d₁ II d₂ daca ( m+1 ) / ( m+1 ) = ( 2 -m) / 2m ; ( 2 -m) / 2m =1
2 -m = 2m ; 2 = 2m +m ; 3m =2 ; m = 2/3
d₁ perpendic.d₂ daca ( m+1 ) ·( m +1 ) + ( 2 -m)· 2m =0
m² +2m +1 + 4m - 2m²=0
-m² + 6m +1 =0
m² - 6m -1 =0
Δ = 36 +4 =40 = 4·10
m₁= ( 6 - 2√10 ) /2 = 3 -√10
m₂= 3+√10