Question

A(1.2) B(0.3) determina ecuatia dreptei

Answer 1

Explicație pas cu pas:

Pentru a determina ecuatia unei dreptei cand stim coordonatele a doua puncte de pe aceasta avem mai multe posibilitati.

Metoda 1 (cu determinant):

$AB: \left|\begin{array}{ccc}1&2&1\\0&3&1\\x&y&1\end{array}\right| =0\\AB: 3+0+2x-3x-y-0=0\\AB: -x-y+3=0\\AB: x+y-3=0$

Metoda 2 (cu formula de determinare a ecuatiei dreptei cand stim coordonatele a doua puncte):

$AB: \frac{x-x_A}{x_B-x_A}=\frac{y-y_A}{y_B-y_A}\\AB: \frac{x-1}{0-1}=\frac{y-2}{3-2}\\AB: (3-2)*(x-1)=(0-1)*(y-2)\\AB: x-1=-y+2\\AB: x+y-1-2=0\\AB: x+y-3=0$

Metoda 3 (gasind vectorul director al dreptei AB si punand conditia ca A sau B sa apartina dreptei):

Vectorul director este:

$\vec{AB}=(x_B-x_A)\vec{i}+(y_B-y_A)\vec{j}=-\vec{i}+\vec{j}$

Coordonatele vectorului director sunt:

$\vec{AB}=(-1,1)$.

Ecuatia dreptei va fi:

$AB: \frac{x-x_A}{x_{\vec{AB}}}=\frac{y-y_A}{y_{\vec{AB}}}\\AB: \frac{x-1}{-1}=\frac{y-2}{1}\\AB: x+y-3=0$

Sau:

$AB: \frac{x-x_B}{x_{\vec{AB}}}=\frac{y-y_B}{y_{\vec{AB}}}\\AB: \frac{x-0}{-1}=\frac{y-3}{1}\\AB: x+y-3=0$