Question

Se considera punctele A(3,2),B(-3,-4),C(1,5). Sa se scrie ecuatia dreptei care trece prin punctul A si este paralela cu dreapta BC.

Answer 1

$\it (BC):\ \dfrac{y-y_B}{y_C-y_B} = \dfrac{x-x_B}{x_C-x_B} \Rightarrow \dfrac{y+4}{5+4}=\dfrac{x+3}{1+3} \Rightarrow \dfrac{y+4}{9} = \dfrac{x+3}{4}\Rightarrow \\ \\ \\ \Rightarrow 4y+16=9x+27|_{-16} \Rightarrow 4y = 9x+11|_{:4} \Rightarrow y = \dfrac{9}{4}x +\dfrac{11}{4} \\ \\ \\ (BC) : y = \dfrac{9}{4}x +\dfrac{11}{4}$

$\it d||BC \Rightarrow (d): y=\dfrac{9}{4}x +n \\ \\ \\ A(3,\ 2) \in d \Rightarrow 2 = \dfrac{9}{4}\cdot3+n \Rightarrow n = 2-\dfrac{27}{4} \Rightarrow n = -\dfrac{19}{4} \\ \\ \\ (d): \ \ y = \dfrac{9}{4}x-\dfrac{19}{4}$