Question

Ex 15 va rog mult de tottt a) si b)

Answer 1

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

a)Fie M,N şi P mijloacele laturilor AB,BC,respectiv AC.

M mijlocul lui AB :

$x_{M}=\frac{x_{A}+x_{B}}{2} = \frac{1 + 3}{2} = \frac{4}{2} = 2$

$y_{M}=\frac{y_{A}+y_{B}}{2} = \frac{3 + ( - 3)}{2} = \frac{3 - 3}{2} = \frac{0}{2} = 0$

$= >M(2,0)$

N mijlocul lui BC :

$x_{N}=\frac{x_{B}+x_{C}}{2} = \frac{3 + ( - 5)}{2} = \frac{3 - 5}{2} = \frac{ - 2}{2} = - 1$

$y_{N}=\frac{y_{B}+y_{C}}{2} = \frac{ - 3 + 1}{2} = \frac{ - 2}{2} = - 1$

$= > N(-1,-1)$

P mijlocul lui AC :

$x_{P}=\frac{x_{A}+x_{C}}{2} = \frac{1 + ( - 5)}{2} = \frac{1 - 5}{2} = \frac{ - 4}{2} = - 2$

$y_{P}=\frac{y_{A}+y_{C}}{2} = \frac{3 + 1}{2} = \frac{4}{2} = 2$

$= > P(-2,2)$

$MN=\sqrt{{(x_{N}-x_{M})}^{2}+{(y_{N}-y_{M})}^{2}}$

$MN = \sqrt{ {( - 1 - 2)}^{2} + {( - 1 - 0)}^{2} }$

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

$MN = \sqrt{9 + 1}$

$MN = \sqrt{10}$

$MP=\sqrt{{(x_{P}-x_{M})}^{2}+{(y_{P}-y_{M})}^{2}}</p><p>$

$MP = \sqrt{ {( - 2 - 2)}^{2} + {(2 - 0)}^{2} }$

$MP = \sqrt{ {( - 4)}^{2} + {2}^{2} }$

$MP = \sqrt{16 + 4}$

$MP = \sqrt{20}$

$MP = 2 \sqrt{5}$

$NP=\sqrt{{(x_{P}-x_{N})}^{2}+{(y_{P}-y_{N})}^{2}}</p><p>$

$NP = \sqrt{ {[ - 2 - ( - 1)]}^{2} + {[2 - ( - 1)]}^{2} }$

$NP = \sqrt{ {( - 2 + 1)}^{2} + {(2 + 1)}^{2} }$

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

$NP = \sqrt{1 + 9}$

$NP = \sqrt{10}$

$MN=NP=>\Delta MNP \: isoscel$

$P=MN+MP+NP = \sqrt{10} + 2 \sqrt{5} + \sqrt{10} = 2 \sqrt{5} + 2 \sqrt{10} = 2( \sqrt{5} + \sqrt{10} )$

b)Fie G centrul de greutate al triunghiului ABC :

$x_{G}=\frac{x_{A}+x_{B}+x_{C}}{3} = \frac{1 + 3 + ( - 5)}{3} = \frac{4 - 5}{3} = - \frac{1}{3}$

$y_{G}=\frac{y_{A}+y_{B}+y_{C}}{3}</p><p> = \frac{3 + ( - 3) + 1}{3} = \frac{3 - 3 + 1}{3} = \frac{1}{3}$

$= > G( - \frac{1}{3} , \frac{1}{3} )$