$\begin{array}{l}13)\quad A(1;0)\quad B(4;5)\\ \to \overrightarrow{AB}=(3;5)\\ \text{Ta có:}\,\,C \in (d): y = 2x + 1\\ \to C(x;2x + 1)\\ \to \overrightarrow{AC} = (x-1;2x + 1)\\ \text{A, B, C thẳng hàng $\Leftrightarrow \overrightarrow{AB},\,\overrightarrow{AC}$ cùng phương}\\ \Leftrightarrow \dfrac{3}{x-1} = \dfrac{5}{2x+1}\\ \Leftrightarrow 3(2x +1) = 5(x-1)\\ \Leftrightarrow 6x + 3 = 5x - 5\\ \Leftrightarrow x =-8\\ Vậy\quad C(-8;-15)\\ 14)\quad A(-3;4)\quad B(1;1)\quad C(9;-5)\\ \to \begin{cases}\overrightarrow{AB} = (4;-3)\\\overrightarrow{AC} =(12;-9)\end{cases}\\ \text{Ta có:}\\ \quad \dfrac{4}{12} = \dfrac{-3}{-9}\\ \to \text{A, B, C thẳng hàng} \end{array}$
