a: Theo đề, ta có:
\(\left\{{}\begin{matrix}x_A+x_B=2\\x_B+x_C=0\\x_C+x_A=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x_A-x_C=2\\x_A+x_C=-2\\x_B+x_C=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x_A=0\\x_C=-2\\x_B=-x_C=2\end{matrix}\right.\)
Theo đề, ta có:
\(\left\{{}\begin{matrix}y_A+y_B=6\\y_B+y_C=-4\\y_A+y_C=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y_A+y_B=6\\y_B-y_A=-8\\y_B+y_C=-4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y_A=7\\y_B=-1\\y_C=-4-y_B=-4+1=-3\end{matrix}\right.\)
=>A(0;7); B(2;-1); C(-2;-3)
a: vecto AB=(2;-8)=(1;-4)
=>VTPT là (4;1)
PT AB là 4(x-0)+1(y-7)=0
=>4x+y-7=0
vecto AC=(-2;-10)=(1;5)
=>VTPT là (-5;1)
=>PT AC là -5(x-0)+1(y-7)=0
=>-5x+y-7=0
vecto BC=(-4;-2)=(2;1)
=>VTPT là (-1;2)
=>Phương trình BC là -1(x-2)+2(y+1)=0
=>-x+2+2y+2=0
=>-x+2y+4=0
