Rozwiązane

Oblicz obwód ABC o wierzchołkach
A=(-3,-2),B=(-1,3),C=(2,1)

Odpowiedź :

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[tex]A=(-3;-2)\\\\B=(-1;3)\\\\C=(2;1)[/tex]

[tex]|AB|=\sqrt{(-1-(-3))^2+(3-(-2))^2}=\\\\=\sqrt{(-1+3)^2+(3+2)^2}=\\\\=\sqrt{2^2+5^2}=\\\\=\sqrt{4+25}=\\\\=\sqrt{29}[/tex]

[tex]|AC|=\sqrt{(2-(-3))^2+(1-(-2))^2}=\\\\=\sqrt{(2+3)^2+(1+2)^2}=\\\\=\sqrt{5^2+3^2}=\\\\=\sqrt{25+9}=\\\\=\sqrt{34}[/tex]

[tex]|BC|=\sqrt{(2-(-1))^2+(1-3)^2}=\\\\=\sqrt{(2+1)^2+(-2)^2}=\\\\=\sqrt{3^2+3}=\\\\=\sqrt{9+4}=\\\\=\sqrt{13}[/tex]

[tex]L=|AB+|AC|+|BC|\\\\L=\sqrt{29}+\sqrt{34}+\sqrt{13}[/tex]

LoczeeQviz

Wzór na długość odcinka

[tex] \sqrt{(x_2 - x_1 {)}^{2} + (y_2 - y_1 {)}^{2} } [/tex]

[tex] |AB| = \sqrt{( (- 1) - ( - 3) {)}^{2} + (3 - ( - 2) {)}^{2} } = \sqrt{( - 1 + 3 {)}^{2} + (3 + 2 {)}^{2} } = \sqrt{ {2}^{2} + {5}^{2} } = \sqrt{4 + 25} = \sqrt{29} [/tex]

[tex] |AC| = \sqrt{(2 - ( - 3) {)}^{2} + (1 - ( - 2) {)}^{2} } = \sqrt{(2 + 3 {)}^{2} + (1 + 2 {)}^{2} } = \sqrt{ {5}^{2} + {3}^{2} } = \sqrt{25 + 9} = \sqrt{34} [/tex]

[tex] |BC| = \sqrt{(2 - ( - 1 ){)}^{2} + (1 - 3 {)}^{2} } = \sqrt{(2 + 1 {)}^{2} + ( - 2 {)}^{2} } = \sqrt{ {3}^{2} + 4} = \sqrt{9 + 4} = \sqrt{13} [/tex]

Obwód

[tex] \huge{\underline{ \sqrt{29} + \sqrt{34} + \sqrt{13} }}[/tex]

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