Răspuns:
[tex]A(1,3),B(-2,5),C(-2,-4)[/tex]
[tex]a)AB = \sqrt{{(x_{B}-x_{A})}^{2}+{(y_{B}-y_{A})}^{2}}[/tex]
[tex]AB = \sqrt{ {( - 2 - 1)}^{2} + {(5 - 3)}^{2} } [/tex]
[tex]AB = \sqrt{ {( - 3)}^{2} + {2}^{2} } [/tex]
[tex]AB = \sqrt{9 + 4} [/tex]
[tex]AB = \sqrt{13} [/tex]
[tex]AC = \sqrt{{(x_{C}-x_{A})}^{2}+{(y_{C}-y_{A})}^{2}}[/tex]
[tex]AC = \sqrt{ {( - 2 - 1)}^{2} + {( - 4 - 3)}^{2} } [/tex]
[tex]AC = \sqrt{ {( - 3)}^{2} + {( - 7)}^{2} } [/tex]
[tex]AC = \sqrt{9 + 49} [/tex]
[tex]AC = \sqrt{58} [/tex]
[tex]BC=\sqrt{{(x_{C}-x_{B})}^{2}+{(y_{C}-y_{B})}^{2}}
[/tex]
[tex]BC=\sqrt{{[-2-(-2)]}^{2}+{(-4-5)}^{2}}[/tex]
[tex]BC=\sqrt{{(-2 + 2)}^{2}+{(-9)}^{2}}[/tex]
[tex]BC=\sqrt{0+81}[/tex]
[tex]BC = \sqrt{81} [/tex]
[tex]BC = 9[/tex]
[tex]P=AB+AC+BC = \sqrt{13} + \sqrt{58} + 9[/tex]
[tex]b)Fie\:AM\:mediana\:din\:A = > M\:mijlocul\:lui\:BC[/tex]
[tex]x_{M}=\frac{{x_{B}+x_{C}}}{2} = \frac{ - 2 - 2}{2} = \frac{ - 4}{2} = - 2[/tex]
[tex]y_{M}=\frac{{y_{B}+y_{C}}}{2} = \frac{5 - 4}{2} = \frac{1}{2} [/tex]
[tex] = > M(-2, \frac{1}{2} )[/tex]
[tex]AM=\sqrt{{(x_{M}-x_{A})}^{2}+{(y_{M}-y_{A})}^{2}}[/tex]
[tex]AM = \sqrt{ {{( - 2 - 1)}^{2} + {( \frac{1}{2} - 3)}^{2} }} [/tex]
[tex]AM = \sqrt{ { ( - 3)}^{2} + {( \frac{1}{2} - \frac{6}{2} )}^{2} } [/tex]
[tex]AM = \sqrt{9 + {( - \frac{5}{2} )}^{2} } [/tex]
[tex]AM = \sqrt{9 + \frac{25}{4} } [/tex]
[tex]AM = \sqrt{ \frac{36}{4} + \frac{25}{4} } [/tex]
[tex]AM = \sqrt{ \frac{61}{4} } [/tex]
[tex]AM = \frac{ \sqrt{61} }{ \sqrt{4} } [/tex]
[tex]AM = \frac{ \sqrt{61} }{2} [/tex]
c)Triunghiul ABC este un triunghi oarecare
