Question

Qual deve ser o valor de (k) para que o ponto (1:1) pertença á recta da equação (k-2)x-4y+20=0

Answer 1

$(k-2)x-4y+20=0\\\\\\\\(k-2)\cdot 1-4\cdot 1+20=0\\\\\\\\(k-2)-4+20=0\\\\\\\\(k-2)+16=0\\\\\\\\k-2=-16\\\\\\\\k=-16+2\\\\\\\\\boxed{k= -14}\\\\\\\\\text{D\'uvidas? Comente.}$

Answer 2

Usa o método de substituição :

Os pontos (1,1):

Substitui entre x,y:

K-2 × x-4y+ 20=0
K-2 ×1-4 ×1 +20=0
K-2 -4 +20=0
K-2 +16=0
K-2 =-16
K=-16+2
K=-14

Precisando de qualquer coisa estou aqui .