If `(-sqrt(2),sqrt2)` are cartesian co-ordinates off the point, then its polar co-ordinates are . . .

Options

A. `(1,(4pi)/(3))`
B. `(2,(3pi)/(4))`
C. `(3,(7pi)/(4))`
D. `(4,(5pi)/(4))`

Answer

B

Explanation

We have, `rcostheta=-sqrt(2) and r sin theta=sqrt(2)`
`thereforer^(2)cos^(2)theta+r^(2)sin^(2)theta=2+2=4`
`impliesr^(2)(cos^(2)theta+sin^(2)theta)=4`
`impliesr=+-2 and theta=tan^(-1)((sqrt(2))/(-sqrt(2)))`
`theta=tan^(-1)(-1)impliestheta=(3pi)/(4)`
`therefore`Required polar co-ordinate is `(r,theta)`
`=(2,(3pi)/(4))`