Question

ATENÇÃO cade os gênios de matemática agora.
Calcule a distância do ponto A(1,2) a reta definida por B(5,7) e C(1,1).

Answer 1

$\bullet~~$ Encontrando a equação da reta que passa pelos pontos $B(5,\;7)$ e $C(1,\;1):$

$r:~\dfrac{y-y_{_{B}}}{x-x_{_{B}}}=\dfrac{y_{_{C}}-y_{_{B}}}{x_{_{C}}-x_{_{B}}}\\\\\\ r:~\dfrac{y-7}{x-5}=\dfrac{1-7}{1-5}\\\\\\ r:~\dfrac{y-7}{x-5}=\dfrac{-6}{-4}\\\\\\ r:~\dfrac{y-7}{x-5}=\dfrac{3}{2}\\\\\\ r:~2\,(y-7)=3\,(x-5)\\\\ r:~2y-14=3x-15$

$r:~0=3x-15-2y+14\\\\ \boxed{\begin{array}{c} r:~3x-2y-1=0 \end{array}}~~\Rightarrow~~\left\{\begin{array}{l}a=3\\b=-2\\c=-1 \end{array} \right.$

__________________________

A distância do ponto $A(1,\;2)$ até a reta $r:$

(fórmula da distância do ponto à reta)

$d_{A,\,r}=\dfrac{|ax_{_{A}}+by_{_{A}}+c|}{\sqrt{a^2+b^2}}\\\\\\ d_{A,\,r}=\dfrac{|3x_{_{A}}-2y_{_{A}}-1|}{\sqrt{(3)^2+(-2)^2}}\\\\\\ d_{A,\,r}=\dfrac{|3x_{_{A}}-2y_{_{A}}-1|}{\sqrt{9+4}}\\\\\\ d_{A,\,r}=\dfrac{|3x_{_{A}}-2y_{_{A}}-1|}{\sqrt{13}}\\\\\\ d_{A,\,r}=\dfrac{|3\cdot 1-2\cdot 2-1|}{\sqrt{13}}\\\\\\ d_{A,\,r}=\dfrac{|3-4-1|}{\sqrt{13}}\\\\\\ d_{A,\,r}=\dfrac{|\!-2|}{\sqrt{13}}\\\\\\ \boxed{\begin{array}{c} d_{A,\,r}=\dfrac{2}{\sqrt{13}}~\mathrm{~u.c.} \end{array}}$