Explicație pas cu pas:
A,B,C coliniare
[tex]\left|\begin{array}{ccc}1&2&1\\2&3&1\\3&a&1\end{array}\right| = 0[/tex]
[tex]1 \cdot (3 - a) - 2 \cdot (2 - 3) + 1 \cdot (2a - 9) = 0 \\ [/tex]
[tex]3 - a + 2 + 2a - 9 = 0[/tex]
[tex]a - 4 = 0 \implies \red {\bf a = 4}[/tex]
sau:
f(x) = ax + b
f(1) = 2 => a + b = 2
f(2) = 3 => 2a + b = 3
2a - a = 3 - 2 <=> a = 1 => b = 1
f(x) = x + 1
C(3,a): f(3) = a
3 + 1 = a => a = 4
