Question

Questão 01: Dada a reta “s” de equação 2x-5y+6= 0 e o ponto A(2;-3), calcule a distância entre

a reta “s” e o ponto A.

Questão 02: Calcule a distância entre os pontos a abaixo:

a) A(1;5) e B( 2;-4)

b) M(0;6) e N(1;5)

não me julguem, pfvr​

Answer 1

01

$s:2x-5y+6=0\ \therefore\ a=2,\ b=-5,\ c=6\\ A(x_A,y_A)=A(2,-3)\\\\ \boxed{d=\dfrac{|ax_A+by_B+c|}{\sqrt{a^2+b^2}}}$

$d_{s,A}=\dfrac{|2(2)-5(-3)+6|}{\sqrt{2^2+(-5)^2}}=\dfrac{|4+15+6|}{\sqrt{4+25}}\ \therefore\ \boxed{d_{s,A}=\dfrac{25}{\sqrt{29}}\ u.c.}$

02

$\boxed{d_{AB}^2=(x_B-x_A)^2+(y_B-y_A)^2}$

a)

$d_{AB}^2=(2-1)^2+(-4-5)^2=1^2+(-9)^2\ \therefore\\\\ d_{AB}=\sqrt{1+81}=\sqrt{82}\ \therefore\ \boxed{d_{AB}=\sqrt{82}\ u.c.}$

b)

$d_{MN}^2=(1-0)^2+(5-6)^2=1^2+(-1)^2\ \therefore\\\\ d_{MN}=\sqrt{1+1}=\sqrt{2}\ \therefore\ \boxed{d_{MN}=\sqrt{2}\ u.c.}$