Convert to Polar y=- square root of 3x

$y = - \sqrt{3} x$

Step 1

Since $x = r \cos (θ)$ , replace $x$ with $r \cos (θ)$ .

$y = - \sqrt{3} (r \cos (θ))$

Step 2

Since $y = r \sin (θ)$ , replace $y$ with $r \sin (θ)$ .

$r \sin (θ) = - \sqrt{3} (r \cos (θ))$

Step 3

Solve for $r$ .

Simplify the right side.

Remove parentheses.

$r \sin (θ) = - \sqrt{3} r \cos (θ)$

Add $\sqrt{3} r \cos (θ)$ to both sides of the equation.

$r \sin (θ) + \sqrt{3} r \cos (θ) = 0$

Factor $r$ out of $r \sin (θ) + \sqrt{3} r \cos (θ)$ .

Factor $r$ out of $r \sin (θ)$ .

$r (\sin (θ)) + \sqrt{3} r \cos (θ) = 0$

Factor $r$ out of $\sqrt{3} r \cos (θ)$ .

$r (\sin (θ)) + r (\sqrt{3} \cos (θ)) = 0$

Factor $r$ out of $r (\sin (θ)) + r (\sqrt{3} \cos (θ))$ .

$r (\sin (θ) + \sqrt{3} \cos (θ)) = 0$

Divide each term in $r (\sin (θ) + \sqrt{3} \cos (θ)) = 0$ by $\sin (θ) + \sqrt{3} \cos (θ)$ and simplify.

Divide each term in $r (\sin (θ) + \sqrt{3} \cos (θ)) = 0$ by $\sin (θ) + \sqrt{3} \cos (θ)$ .

$\frac{r (\sin (θ) + \sqrt{3} \cos (θ))}{\sin (θ) + \sqrt{3} \cos (θ)} = \frac{0}{\sin (θ) + \sqrt{3} \cos (θ)}$

Simplify the left side.

Cancel the common factor of $\sin (θ) + \sqrt{3} \cos (θ)$ .

Cancel the common factor.

$\frac{r (\sin (θ) + \sqrt{3} \cos (θ))}{\sin (θ) + \sqrt{3} \cos (θ)} = \frac{0}{\sin (θ) + \sqrt{3} \cos (θ)}$

Divide $r$ by $1$ .

$r = \frac{0}{\sin (θ) + \sqrt{3} \cos (θ)}$

Simplify the right side.

Divide $0$ by $\sin (θ) + \sqrt{3} \cos (θ)$ .

$r = 0$