A = (1, 2), B = (6, –10) locus of P such that PA – PB = 13 is

$$\begin{gathered} \mathrm{Given} \mathrm{A}=(1, 2), \mathrm{B}=(6, 10) \\ \mathrm{Now} \mathrm{AB}=\sqrt{{(6-1)}^2+{(-10-2)}^2} \\ =\sqrt{25+144} \\ =\sqrt{169} \\ \mathrm{AB}=13 \\ \mathrm{Given} \mathrm{that} \mathrm{PA}-\mathrm{PB}=13 \\ \Rightarrow \mathrm{PA}-\mathrm{PB}=AB \\ \therefore \mathrm{locus} \mathrm{of} \text{'}\mathrm{P}\text{'} \mathrm{respresents} \mathrm{a} \mathrm{ray} \mathrm{with} \mathrm{vertex} \mathrm{at} \mathrm{B} \mathrm{not} \mathrm{containing} \mathrm{A}. \end{gathered}$$