A line with parametric polar

Question

A line has the parametric polar (ρ, θ) description as $2 = x \cos 60° + y \sin 60°$.

What is the parametric polar equation for a line that is perpendicular to the given line and passes through the point $(5,5)$?

Answer 1

I assume that the line has equation $r:x \cos 60° + y \sin 60°=2$ that is

$$r:y=-\frac{x}{\sqrt{3}}+\frac{4}{\sqrt{3}}$$ Its slope is $m=-\dfrac{1}{\sqrt 3}$

Thus the slope of the perpendicular is $m^{\perp}=\sqrt 3$

The line perpendicular to $r$ passing through $(5,5)$ has equation

$y-5=\sqrt{3}(x-5)$

To convert into polar form we substitute

$x=\rho\cos \theta;\;y=\rho\sin\theta$

$\rho\sin\theta-5=\sqrt{3}(\rho\cos \theta-5)$

Now we solve for $\rho$

$\rho\sin\theta-5=\sqrt{3}\rho\cos \theta-5\sqrt{3}$

$\rho\sin\theta-\sqrt{3}\rho\cos \theta=5-5\sqrt{3}$

$\rho(\sin\theta-\sqrt{3}\cos \theta)=5-5\sqrt{3}$

Hope this can be useful $$\rho=\frac{5-5\sqrt{3}}{\sin\theta-\sqrt{3}\cos \theta}$$