$\begin{array}{l}A(-4;1)\quad B(2;4)\quad C(2;-3)\\ \to \begin{cases}AB = \sqrt{(2+4)^2 + (4-1)^2} = 3\sqrt6\\ AC=\sqrt{(2+4)^2 + (-3-1)^2} = 2\sqrt{13}\\ BC = \sqrt{(2-2)^2 + (-3-4)^2} = 7\end{cases}\\ \to P_{ABC} = AB + AC + BC = 3\sqrt6 + 2\sqrt{13} + 7\\ Gọi \,\,N(a;0)\in Oy\\ \to \begin{cases}\overrightarrow{AN} = (a+4;-1)\\\overrightarrow{BN} = (a-2;-4)\end{cases}\\ \text{A,B,N thẳng hàng} \,\,\Leftrightarrow \overrightarrow{AN},\,\overrightarrow{BN}\,\text{cùng phương}\\ \Leftrightarrow \dfrac{a+4}{a-2} = \dfrac{-1}{-4}\\ \Leftrightarrow 4(a+4) = a-2\\ \Leftrightarrow 3a = -18\\ \Leftrightarrow a = -6\\ Vậy\,\,N(-6;0) \end{array}$
