Cho (d1) y=2x , (d2) y = 1/2x , (d3) y= -x +3 . (d3) cat ( d1) tai A , (d3) cat (d2) tai B a) Vẽ (d1) , (d2) , (d3) trên cùng hệ trực. Xác định tọa độ A , B b) CM : tam giác AOB cân c) Tính chu vi và diện tích tam giác OAB Lam On Giai That Day Du Giup MINH Voi

\(\begin{array}{l} {d_1}:\,\,\,y = 2x;\,\,{d_2}:\,\,\,y = \frac{1}{2}x;\,\,\,{d_3}:\,\,\,y = - x + 3\\ {d_3} \cap {d_1} = \left\{ A \right\};\,\,\,{d_3} \cap {d_2} = \left\{ B \right\}\\ a)\,\,\,Toa\,\,\,do\,\,\,diem\,\,\,A\,\,\,la\,\,\,nghiem\,\,\,cua\,\,he\,\,\,pt:\\ \left\{ \begin{array}{l} y = - x + 3\\ y = 2x \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} x = - 3\\ y = - 6 \end{array} \right. \Rightarrow A\left( {1;\,\,2} \right)\\ Toa\,\,\,do\,\,\,diem\,\,\,B\,\,\,la\,\,\,nghiem\,\,\,cua\,\,he\,\,\,pt:\\ \left\{ \begin{array}{l} y = - x + 3\\ y = \frac{1}{2}x \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} x = 2\\ y = 1 \end{array} \right. \Rightarrow B\left( {2;\,\,1} \right).\\ b)\,\,\,Ta\,\,co:\,\\ OA = \sqrt {{{\left( {{x_A} - {x_O}} \right)}^2} + {{\left( {{y_A} - {y_O}} \right)}^2}} = \sqrt {{1^2} + {2^2}} = \sqrt 5 .\\ OB = \sqrt {{{\left( {{x_B} - {x_O}} \right)}^2} + {{\left( {{y_B} - {y_O}} \right)}^2}} = \sqrt {{2^2} + {1^2}} = \sqrt 5 .\\ \Rightarrow OA = OB \Rightarrow \Delta AOB\,\,\,can\,\,tai\,\,\,O.\\ c)\,\,\,Ta\,\,\,co:\\ AB = \sqrt {{{\left( {{x_B} - {x_A}} \right)}^2} + {{\left( {{y_B} - {y_A}} \right)}^2}} = \sqrt {{{\left( {2 - 1} \right)}^2} + {{\left( {1 - 2} \right)}^2}} = \sqrt 2 \\ \Rightarrow Chu\,\,vi\,\,tam\,\,giac\,\,AOB\,\,la:\,\,\,\\ OA + OB + AB = 2\sqrt 5 + \sqrt 2 \,\,\,\,\,\left( {dvdd} \right).\\ Ke\,\,\,AH \bot Oy,\,\,\,BK\, \bot Ox\,\,\,nhu\,\,hinh\,\,ve.\\ Khi\,\,\,do\,\,ta\,\,co:\,\,\,\\ {S_{AOB}} = {S_{OMN}} - {S_{AOM}} - {S_{BON}}\\ = \frac{1}{2}OM.ON - \frac{1}{2}.AH.OM - \frac{1}{2}BK.ON\\ = \frac{1}{2}.3.3 - \frac{1}{2}.1.3 - \frac{1}{2}.3.1 = \frac{3}{2}\,\,\left( {dvdt} \right). \end{array}\)