Bălţătescu Ştefana I.PROBLEME ALGEBRA Ex. 160./ e)A(1,3), B(3,6);(AB)=? ( AB ) = | | x y 1 1 3 1 3 6 1 | =0 ↔ | x y 1 1− x 3− y 0 3− x 6− y 0 | | =0 ( 1−x )( 6−y ) −( 3−y )( 3−x )=0 ↔ 6−y−6 x+xy −9+ 3 x+ 3 y−xy=0 ↔ ( AB ) =−3 x +2 y−3=0 Ex.169/d)A ( −1,1) ;B ( 0,3 ) ;C ( 4,1 ) ;D ( 0 ,−2) ;S=? S ABCD=S ABC + S ACD S ABC = 1 2 ‖ −1 1 1 0 3 1 4 1 1 ‖ = ‖ −1 0 0 0 3 1 4 5 5 ‖ = 1 2 | (−1 ) 10 | =5 S ACD=¿ 1 2 ‖ −1 1 1 4 3 1 0 −2 1 ‖ = 1 2‖ −1 0 1 4 0 1 0 −3 1 ‖ = 1 2 | 3 | −1 1 4 1 | | = 3(−7) 2 = −21 2 ¿ S ABCD=5− 21 2 = 31 5 Ex. 162./ d) A ( a, 2) ,B ( 1 , 1 +a) ,C ( 3,4 ) : puncte distinctesi coliniare ; a=? | a 2 1 1 1+ a 1 3 4 1 | =0 ↔ | a 2 1 1−a a−1 0 3−a 2 0 | =0 ↔ 2 ( 1−a )−( 3−a)( a−1 ) → 2−2 a−3 a +3+a 2 −a=0 ↔a 2 −6 a+5=0 ∆ =36−20 =16 a 1 = 6 +4 3 =5 a 2 = 6−4 2 =1 Ex. 163. A(1,1); (d)=? Clasa a XI-A RD
