Question

Determine a distância do ponto Q=(1,1) á reta t, cuja equação é x + y -3 = 0

Answer 1

                           $d=\dfrac{|1+1-3|}{\sqrt{1^2+1^2}}=\dfrac{1}{\sqrt{2}}$

Answer 2

$\mathrm{\Rightarrow Q(x_Q,y_Q)\ \to\ Q(1,1)}\\ \mathrm{\Rightarrow t:ax+by+c=0\ \to\ x+y-3=0}\\\\ \mathrm{d_{Q;t}=\dfrac{|ax_Q+by_Q+c|}{\sqrt{a^2+b^2}}\ \to\ d_{Q;t}=\dfrac{|1.1+1.1-3|}{\sqrt{1^2+1^2}}\ \to}\\\\\\ \mathrm{\to\ d_{Q;t}=\dfrac{|2-3|}{\sqrt{2}}\ \to\ d_{Q;t}=\dfrac{|-1|}{\sqrt{2}}\ \to\ d_{Q;t}=\dfrac{1}{\sqrt{2}}}\\\\\\ \mathrm{d_{Q;t}=\dfrac{1}{\sqrt{2}}\times\dfrac{\sqrt{2}}{\sqrt{2}}}\ \to\ \boxed{\mathbf{d_{Q;t}=\dfrac{\sqrt{2}}{2}\ u.c.}}$