Question

Determinati ecuatia dreptei ce trece prin punctul A(2 , -3) , si are panta m=2/3 (geometria clasa a X-a)
VA ROG,AM NEVOIE DE AJUTOR RAPID!!!!!

Answer 1

Răspuns:

Explicație pas cu pas:

y-(-3) = 2/3 (x-2)

y+3 = 2x/3 - 4/3

y = 2x/3 - 4/3 - 3

y = 2x/3 - 13/3

y = 2x/3 - 4 1/3

Answer 2

$\it d:\ y=mx+n\\ \\ m=\dfrac{2}{3} \Rightarrow y= \dfrac{2}{3}x+n\ \ \ \ \ \ (1)\\ \\ \\ A(2,\ -3)\in d \Rightarrow -3=\dfrac{2}{3}\cdot2+n \Rightarrow -3= \dfrac{4}{3}+n \Rightarrow n=-3-\dfrac{4}{3} \Rightarrow \\ \\ \\ \Rightarrow n=-\dfrac{13}{3}\ \ \ \ \ \ (2)\\ \\ \\ (1),\ (2) \Rightarrow d:\ \ y=\dfrac{2}{3}x-\dfrac{13}{3}|_{\cdot3} \Rightarrow 3y=2x-13 \Rightarrow 2x-3y-13=0$