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 $m_2?$
$r\cos (\theta + \alpha_1) = k_1$
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.