Question

Im reperul cartezian xOy se considera punctele A(2,-1) si B(-2,a),a€R. Sa se determine numarul real astfel incat dreapta AB sa contina punctul O(0,0)

Answer 1

$A(2,-1)\\ B(-2,a) \ \ ; \ a \in R \\\\\\ AB \ \ : \ \ \frac{x-x_1}{x_2-x_1}=\frac{y-y_1}{y_2-y_1} \\\\ AB \ : \ \frac{x-2}{-2-2}=\frac{y+1}{a+1} \\\\ (x-2)(a+1)= -4y-4 \\\\ xa+x-2a-2+4y+4=0 \\\\\\ \underline{AB \ : \ x(a+1)+4y+2-2a=0} \\\\\\ O(0;0) \in AB \ ==\ \textgreater \ 0*(a+1) + 4*0 +2-2a=0 \\\\ 2-2a=0 \\\\ \boxed{a=1}$

Answer 2

Aceasta este rezolvarea mea: