Fie M mijlocul lui BC.
[tex] x_{M} = \frac{ x_{B+ x_{C} } }{2}= \frac{9+3}{2} =6[/tex]
[tex] y_{M} = \frac{ y_{B} + y_{C} }{2} = \frac{8+6}{2} = 7[/tex]
⇒ M ( 6,7 )
d(A,M) = [tex] \sqrt{ ( y_{M} - y_{A} )^{2} + ( x_{M} - x_{A} )^{2} } = \sqrt{ (7-6)^{2}+ (6+1)^{2} } = \sqrt{50} =5 \sqrt{2} [/tex]
