Question

Determine a distâncias entre P (2,1) e (r) x+2y-14=0

Answer 1

$\text{Dist\^ancia entre um ponto e uma reta.}\\\\\text{Equa\c{c}\~ao geral da reta:}~~Ax+By+C=0\\\\\text{A reta r do exerc\'icio j\'a est\'a em sua forma geral:}~~1x+2y-14=0\\\\\text{Coordenadas do ponto fornecido pelo exerc\'icio:}~~P~(2,1)\\\\\text{F\'ormula para se calcular a dist\^ancia de um ponto a uma reta:}\\\\\\d=\begin{vmatrix} \dfrac{Ax_0+By_0+C}{\sqrt{A^2+B^2}} \end{vmatrix}\\\\\\\text{Onde}~x_0~\text{e}~y_0~\text{s\~ao as coordenadas do ponto.}$


$\text{Substituindo os dados e resolvendo:}\\\\\\d=\begin{vmatrix} \dfrac{1\cdot 2+2\cdot 1+(-14)}{\sqrt{1^2+2^2}} \end{vmatrix}\\\\\\d=\begin{vmatrix} \dfrac{2+2-14)}{\sqrt{1+4}} \end{vmatrix}\\\\\\d=\begin{vmatrix} \dfrac{-10}{\sqrt{5}} \end{vmatrix}\\\\\\d=\begin{vmatrix} \dfrac{-10}{\sqrt{5}}\cdot\dfrac{\sqrt{5}}{\sqrt{5}}\end{vmatrix}\\\\\\d=\begin{vmatrix} \dfrac{-10\sqrt{5}}{\sqrt{5^2}}\end{vmatrix}\\\\\\d=\begin{vmatrix} \dfrac{-10 \sqrt{5}}{5}\end{vmatrix}$

$d=\begin{vmatrix} -2\sqrt{5} \end{vmatrix}\\\\\large\fbox{\fbox{ d=2\sqrt{5} }}$