Daca afixele celor doua varfuri sunt 1+i si 3+2i, atunci coordonatele lor sunt
A(1, 1) si B(3, 2)
Trebuie sa-l aflam pe C(a, b)
ABC - echilateral ==> AB = BC = AC
[tex]AB=\sqrt{(x_B-x_A)^2+(y_B-y_A)^2}=\sqrt{2^2+1^2}=\sqrt{5}[/tex]
[tex]BC=\sqrt{(x_C-x_B)^2+(y_C-y_B)^2}=\sqrt{(a-3)^2+(b-2)^2}[/tex]
[tex]AC=\sqrt{(x_C-x_A)^2+(y_C-y_A)^2}=\sqrt{(a-1)^2+(b-1)^2}[/tex]
[tex]\sqrt{(a-1)^2+(b-1)^2}=\sqrt{(a-3)^2+(b-2)^2}=\sqrt{5}\\ (a-1)^2+(b-1)^2=(a-3)^2+(b-2)^2=5\\ (a^2-2a+1)+(b^2-2b+1)=(a^2-6a+9)+(b^2-4b+4)\\\\ 4a+2b=11\rightarrow b=\frac{11-4a}{2} \\\\ (a-1)^2+((\frac{11-4a}{2})-1)^2=5\\\\
(a^2-2a+1)+(\frac{16a^2-88a+121}{4}-(11-4a)+1)=5\\
(4a^2-8a+4)+(16a^2-88a+121)+16a-40=5\cdot4\\
20a^2-80a+65=0\\
4a^2-16a+13=0\\
\Delta=256-4\cdot4\cdot13=48\\
a_{1,2}=\frac{16\pm4\sqrt{3}}{8}=\boxed{\frac{4\pm\sqrt{3}}{2}}\\
[/tex]
[tex] b=\frac{11-4a}{2}=\boxed{\frac{3\mp2\sqrt{3}}{2}\\}
[/tex]
[tex]C:\frac{4\pm\sqrt{3}}{2}+\frac{3\mp2\sqrt{3}}{2}i[/tex]
