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Geometria analityczna

Geometria Analityczna class=

Odpowiedź :

1.

[tex]|AB|=\sqrt{\left(x_A-x_B\right)^2+\left(y_A-y_B\right)^2}\\|AB|=\sqrt{\left(4-(-1)\right)^2+\left(-3-5\right)^2}=\sqrt{5^2+\left(-8\right)^2}=\sqrt{25+64}=\sqrt{89}[/tex]

2.

[tex]|OM|=\sqrt{(-7)^2+6^2}=\sqrt{49+36}=\sqrt{85}[/tex]

13.

[tex]S=(5,5)\\A=(m,3)\\B=(4,k-3)[/tex]

[tex]S=\left(\cfrac{x_1+x_2}{2},\cfrac{y_1+y_2}{2}\right)\\(5,5)=\left(\cfrac{m+4}{2},\cfrac{3+k-3}{2}\right)\\5=\cfrac{m+4}{2}\\10=m+4\\m=6\\5=\cfrac{3+k-3}{2}\\k=10[/tex]

14.

[tex]a=|AB|=\sqrt{\left(-5-4\right)^2+\left(3-(-1)\right)^2}=\sqrt{\left(-9\right)^2+4^2}=\sqrt{81+16}=\sqrt{97}\\l=3a=3\sqrt{97}\\S=\cfrac{a^2\sqrt{3}}{4}=\cfrac{\left(\sqrt{97}\right)^2\sqrt{3}}{4}=\cfrac{97\sqrt{3}}{4}[/tex]

15.

[tex]a=|AB|=\sqrt{\left(-5-1\right)^2+\left(11-(-3)\right)^2}=\sqrt{\left(-6\right)^2+14^2}=\sqrt{36+196}\\=\sqrt{232}=2\sqrt{58}\\d=a\sqrt{2}=2\sqrt{58}*\sqrt{2}=2\sqrt{116}=4\sqrt{29}[/tex]

16.

[tex]|AB|=\sqrt{\left(x_A-x_B\right)^2+\left(y_A-y_B\right)^2}\\|AB|=\sqrt{\left(-5-(-3)\right)^2+\left(-1-5\right)^2}=\sqrt{\left(-2\right)^2+\left(-6\right)^2}=\sqrt{4+36}\\=\sqrt{40}=2\sqrt{10}[/tex]