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Rozwiązane

4. Dane są punkty A = (-2;0) i B = (6;4 ). Długość odcinka AB to
: a.
[tex]2 \sqrt{6} [/tex]

b.80
c.10
d.
[tex]4 \sqrt{5} [/tex]
​

Odpowiedź :

ZbiorJviz

[tex]A=(-2,0),~~B=(6,4)\\\\\mid AB \mid = \sqrt{(6-(-2))^{2} +(4-0)^{2} } =\sqrt{(6+2)^{2} +4^{2} }=\sqrt{8^{2} +4^{2} }=\sqrt{64+16} =\sqrt{80} =\sqrt{16\cdot 5 } =4\sqrt{5} \\\\\\Odp:~~D.~~4\sqrt{5}[/tex]

Basetlaviz

[tex]A = (-2,0) \ \ \rightarrow \ \ x_{A} = -2, \ \ y_{A} = 0\\B = (6,4) \ \ \ \rightarrow \ \ x_{B} = 6, \ \ y_{B} = 4\\\\\\|AB| = \sqrt{(x_{B}-x_{A})^{2}+(y_{B}-y_{A})^{2}}}\\\\\\|AB| = \sqrt{(6-(-2))^{2}+(4-0)^{2}}} = \sqrt{8^{2}+4^{2}} = \sqrt{64+16} = \sqrt{80}=\\\\=\sqrt{16\cdot5} = 4\sqrt{5}\\\\\\\boxed{d) \ |AB| = 4\sqrt{5}}[/tex]

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