Question

Calcule o valor de x sabendo que os pontos A(1/2, x) + B(- 1, 6) e C(2/3, 0) são colineares.

Answer 1

Explicação passo-a-passo:

Para três pontos serem colineares:

$A=\left[\begin{array}{ccc}x_{a} &y_{a}&1\\x_{b}&y_{b}&1\\x_{b}&y_{c}&1\end{array}\right]$

$detA=0$

Temos que A(1/2, x) , B(- 1, 6) e C(2/3, 0)

$A=\left[\begin{array}{ccc}\frac{1}{2} &x&1\\-1&6&1\\\frac{2}{3} &0&1\end{array}\right]$

$detA=0\\\\\frac{1}{2} .6.1+x.1.\frac{2}{3}+1.(-1).0-1.6.\frac{2}{3}-1.0.\frac{1}{2}-x.(-1).1=0\\\\3+\frac{2x}{3} -4+x=0\\\\ \frac{5x}{3}-1=0\\ \\\frac{5x}{3}=1\\\\5x=3\\\\x=\frac{3}{5}$