Polar coordinates=(6sqrt 2, 3pi/4)=(r, theta)→r=6 sqrt 2, theta=3pi/4
Rectangular coordinates=(x,y)=?
x=r cos theta=(6 sqrt 2) cos(3pi/4)=(6 sqrt 2)(-sqrt 2 / 2)
x=-3 (sqrt 2)^2=-3(2)→x=-6
y=r sin theta=(6 sqrt 2) sin (3pi/4)=(6 sqrt 2)(sqrt 2 / 2)
y=3 (sqrt 2)^2=3(2)→y=6
Rectangular coordinates of the point = (x,y) =(-6,6)
Answer: Option a. (-6,6)