Question

para que as retas 2y-x-3=0 e 3y+kx-2=0 sejam perpendiculares, o valor de k deve ser:
(a)-2/3 (d)6
(b)2/3 (e)-6
(c)5

Answer 1

$2y-x-3=0\\y=\frac{x+3}{2}\\m1=\frac{1}{2}\\3y+kx-2=0\\y=\frac{-kx+2}{3}\\m2=\frac{-k}{3}\\m1.m2=-1\\\frac{1}{2}.\frac{-k}{3}=-1\\k=6$