Punctele A,B,C sunt coliniare daca toate 3 apartin aceleiasi drepte.
Metoda I
Aflam ecuatia dreptei AB si punem conditia ca punctul C sa se afle pe dreapta AB ----> punctul C trebuie sa verifice ecuatia dreptei.
Ecuatia dreptei AB se afla cu formula:
[tex]d: \ \frac{x-x_1}{x_2-x_1}=\frac{y-y_1}{y_2-y_1}[/tex]
Avem punctele A(2;3) si B(4;5) deci ecuatia dreptei este:
[tex]AB: \ \ \frac{x-2}{4-2}=\frac{y-3}{5-3}\ \textless \ =\ \textgreater \ \\\\ \ \textless \ =\ \textgreater \ \frac{x-2}{2}=\frac{y-3}{2} \ \textless \ =\ \textgreater \ \\\\ \ \textless \ =\ \textgreater \ x-2=y-3 \\\\\\ \underline{AB: \ x-y+1=0}[/tex]
Punem conditia ca punctul C sa verifice ecuatia dreptei:
[tex]C (m+1, m ^2) \in AB \Longrightarrow (m+1)-m^2+1=0 \\\\ -m^2+m+2=0 \\\\ \Delta=1^2-4*2*(-1)\to 1+8=9 \\\\ m_1=\frac{-1+3}{-2}\to -1 \\\\ m_2=\frac{-1-3}{-2}\to 2[/tex]
Obtinem solutia: [tex]S \in \{-1;2 \}[/tex]
Metoda II
Pentru a stabili daca punctele A,B,C sunt coliniare se verifica egalitatea folosind regula lui Sarrus:
[tex] \left[\begin{array}{ccc}x_1&y_1&1\\x_2&y_2&1\\x_3&y_3&1\end{array}\right] =0 \\\\ .\ \ \ x_1 \ \ \ y_1 \ \ \ 1 \\ . \ \ \ x_2 \ \ \ y_2 \ \ \ 1 [/tex]
Se aplica pentru exercitiul nostru formula:
[tex]x_1*y_2*1+x_2*y_3*1+x_3*y_1*1-1*y_2*x_3-1*y_3*x_1- \\ -1*y_1*x_2=0 \\\\ 2*5+4*m^2+(m+1)*3-5*(m+1)-m^2*2-3*4=0 \\\\ 10+4m^2+3m+3-5m-5-2m^2-12=0 \\\\ 2m^2-2m-4=0 \ \ |:2 \\\\ m^2-m-2=0 \\\\ \Delta=(-1)^2-4*(-2)\to 1+8 \to 9 \\\\ m_1= \frac{1+3}{2}\to 2 \\\\ m_2=\frac{1-3}{2}\to -1[/tex]
Si se obtine solutia anterioara.Raspuns final:
[tex]\boxed{\boxed{S \in \{-1;2 \}}}[/tex]
