a:
(d'): x-2y=3
=>2y=x-3
=>\(y=\dfrac{1}{2}x-\dfrac{3}{2}\)
Tọa độ giao điểm của 2 đường thẳng trên là:
\(\left\{{}\begin{matrix}-3x+2=\dfrac{1}{2}x-\dfrac{3}{2}\\y=-3x+2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}-\dfrac{7}{2}x=-\dfrac{7}{2}\\y=-3x+2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=1\\y=-3\cdot1+2=-1\end{matrix}\right.\)
vậy: A(1;-1)
b: Tọa độ B là:
\(\left\{{}\begin{matrix}y=0\\-3x+2=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=\dfrac{2}{3}\\y=0\end{matrix}\right.\)
Vậy: O(0;0); A(1;-1); \(B\left(\dfrac{2}{3};0\right)\)
\(OA=\sqrt{\left(1-0\right)^2+\left(-1-0\right)^2}=\sqrt{2}\)
\(OB=\sqrt{\left(\dfrac{2}{3}-0\right)^2+\left(0-0\right)^2}=\dfrac{2}{3}\)
\(AB=\sqrt{\left(\dfrac{2}{3}-1\right)^2+\left(0+1\right)^2}=\dfrac{\sqrt{10}}{3}\)
Xét ΔOAB có \(cosAOB=\dfrac{OA^2+OB^2-AB^2}{2\cdot OA\cdot OB}=\dfrac{2+\dfrac{4}{9}-\dfrac{10}{9}}{2\cdot\sqrt{2}\cdot\dfrac{2}{3}}=\dfrac{\sqrt{2}}{2}\)
=>\(sinAOB=\sqrt{1-\left(\dfrac{\sqrt{2}}{2}\right)^2}=\dfrac{\sqrt{2}}{2}\)
Diện tích tam giác AOB là:
\(S_{AOB}=\dfrac{1}{2}\cdot OA\cdot OB\cdot sinAOB\)
\(=\dfrac{1}{2}\cdot\sqrt{2}\cdot\dfrac{2}{3}\cdot\dfrac{\sqrt{2}}{2}=\dfrac{1}{3}\)
c: Thay x=1 và y=-1 vào (d''), ta được:
\(1\left(-2m+1\right)+m+1=-1\)
=>-2m+1+m+1=-1
=>2-m=-1
=>m=3
