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)