Odpowiedź :
Wystarczy skorzystać z tw. Pitagorasa:
|-4|²+|8|²=x²
16+64=x²
80=x²
[tex]x=\sqrt{80}=\sqrt{5\cdot16}=4\sqrt{5}[/tex]
[tex]P = (-4,8) \ \ \rightarrow \ \ x_{P} = -4, \ \ y_{P} = 8\\P_{o} = (0,0) \ \ \rightarrow \ \ \ x_{P_{o}} = 0, \ \ y_{P_{o}} = 0[/tex]
Długość odcinka:
[tex]d = \sqrt{(x_{P_{o}}-x_{P})^{2}+(y_{P_{o}}-y_{P})^{2}[/tex]
[tex]d = \sqrt{(0-(-4)^{2}+(0-8)^{2}}}=\sqrt{4^{2}+(-8)^{2}}} = \sqrt{16+64} = \sqrt{80} = \sqrt{16\cdot5} =\\\\= 8\sqrt{5}[/tex]
[tex]\boxed{Odp. \ C. \ 4\sqrt{5}}[/tex]