[tex]A(2,1),\quad O(0,0) \\ \\d:~ax+by+c = 0 \\ O(0,0)\in d\Rightarrow 0 = 0+0+c \Rightarrow c = 0\Rightarrow ax+by+0 = 0\\ \\ d(A,d) = \dfrac{|a\cdot 2+b\cdot 1+0|}{\sqrt{a^2+b^2}} = 2 \Rightarrow\\ \\ \Rightarrow \dfrac{|2a+b|}{\sqrt{a^2+b^2}} = 2 \Rightarrow \dfrac{\sqrt{(2a+b)^2}}{\sqrt{a^2+b^2}}=2 \Rightarrow \dfrac{(2a+b)^2}{a^2+b^2}=4 \Rightarrow \\ \\\Rightarrow 4a^2+4ab+b^2 =4a^2+4b^2 \Rightarrow 3b^2 -4ab = 0[/tex]
[tex]\Rightarrow b = 0\quad sau\quad 3b-4a = 0 \\ \Rightarrow ax+0+0 = 0 \Rightarrow \boxed{d:~~x=0} \quad sau\quad 4a = 3b\Rightarrow \dfrac{a}{b} = \dfrac{3}{4} \\ \\ ax+by = 0 \Rightarrow by = -ax \Rightarrow y = -\dfrac{a}{b}x \Rightarrow y = -\dfrac{3}{4}x \Rightarrow\\ \\ \Rightarrow \boxed{d:\quad 3x+4y=0} \\ \\ \Rightarrow \text{Raspuns corect: }A)~~\text{alt raspuns}[/tex]
[tex]\text{M-am folosit de formula:} \\ \\ A(x_0,y_0),\quad d:~~ax+by+c = 0 \\ \\ d(A,d)= \dfrac{|a\cdot x_0+b\cdot y_0+c|}{\sqrt{a^2+b^2}}[/tex]
