Daca 3 puncte sunt coliniare atunci determinantul este egal cu zero.
[tex]\Delta=0\\\\\Delta=\left|\begin{array}{ccc}1&a&1\\3&2&1\\2&1&1\end{array}\right|[/tex]
1 a 1
3 2 1
Δ=2+3+2a-(4+1+3a)=5+2a-5-3a=-a
-a=0
a=0
Sau
Aflam ecuatia dreptei AC, iar B trebuie sa apartina dreptei
[tex]ecuatia\ dreptei\ AC:\\\\\frac{x-x_A}{x_C-x_A} =\frac{y-y_A}{y_C-y_A} \\\\\frac{x-1}{2-1} =\frac{y-a}{2-a} \\\\(x-1)(1-a)=y-a\\\\y=(x-1)(1-a)+a\\\\f(x)=(x-1)(1-a)+a[/tex]
f(3)=2
2(1-a)+a=2
2-2a+a=2
-a=0
a=0
Un alt exercitiu asemanator gasesti aici: https://brainly.ro/tema/82872
#SPJ5
