Răspuns :
1) M-mijlocul lui [BC] , P-mijlocul lui [AB] N-mijlocul lui [AC] -> xM= xB+xC/2 yM= yB+yC /2 ->M(-3+4/2 , -1+5 /2 ) ->M(1/2 , 2) N(xA+xC/2 , yA+yC/2) -> N(1+4/2 , 9+5/2) -> N(5/2 , 7) P(xA+xB/2 , yA+yB/2) -> P(1-3/2 , 9-1/2 ) -> P(-1,4). 2) Fiindca e echilateral punem conditia ca : AB=AC=BC C(x,y). Acum calculam lungimile.. AB=[tex] \sqrt{(4-0) ^{2} + (1-1)^{2} } = \sqrt{16} [/tex] = 4 AC=[tex] \sqrt{(X-0)^{2}+(Y-1)^{2}} = \sqrt{ x^{2} + (Y-1)^{2} } [/tex] BC= [tex] \sqrt{(X-4)^{2}+(Y-1)^{2}} [/tex] Acum facem sistem. [tex] \left \{ {{AB=AC \atop {AB=BC}} \right. [/tex] [tex] \left \{ {{4= \sqrt{ x^{2} +(Y-1)^{2}} } \atop \ {4= \sqrt{(x-4)^{2}+(y-1)^{2} [/tex] Acum. luam si rezolvam AC=BC -> [tex] \sqrt{ x^{2} +(y-1)^{2}} = \sqrt{(x-4)^{2}+(y-1)^{2} [/tex] | ridicam la patrat. -> [tex] x^{2} [/tex] + [tex] y^{2} [/tex] -2y +1 = [tex] x^{2} [/tex] -8x+16 + [tex] y^{2} [/tex] -2y +1 |reducem termenii... astfel o sa ramana doar -> 1 = -8x+16+1 -> 8x=16 -> X=2. Am aflat prima necunoscuta care reprezinta coordonatul x. Acum luam AB=AC -> 4 = [tex] \sqrt{ x^{2} +(y-1) ^{2} } [/tex] | ridicam la patrat ->>>>> 16 = [tex] x^{2} [/tex] + [tex] y^{2} [/tex] -2y +1 -> [tex] x^{2} + y^{2} [/tex] -2y =16-1 -> [tex] x^{2} + y^{2} [/tex] -2y =15 -> stiim ca x =2 -> vom inlocui x-ul cu 2. ->>> [tex]2 ^{2} [/tex] + [tex] y^{2} [/tex] -2y = 15 ---> [tex] y^{2} [/tex] - 2y = 15-4 ->>>>>> [tex] y^{2} [/tex] -2y =11 ->>>> [tex] y^{2} [/tex] -2y -11 = 0 . Aceasta este o ecuatie de gradul 2. deci vom folosi formulele acesteia. Δ = b²-4ac. √ = 4+44 , Δ =48 y1,2 = [tex]y1= \frac{-b+4 \sqrt{3} }{2a} [/tex] y2=[tex] \frac{-b-4 \sqrt{3} }{2a} [/tex] y1 = [tex] \frac{2+4 \sqrt{3} }{2 } [/tex] si y2 = [tex] \frac{2-4 \sqrt{3} }{2} [/tex] y1 = 1+ 2√3 si y2 = 1 - 2√3 Deci C are 2 solutii si anume : C1(2, 1+2√3) si C2(2, 1 -2√3)
