Bonjour ;
1)
a) (k² + 1) AGk + k BGk - k CGk = 0
donc : (k² + 1) AGk = - k BGk + k CGk = - k (BA + AGk) + k (CA +AGk)
= - k BA - k AGk + k CA + k AGk
= - k BA + k CA = k AB - k AC ,
donc : AGk = k/(k² + 1) AB - 5/(k² + 1) AC ,
donc : ∀ x ∈ R , Gk ∈ (ABC) .
b) On a : AGk = k/(k² + 1) AB - 5/(k² + 1) AC = k/(k² + 1) (AB - AC)
= k/(k² + 1) (AB + CA) = k/(k² + 1) CB = - k/(k² + 1) BC .
