How can I check that two lines are perpendicular and parallel in polar coordinates? Given two lines $r\cos(\theta-\alpha_1)=k_1$ and $r\cos(\theta-\alpha_2)=k_2$, how can I prove that they are:

*

*Perpendicular $\iff$ $\sin\alpha_1\sin\alpha_2+\cos\alpha_1\cos\alpha_2=0$

*Parallel $\iff$ $\sin\alpha_1\cos\alpha_2-\cos\alpha_1\sin\alpha_2=0$
I know that $r=\pm\sqrt{x^2+y^2}$ and $x=r\cos\theta$, $y=r\sin\theta$
Also, in eucledian coordinates, two lines are perpendicular if $m_1\cdot m_2=-1$ and parallel if $m_1=m_2$
How can I find $m_1$ and $m_2?$
 A: The equation of a straight line in polar coordinates is given by $$r=\frac{k}{\sin(\alpha-\theta)}$$ where $k$ is perpendicular distance from the origin to the line and $\alpha$ is the angle made by the line with the polar axis.
Now, for two lines to be perpendicular, we have $$\alpha_2=\alpha_1\pm(2n+1)\frac{π}{2}$$ $$ \implies \alpha_2-\alpha_1=\pm(2n+1)\frac{π}{2}\tag 1$$
For the first part, we have $$\begin{align}LHS&=\sin\alpha_1\sin\alpha_2+\cos\alpha_1\cos\alpha_2\\&=\cos(\alpha_2-\alpha_1)\\ &=0=RHS\end{align}$$
Similarly you can proceed for the second part.
A: $r\cos (\theta + \alpha_1) = k_1$
By angle addition identities.
$r\cos \theta\cos \alpha_1 - r\sin\theta\sin \alpha_1 = k_1$
$r\cos\theta = x\\
r\sin\theta = y$
$x\cos \alpha_1 - y\sin \alpha_1 = k_1\\
y = \cot \alpha_1 - k_1\csc \alpha_1\\
y = m_1 x + b_1$
$m_1 = \cot \alpha_1.$ Similarly, $m_2 = \cot \alpha_2.$
$m_1 = m_2 \implies  \frac {\sin \alpha_1}{\cos\alpha_1} = \frac {\sin \alpha_2}{\cos\alpha_2}$
$m_1m_2 = 1 \implies  \frac {\sin \alpha_1}{\cos\alpha_1} \frac {\sin \alpha_2}{\cos\alpha_2} = 1$
Cross multiply and simplify.
