Question

se dau punctele A(1,3);B(-2,5);C(-2,-4)
a)calculează perimetru triunghiulABC
b)calculează lungimea medianei din A
c)natura ABC​

Answer 1

Răspuns:

$A(1,3),B(-2,5),C(-2,-4)$

$a)AB = \sqrt{{(x_{B}-x_{A})}^{2}+{(y_{B}-y_{A})}^{2}}$

$AB = \sqrt{ {( - 2 - 1)}^{2} + {(5 - 3)}^{2} }$

$AB = \sqrt{ {( - 3)}^{2} + {2}^{2} }$

$AB = \sqrt{9 + 4}$

$AB = \sqrt{13}$

$AC = \sqrt{{(x_{C}-x_{A})}^{2}+{(y_{C}-y_{A})}^{2}}$

$AC = \sqrt{ {( - 2 - 1)}^{2} + {( - 4 - 3)}^{2} }$

$AC = \sqrt{ {( - 3)}^{2} + {( - 7)}^{2} }$

$AC = \sqrt{9 + 49}$

$AC = \sqrt{58}$

$BC=\sqrt{{(x_{C}-x_{B})}^{2}+{(y_{C}-y_{B})}^{2}}</p><p>$

$BC=\sqrt{{[-2-(-2)]}^{2}+{(-4-5)}^{2}}$

$BC=\sqrt{{(-2 + 2)}^{2}+{(-9)}^{2}}$

$BC=\sqrt{0+81}$

$BC = \sqrt{81}$

$BC = 9$

$P=AB+AC+BC = \sqrt{13} + \sqrt{58} + 9$

$b)Fie\:AM\:mediana\:din\:A = > M\:mijlocul\:lui\:BC$

$x_{M}=\frac{{x_{B}+x_{C}}}{2} = \frac{ - 2 - 2}{2} = \frac{ - 4}{2} = - 2$

$y_{M}=\frac{{y_{B}+y_{C}}}{2} = \frac{5 - 4}{2} = \frac{1}{2}$

$= > M(-2, \frac{1}{2} )$

$AM=\sqrt{{(x_{M}-x_{A})}^{2}+{(y_{M}-y_{A})}^{2}}$

$AM = \sqrt{ {{( - 2 - 1)}^{2} + {( \frac{1}{2} - 3)}^{2} }}$

$AM = \sqrt{ { ( - 3)}^{2} + {( \frac{1}{2} - \frac{6}{2} )}^{2} }$

$AM = \sqrt{9 + {( - \frac{5}{2} )}^{2} }$

$AM = \sqrt{9 + \frac{25}{4} }$

$AM = \sqrt{ \frac{36}{4} + \frac{25}{4} }$

$AM = \sqrt{ \frac{61}{4} }$

$AM = \frac{ \sqrt{61} }{ \sqrt{4} }$

$AM = \frac{ \sqrt{61} }{2}$

c)Triunghiul ABC este un triunghi oarecare