What is the polar form of #( -18,-6 )#?

Question

Emma Aldridge · Accepted Answer

#(18.974, 3.463)#

We're asked to find the polar form of a rectangular coordinate.

We can do so by using the equations

#r = sqrt(x^2 + y^2)#

#theta = arctan(y/x)#

The #x#-coordinate is #-18#, and the #y#-coordinate is #-6#, so

#r = sqrt((-18)^2 + (-6)^2) = color(red)(18.974#

#theta = arctan((-6)/(-18)) = 0.322 + pi = color(blue)(3.463#

The #pi# was added to fix the calculator error, the coordinate is located in quadrant #III#. (Remember the angle #theta# is in radians.)

The polar form of this coordinate is thus

#(color(red)(18.974), color(blue)(3.463))#

Isaiah Allen · Answer

undefined The polar form of the point $(-18, -6)$ can be found using the conversion formulas:

$$ r = \sqrt{x^2 + y^2} $$
$$ \theta = \arctan\left(\frac{y}{x}\right) $$

Substituting the given values, we get:

$$ r = \sqrt{(-18)^2 + (-6)^2} = \sqrt{324 + 36} = \sqrt{360} = 6\sqrt{10} $$
$$ \theta = \arctan\left(\frac{-6}{-18}\right) = \arctan\left(\frac{1}{3}\right) $$

Therefore, the polar form of the point $(-18, -6)$ is $6\sqrt{10} \, \text{cis}(\arctan(1/3))$.