Question

   In reperul cartezian xOy se considera punctele  A(1,2), B(5,6), C(-1,1). Sa se determine ecuatia medianei duse din varful C al triunghiului ABC.

Answer 1

ΔABC
M∈ [AB]
M-mij lui [AB]  <=> [AM]=[MB]
[CM-mediana

M(x;y) 

$x_M =\frac{x_A+x_B}{2} = \frac{1+5}{2}= \frac{6}{2}=3 \\ y_M= \frac{y_A+Y_B}{2}= \frac{2+6}{2}= \frac{8}{2}= 4 =>M(3;4) \\ C(-1;1) \ & \ M(3;4) \\ CM= \sqrt{(x_M -x_C)^2 +(y_M-y_C)^2} = \\ CM= \sqrt{(3+1)^2 +(4-1)^2 } =\\ CM= \sqrt{4^2+3^2}= \sqrt{16+9}= \sqrt{25}=5$
$\left[\begin{array}{ccc}X&Y&1\\-1&1&1\\3&4&1\end{array}\right] =0 ; \ X_C=-1 \ ; \ X_M=3 \ ; \ Y_C=1 \ ; \ Y_M=4 \\ X*1*1 +(-1)*4*1+3*Y*1 - 3 -1*4*X-1*Y*(-1)=0 \\ (1-4)X +(3+1)Y+(-1)*4-1*3=0 => \ecuatia \ dreptei \ este \ : \\ -3x+4y-7=0 |-1 \\ 3x-4y+7=0$

Answer 2

 1)Avem Δ ABC,unde M∈ [AB]
M-mijlocul lui [AB] , [CM]-mediana
M(x,y)
xM=$\frac{xa}{xb}$/2=6/2=3
yM =$\frac{ya}{yb} /2$=8/2=4
M(3,4)
CM=√(xM-Xc)^2+(yM-yC)^2=√(3+1)+(4-1)^2=√25 =5
Xc=-1
XM=3
Yc=1
YM=4
 (1-4)x+(3+1)y-4-3=0
3x-4y+7=0 - ecuatia medianei