Odpowiedź:
Szczegółowe wyjaśnienie:
Wzór na odległość punktu od prostej:
[tex]Q(x_Q,y_Q),\ Ax+By+C=0\\\\d=\dfrac{|Ax_Q+By_Q+C|}{\sqrt{A^2+B^2}}[/tex]
[tex]A(-1,\ 2)\\x-y+2=0\\\\d_1=\dfrac{|1\cdot(-1)-1\cdot2+2|}{\sqrt{(-1)^2+2^2}}=\dfrac{|-1-2+2|}{\sqrt{1+4}}=\dfrac{|-1|}{\sqrt5}\cdot\dfrac{\sqrt5}{5}=\dfrac{\sqrt5}{5}\\\\A(-1,\ 2)\\x+y=0\\\\d_2=\dfrac{|1\cdot(-1)+1\cdot2+0|}{\sqrt{1^2+1^2}}=\dfrac{|-1+2|}{\sqrt{1+1}}=\dfrac{|1|}{\sqrt2}\cdot\dfrac{\sqrt2}{\sqrt2}=\dfrac{\sqrt2}{2}[/tex]
[tex]\dfrac{\sqrt5}{5}<\dfrac{\sqrt2}{2}[/tex]
