Đáp án:
$\displaystyle \begin{array}{{>{\displaystyle}l}} a.\ a=-2;\ b=3\\ ( d) :\ y=-2x+3\\ b.Chu\ vi\ \Delta OBC:\ \frac{9+3\sqrt{5}}{2}\\ S_{OBC} =\frac{9}{4}\\ c.Chu\ vi\ \Delta OBD:2+\sqrt{2}\\ S_{OBC} =\frac{1}{2} \end{array}$
Giải thích các bước giải:
$\displaystyle \begin{array}{{>{\displaystyle}l}} a.\ A\in ( d) \Rightarrow 2a+b=-1\ ( 1)\\ \ \ \ \ B\in ( d) \Rightarrow \frac{3}{2} a+b=0\ ( 2)\\ Từ\ ( 1)( 2) \Rightarrow a=-2;\ b=3\\ \Rightarrow ( d) :\ y=-2x+3\\ b.( d) \cap Oy=\{C\}\\ \Rightarrow -2.0+3=y\\ \Rightarrow y=3\\ \Rightarrow C( 0;3)\\ Ta\ \ có\ OB=\frac{3}{2} ;\ OC=3\\ \Rightarrow BC=\sqrt{\left(\frac{3}{2}\right)^{2} +3^{2}} =\frac{3\sqrt{5}}{2}\\ Chu\ vi\ \Delta OBC:\ \frac{3}{2} +3+\frac{3\sqrt{5}}{2} =\frac{9+3\sqrt{5}}{2}\\ S_{OBC} =\frac{1}{2} .\frac{3}{2} .3=\frac{9}{4}\\ c.\ Xét\ \ PT\ \ hoành\ \ độ\ \ giao\ điểm:\\ -2x+3=x\\ \Leftrightarrow x=1\Rightarrow y=1\\ \Rightarrow D( 1;1)\\ Ta\ có:\ OB=1;\ BD=1\\ \Rightarrow OD=\sqrt{1+1} =\sqrt{2}\\ Chu\ vi\ \Delta OBD:\ 1+1+\sqrt{2} =2+\sqrt{2}\\ S_{OBC} =\frac{1}{2} .1.1=\frac{1}{2} \end{array}$
