Question

Determine a equação geral da reta s sabendo que ela é perpendicular a reta r: y = √ 2 x – 2 e que a reta s passa pelo ponto (-1, 0).

Answer 1

Temos a reta :

$\text r : \text y =\sqrt 2.\text x-2$

Reta perpendicular passando no ponto (-1,0)  :

$\text s: \text y- \text y_o=\text m(\text x-\text x_o) \\\\ \text s: \text y -0=\text m(\text x-(-1)) \\\\ \text s: \text y = \text m(\text x+1)$

se duas são perpendiculares, o produto de seus coeficientes angulares é igual a -1.

Achando o coeficiente angular :

$\displaystyle \text m.\sqrt{2} = -1 \\\\ \text m = \frac{-1}{\sqrt2} \\\\\\ \underline{\text m = \frac{-\sqrt 2}{2}}$

substituindo na equação da reta s :

$\displaystyle \text y = \frac{-\sqrt{2}}{2}(\text x+1) \\\\\\ \text y = \frac{-\sqrt{2}.\text x}{2} - \frac{\sqrt2}{2} \\\\\\ \frac{\sqrt{2}.\text x}{2} + \text y +\frac{\sqrt2}{2} =0\\\\\\ \huge\boxed{\text s: \sqrt2.\text x+ 2\text y+\sqrt 2 = 0 }\checkmark$