Step 1 :1) \( \overrightarrow{AB} = \begin{pmatrix} -1 \\ 1 \end{pmatrix}, \overrightarrow{AC} = \begin{pmatrix} 1\\ 1 \end{pmatrix}, \overrightarrow{BC} = \begin{pmatrix} 2\\ 0 \end{pmatrix} \)
Step 2 :2) \( AB = \sqrt{(-1)^2 + 1^2} = \sqrt{2}, AC = \sqrt{1^2 + 1^2} = \sqrt{2}, BC = \sqrt{2^2 + 0^2} = 2 \)
Step 3 :3) (AB) : \( y - 1 = 1(x - 2) \Rightarrow y = x - 1 \), (AC) : \( y - 1 = -1(x - 2) \Rightarrow y = -x + 3 \)
Step 4 :4) (AB) : \( y = x - 1 \), (AC) : \( y = -x + 3 \)
Step 5 :5) Since the slopes of (AB) and (AC) are different, the lines are not parallel, and they intersect at the point A.