Question

How do you convert `(-sqrt(6), -sqrt(2))` to polar form?

Answer 1

`(2sqrt2 , (7pi)/6)`

Explanation:

Use the formulae that links Cartesian to Polar coordinates.

`• r^2 = x^2 + y^2`

`• theta = tan^-1(y/x)`

here `x = -sqrt6 " and " y = -sqrt2`

`r^2 = (-sqrt6)^2 + (-sqrt2)^2 = 6 + 2 = 8`

`r^2 = 8 rArr r = sqrt8 = 2sqrt2`

The point `(-sqrt6,-sqrt2)" is in 3rd quadrant" .`

care must be taken to ensure that `theta" is in 3rd quadrant"`

`theta = tan^-1((-sqrt2)/(-sqrt6)) = 30^@ orpi/6" radians "`

This is the 'related' angle in the 1st quadrant , require 3rd.

`rArr theta = (180+30)^@ = 210^@ or (7pi)/6" radians "`