Ma ajuta cineva la exercitiu 5?va rog e urgent❤️

Răspuns:

x+y-2=0

AB=√(xA-xB)²+(ya-yB)²

Sistem

{2√2=√(0-xB)²+(2-yB)²

{xB+yB-2=0=>xB=2-yB

Inlocuiesti  pe  yB  sub  radical

2√2=√xB²+xB²

2√2=√2xB²

2√2=lxlB√2=> xB=±2

Inlocuiesti  in  ecuatia   dreptei  pe  xB  si-l  aflii  pe yB

-2+yB-2=0

yB-4=0   yB=4

si

2+yB-2=0

yB=0

DEci  B(-2,4)  si  y`B(2,0)

Explicație pas cu pas:

[tex]\it d:\ \ x+y-2=0 \Rightarrow y=2-x\ \ \ \ \ (1)\\ \\ B(x,\ y)\in d\ \stackrel{(1)}{\Longrightarrow} B(x,\ 2-x)\ \ \ \ \ \ (2)\\ \\ A(0,\ 2)\in d\\ \\ AB=2\sqrt2 \Rightarrow AB^2=8 \Rightarrow (x-0)^2+(2-x-2)^2=8 \Rightarrow x^2+x^2=8 \Rightarrow \\ \\ \Rightarrow 2x^2=8|_{:2} \Rightarrow x^2=4 \Rightarrow \sqrt{x^2}=\sqrt4 \Rightarrow |x|=2 \Rightarrow x=\pm2[/tex]

[tex]\it \ x=-2\ \stackrel{(2)}{\Longrightarrow}\ B(-2,\ 4)\\ \\ x=2\ \stackrel{(2)}{\Longrightarrow}\ B(2,\ 0)[/tex]