[tex]The\ slope-point\ form:y-y_0=m(x-x_0)\ where\ m\ is\ a\ slope\\\\If\ k:y=m_1x+b_1\ and\ l:y=m_2x+b_2\\\\k\perp l\iff m_1m_2=-1\to m_2=-\dfrac{1}{m_1}\\\\therefore\\the\ slope\ of\ line\ CD:y-3=-2(x-4)\ is\ m_1=-2\ and\ the\ slope\\of\ a\ line\ prpendicular\ to\ CD\ is\ m_2=-\dfrac{1}{-2}=\dfrac{1}{2}[/tex]
Answer: 1/2
