Question

Se considera punctele A(5,2); B(-3,7); C(1,-5)
Determinati ecuatia dreptei AB.

Answer 1

[tex] \frac{x-x1}{x2-x1} = \frac{y-y1}{y2-y1} [/tex]

[tex] \frac{x-5}{-3-5}= \frac{y-2}{7-2} [/tex]

[tex] \frac{x-5}{-8}= \frac{y-2}{5} 5(x-5)=-8(y-2) 5x-25=-8y+16 5x-25+8y-16=0 5x+8y-41=0[/tex]

Metoda 2 cu determinant

$\left[\begin{array}{ccc}1&1&1\\1&x1&x2\\y&y1&y2\end{array}\right]$

[tex] \left[\begin{array}{ccc}1&1&1\\x&5&-3\\y&2&7\end{array}\right]= 35-3y+2x-(5y-6+7x)=0 35-3y+2x-5y+6-7x=0 -5x-8y+41=0 /(-1) 5x+8y-41=0 [/tex]