Question

Descrição da imagem não disponível Qual a distância do ponto P(2,6) à reta y=-3x+10 ?​​​​​​​

Answer 1

Resposta:

D=√10/5

Explicação passo-a-passo:

P(2,6) à reta y=-3x+10 ?

3x+y-10=0

D=| 3.(2)+6-10|/√(3)²+(1)²

D=|6-4|/√9+1

D=|2|/√10

D=2/√10 ( racionalizando)

D=2√10/10 ( simplificando)

D=√10/5

Answer 2

$\boxed{\begin{array}{l}\underline{\sf Dist\hat ancia~do~ponto~P(x_P,y_P)}\\\underline{\sf\grave a~reta~r:ax+by+c=0}.\\\rm D_{P,r}=\dfrac{| ax_P+y_P+c|}{\sqrt{a^2+b^2}}\\\rm y=-3x+10\\\rm r: 3x+y-10=0\\\rm P(2,6)\begin{cases}\rm x_P=2\\\rm y_P=6\\\rm D_{P,r}=?\end{cases}\end{array}}$

$\boxed{\begin{array}{l}\rm D_{P,r}=\dfrac{|ax_P+by_P+c|}{\sqrt{a^2+b^2}}\\\\\rm D_{P,r}=\dfrac{|3\cdot2+1\cdot6-10|}{\sqrt{3^2+1^2}}\\\\\rm D_{P,r}=\dfrac{|6+6-10|}{\sqrt{9+1}}\\\\\rm D_{P,r}=\dfrac{2}{\sqrt{10}}\cdot\dfrac{\sqrt{10}}{\sqrt{10}}\\\\\rm D_{P,r}=\dfrac{2\sqrt{10}}{\sqrt{10^2}}\\\\\rm D_{P,r}=\dfrac{\diagdown\!\!\!\!2\sqrt{10}}{\diagdown\!\!\!\!\!\!10}\end{array}}$

$\boxed{\begin{array}{l}\huge\boxed{\boxed{\boxed{\boxed{\rm D_{P,r}=\dfrac{\sqrt{10}}{5}}}}}\end{array}}$