Resposta:
[tex]\boxed{d_{A,r}=\dfrac{1}{\sqrt{2}}\ u.c.}[/tex]
Explicação passo-a-passo:
[tex]r:x+y-1=0\ \therefore\ (ax+by+c)\equiv (x+y-1)[/tex]
[tex]A(x_A,y_A)=A(0,2)[/tex]
[tex]\boxed{d_{A,r}=\dfrac{|ax_A+by_A+c|}{\sqrt{a^2+b^2}}}[/tex]
[tex]d_{A,r}=\dfrac{|1(0)+1(2)-1|}{\sqrt{1^2+1^2}}\ \therefore\ \boxed{d_{A,r}=\dfrac{1}{\sqrt{2}}\ u.c.}[/tex]
