and B must be five times the distance between A' and C

The vector to go from B to A' is #{-13,-1}#

1/4 of that vector is #{-13/4,-1/4}#

Apply that to A' to get #(-6-13/4,4-1/4)=(-37/4,15/4)#

Jan 19, 2018

#C=(-37/4,15/4)#

Explanation:

#"under a counterclockwise rotation about the origin of "pi/2#

#• " a point "(x,y)to(-y,x)#

#rArrA(4,6)toA'(-6,4)" where A' is the image of A"#

#rArrvec(CB)=color(red)(5)vec(CA')#

#rArrulb-ulc=5(ula'-ulc)#

#rArrulb-ulc=5ula'-5ulc#

#rArr4ulc=5ula'-ulb#

#color(white)(rArr4ulc)=5((-6),(4))-((7),(5))#

#color(white)(rArrulc)=((-30),(20))-((7),(5))=((-37),(15))#

#rArrulc=1/4((-37),(15))=((-37/4),(15/4))#

#rArrC=(-37/4,15/4)#