Question

Geometria Analítica : Distância do Ponto à reta:

Determine os pontos da retas: 2x-y-1=0 cuja distância à reta r: x-2y+3=0 é igual a 2

Answer 1

Existem um ponto $\text x_o,\text y_o$ da reta s cuja a distância até a reta r é igual a 2.

Reta s

$\text{s :}\ 2\text x -\text y-1=0$

substituindo as coordenadas :

$\text s: 2\text x_o-\text y_o-1=0 } \\\\\text s: \boxed{\text y_o = 2\text x_o-1 }$

Temos a reta r :

$\text{r : x}-2\text y+3$

Distância do ponto à reta :

$\displaystyle \text D = \frac{|\text{a.x}_o+\text {b.y}_o+\text c|}{\sqrt{\text a^2+\text b^2}}$

Substituindo os respectivos valores :

$\displaystyle 2 = \frac{|1.\text x_o-2.\text y_o+3|}{\sqrt{1^2+(-2)^2}} \\\\\\ 2=\frac{|\text x_o-2\text y_o+3|}{\sqrt{5}} \\\\\\ |\text x_o-2\text y_o+3| = 2\sqrt{5}$

porém, vimos que : $\text y_o=2\text x_o-1$

Substituindo :

$|\text x_o-2( 2\text x_o-1)+3| = 2\sqrt 5$

$|\text x_o-4\text x_o+2+3| = 2\sqrt 5 \\\\ |-3\text x_o+5| = 2\sqrt5$

Aplicando a definição de módulo temos :

$\displaystyle -3\text x_o+5 = 2\sqrt5 \ \ \ \ \ \ \ \ \ \text{ou} \ \ \ \ \ \ \ \ -3\text x_o+5=-2\sqrt 5 \\\\-3\text x_o = 2\sqrt{5}-5 \ \ \ \ \ \ \ \ \text{ou} \ \ \ \ \ \ \ \ -3\text x_o=-5-2\sqrt{5} \\\\\boxed{\text x_o = \frac{5-2\sqrt5}{3}} \ \ \ \ \ \ \ \ \ \text{ou} \ \ \ \ \ \ \ \ \ \boxed{\text x_o=\frac{5+2\sqrt{5}}{3} }$

Então, temos que  :

$\displaystyle \text y_o=2\text x_o-1 \\\\ \text y_o=2.(\frac{5-2\sqrt5}{3}) -1 \ \ \ \ \ \ \ \text{ou} \ \ \ \ \ \ \ \text y_o = 2(\frac{5+2\sqrt5}{3})-1\\\\\\ \text y_o = \frac{10-4\sqrt5-3}{3} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{ou} \ \ \ \ \ \ \ \ \ \ \text y_o = \frac{10+4\sqrt5-3}{3} \\\\\\\boxed{\text y_o = \frac{7-4\sqrt5}{3}} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text {ou} \ \ \ \ \ \ \ \ \boxed{\text y_o = \frac{7+4\sqrt5}{3}}$

Soluções :

$\huge\boxed{\text x_\text o = \frac{5-2\sqrt5}{3} \ \ ; \ \ \text y_\text o = \frac{7-4\sqrt{5}}{3}\ }\checkmark$

E

$\huge\boxed{\text x_\text o = \frac{5+2\sqrt5}{3} \ \ ; \ \ \text y_\text o = \frac{7+4\sqrt{5}}{3}\ }\checkmark$