Odpowiedź:
[tex]a)\\\\A=(3,2)\ \ ,\ \ B=(3,-4)\\\\|AB|=\sqrt{(x_{B}-x_{A})^2+(y_{B}-y_{A})^2}=\sqrt{(3-3)^2+(-4-2)^2}=\sqrt{0^2+(-6)^2}=\\=\sqrt{0+36}=\sqrt{36}=6\\\\Odp.Dlugo\'s\'c\ \ odcinka\ \ |AB|\ \ jest\ \ r\'owna\ \ 6\\\\\\b)\\\\A=(-4,-3)\ \ ,\ \ B=(2,-1)\\\\|AB|=\sqrt{(x_{B}-x_{A})^2+(y_{B}-y_{A})^2}=\sqrt{(2-(-4))^2+(-1-(-3))^2}=\\\\=\sqrt{(2+4)^2+(-1+3)^2}=\sqrt{6^2+2^2}=\sqrt{36+4}=\sqrt{40}=\sqrt{4\cdot10}=2\sqrt{10}\\\\Odp.Dlugo\'s\'c\ \ odcinka\ \ AB\ \ jest\ \ r\'owna\ \ 2\sqrt{10}[/tex]
