a: y=(2m+5)x-3
=>(2m+5)x-y-3=0
\(d\left(O;d\right)=\dfrac{\left|\left(2m+5\right)\cdot0+\left(-1\right)\cdot0-3\right|}{\sqrt{\left(2m+5\right)^2+1}}=\dfrac{3}{\sqrt{\left(2m+5\right)^2+1}}\)
Để d=3 thì \(\sqrt{\left(2m+5\right)^2+1}=1\)
=>m=-5/2
b: Tọa độ A là:
\(\left\{{}\begin{matrix}y=0\\x=\dfrac{3}{2m+5}\end{matrix}\right.\)
=>OA=3/|2m+5|
Tọa độ B là:
\(\left\{{}\begin{matrix}x=0\\y=-3\end{matrix}\right.\)
=>OB=3
Theo đề, ta có: 1/2*OA*OB=2
=>\(\dfrac{9}{\left|2m+5\right|}\cdot\dfrac{1}{2}=2\)
=>|2m+5|*2=9/2
=>|2m+5|=9/4
=>2m+5=9/4 hoặc 2m+5=-9/4
=>m=-11/8 hoặc m=-29/8