How do you change the rectangular coordinate #(-6, 6sqrt3)# into polar coordinates?

Question
Answer 1

#(-6,6sqrt3)to(12,(2pi)/3)#

To convert from #color(blue)"rectangular to polar coordinates"#

That is #(x,y)to(r,theta)#

#color(orange)"Reminder"#

#color(red)(|bar(ul(color(white)(a/a)color(black)(r=sqrt(x^2+y^2))color(white)(a/a)|)))" and " color(red)(|bar(ul(color(white)(a/a)color(black)(theta=tan^-1(y/x))color(white)(a/a)|)))#

here x = - 6 and y #=6sqrt3#

#rArrr=sqrt(6^2+(6sqrt3)^2)=sqrt(36+108)=sqrt144=12#

Now #(-6,6sqrt3)# is in the 2nd quadrant so we must ensure that #theta# is in the 2nd quadrant.

#theta=tan^-1((6sqrt3)/-6)=tan^-1(-sqrt3)#

#=-pi/3" in 4th quadrant"#

#rArrtheta=(pi-pi/3)=(2pi)/3" in 2nd quadrant"#

#rArr(-6,6sqrt3)to(12,(2pi)/3)#

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