Question

Determine o valor de k, para que as retas (r) (k +1)x - 3y - 5 = 0 e (s) 2x - ky +10 = 0 sejam parelelas.

Answer 1

Resposta:

Alternativa e)

Explicação passo-a-passo:

(r): (k +1)x - 3y - 5 = 0 => 3y=(k+1)x-5 => y=(k+1)x/3-5/3 => mr=(k+1)/3

(s): 2x - ky +10 = 0 => ky=2x+10 => y=2x/k+10/k => ms=2/k

Para que as retas r e s sejam paralelas:

mr=ms

(k+1)/3=2/k

k(k+1)=6

k²+k-6=0

$\displaystyle Aplicando~a~f\'{o}rmula~de~Bhaskara~para~k^{2}+k-6=0~~\\e~comparando~com~(a)k^{2}+(b)k+(c)=0,~temos~a=1{;}~b=1~e~c=-6\\\\\Delta=(b)^{2}-4(a)(c)=(1)^{2}-4(1)(-6)=1-(-24)=25\\\\k^{'}=\frac{-(b)-\sqrt{\Delta}}{2(a)}=\frac{-(1)-\sqrt{25}}{2(1)}=\frac{-1-5}{2}=\frac{-6}{2}=-3\\\\k^{''}=\frac{-(b)+\sqrt{\Delta}}{2(a)}=\frac{-(1)+\sqrt{25}}{2(1)}=\frac{-1+5}{2}=\frac{4}{2}=2\\\\S=\{-3,~2\}$