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 `r cos theta = - sqrt(2)` and `r sin theta = sqrt(2)`
`therefore r^(2) cos ^(2) theta + r^(2) sin theta = 2 + 2 = 4`
`rArr^(2) )(cos ^(2) theta + sin^(2) theta) = 4`
` rArr r = pm and theta = tan^(-1)((sqrt(2))/(-sqrt(2)))`
`theta = tan^(-1) (-1)rArr theta = (3pi)/(4)`
`therefore ` Requrired polar co-ordinates is `( r , theta)`
` =(2,(3pi)/(4))`