A(-2;4); B(2;2); C(-1;-1)
M mijlocul lui BC
M(xM; yM);
xM=(xB+xC)/2=(2-1)/2=1/2
yM=(yB+yC)/2=(2-1)/1=1/2
=> M(1/2; 1/2)
AM=√[(xM-xA)²+(yM-yA)²]
AM=√[(1/2+2)²+(1/2-4)²]= √(25/4+49/4)=√(74/4)=√74 / 2 u
D. √74 / 2 u
