$\left \{ {{a+b=4} \atop {3a+b=5}} \right.$$\Leftrightarrow$$\left \{ {{a=\dfrac{1}{2}} \atop {b=\dfrac{7}{2}}} \right.$ $\Rightarrow AB: y=\dfrac{1}{2}x+\dfrac{7}{2}\\ hay\ AB:x-2y+7=0$
$\left \{ {{3a+b=5} \atop {6a+b=4}} \right.$$\Leftrightarrow$$\left \{ {{a=\dfrac{-1}{3}} \atop {b=6}} \right.$$\Rightarrow BC:y=\dfrac{-1}{3}x+6\\ hay\ BC:x+3y-18=0$
$\left \{ {{6a+b=4} \atop {2a+b=2}} \right.$ $\Leftrightarrow$$\left \{ {{a=\dfrac{1}{2}} \atop {b=1}} \right.$$\Rightarrow CD:y=\dfrac{1}{2}x+1\\hay\ CD:x-2y+2=0$
$\left \{ {{2a+b=2} \atop {1a+b=4}} \right.$$\Leftrightarrow$$\left \{ {{a=-2} \atop {b=6}} \right.$$\Rightarrow AD:y=-2x+6\\ hay\ AD:2x+y-6=0$
Ta thấy
$\dfrac{a_1}{a_3}=\dfrac{b_1}{b_3}\ne\dfrac{c_1}{c_3}\Rightarrow AB//CD\ (1)$
$\dfrac{a_1}{a_4}=\dfrac{-b_4}{b_1}\Rightarrow AB\perp AD\Rightarrow CD\perp AD\ (2)$
Từ (1) và (2) suy ra tứ giác ABCD là hình thang vuông tại A và D
$AB=\sqrt{(3-1)^2+(5-4)^2}=\sqrt{5}\\ BC=\sqrt{(6-3)^2+(4-5)^2}=\sqrt{10}\\ CD=\sqrt{(2-6)^2+(2-4)^2}=2\sqrt{5}\\ AD=\sqrt{(2-1)^2+(2-4)^2}=\sqrt{5}$ Chu vi ABCD: $C_{ABCD}=AB+BC+CD+AD\\=\sqrt{5}+\sqrt{10}+2\sqrt{5}+\sqrt{5}=4\sqrt{5}+\sqrt{10}\ (đvđd)$ Diện tích ABCD: $S_{ABCD}=\dfrac{1}{2}.(AB+CD).AD\\ =\dfrac{1}{2}(\sqrt{5}+2\sqrt{5}).\sqrt{5}=\dfrac{15}{2}\ (đvdt)$
