Question

Grupo 10 distância de um ponto a uma reta
A) calcule o valor absoluto da distância dos pontos às retas indicados abaixo :
B) resolva ​

Answer 1

$\Large\boxed{\begin{array}{l}\rm A)\\\tt a)~\sf P(3,4)~~r:4x+3y-36=0\\\sf D_{P,r}=\dfrac{|4\cdot3+3\cdot4-36|}{\sqrt{4^2+3^2}}\\\\\sf D_{P,r}=\dfrac{|12+12-36|}{\sqrt{16+9}}\\\\\sf D_{P,r}=\dfrac{12}{\sqrt{25}}\\\\\huge\boxed{\boxed{\boxed{\boxed{\sf D_{P,r}=\dfrac{12}{5}}}}} \end{array}}$

$\Large\boxed{\begin{array}{l}\tt b)~\sf P(2,-3)~~r:2x-4y+9=0\\\sf D_{P,r}=\dfrac{|2\cdot2-4\cdot(-3)+9|}{\sqrt{2^2+(-4)^2}}\\\\\sf D_{P,r}=\dfrac{29}{\sqrt{4+16}}\\\\\sf D_{P,r}=\dfrac{29}{\sqrt{20}}=\dfrac{29}{\sqrt{2^2\cdot5}}\\\\\sf D_{P,r}=\dfrac{29}{2\sqrt{5}}\cdot\dfrac{\sqrt{5}}{\sqrt{5}}\\\\\huge\boxed{\boxed{\boxed{\boxed{\sf D_{P,r}=\dfrac{29\sqrt{5}}{10}}}}}\end{array}}$

$\Large\boxed{\begin{array}{l}\tt c)~\sf P(0,0)~~r:3x-4y+15=0\\\sf D_{P,r}=\dfrac{15}{\sqrt{3^2+(-4)^2}}\\\\\sf D_{P,r}=\dfrac{15}{\sqrt{9+16}}\\\\\sf D_{P,r}=\dfrac{15}{\sqrt{25}}\\\\\sf D_{P,r}=\dfrac{15}{5}\\\\\huge\boxed{\boxed{\boxed{\boxed{\sf D_{P,r}=3}}}}\end{array}}$

$\Large\boxed{\begin{array}{l}\tt d)~\sf P(5,5)~~x-y+10=0\\\sf D_{P,r}=\dfrac{|1\cdot5-1\cdot5+10|}{\sqrt{1^2+(-1)^2}}\\\\\sf D_{P,r}=\dfrac{|\backslash\!\!\!5-\backslash\!\!\!5+10|}{\sqrt{1+1}}\\\\\sf D_{P,r}=\dfrac{10}{\sqrt{2}}\cdot\dfrac{\sqrt{2}}{\sqrt{2}}\\\\\sf D_{P,r}=\dfrac{10\sqrt{2}}{2}\\\\\huge\boxed{\boxed{\boxed{\boxed{\sf D_{P,r}=5\sqrt{2}}}}}\end{array}}$

$\Large\boxed{\begin{array}{l}\rm B\\\tt a)~\sf m=\dfrac{7-5}{1-6}=-\dfrac{2}{5}\\\\\sf y=7+\bigg(-\dfrac{2}{5}\bigg)(x-1)\\\\\sf y=7-\dfrac{2}{5}x+\dfrac{2}{5}\\\sf 5y=35-2x+2\\\sf 2x+5y-35-2=0\\\sf r:2x+5y-37=0~~A(3,1)\\\sf D_{A,r}=\dfrac{|2\cdot3+5\cdot1-37|}{\sqrt{2^2+5^2}}\\\\\sf D_{A,r}=\dfrac{26}{\sqrt{4+25}}\\\\\sf D_{A,r}=\dfrac{26}{\sqrt{29}}\cdot\dfrac{\sqrt{29}}{\sqrt{29}}\\\\\sf D_{A,r}=\dfrac{26\sqrt{29}}{29}\end{array}}$

$\Large\boxed{\begin{array}{l}\tt b)~\sf u:5x-2y-7=0~~v:3x+y-13=0\\\begin{cases}\sf 5x-2y=7\\\sf 3x+y=13\cdot(2)\end{cases}\\\\+\underline{\begin{cases}\sf5x-\diagdown\!\!\!\!\!\!2y=7\\\sf 6x+\diagdown\!\!\!\!\!\!2y=26\end{cases}}\\\sf 11x=33\\\sf x=\dfrac{33}{11}\\\\\sf x=3\\\sf3x+y=13\\\sf 3\cdot3+y=13\\\sf 9+y=13\\\sf y=13-9\\\sf y=4\\\sf u\cap v= P(3,4)~~r:2x-3y+12=0\\\sf D_{P,r}=\dfrac{|2\cdot3-3\cdot4+12|}{\sqrt{2^2+(-3)^2}}\\\\\sf D_{P,r}=\dfrac{6}{\sqrt{4+9}}=\dfrac{6}{\sqrt{13}} \end{array}}$

$\Huge\boxed{\boxed{\boxed{\boxed{\sf D_{P,r}=\dfrac{6\sqrt{13}}{13}}}}}$