Question

2) A distância entre o ponto P(2, 4) e a reta (r) 3x – 4y + 1 = 0 é:

Answer 1

Explicação passo-a-passo:

$\sf d=\dfrac{|a\cdot x_0+b\cdot y_0+c|}{\sqrt{a^2+b^2}}$

$\sf d=\dfrac{|3\cdot2+(-4)\cdot4+1|}{\sqrt{3^2+(-4)^2}}$

$\sf d=\dfrac{|6-16+1|}{\sqrt{9+16}}$

$\sf d=\dfrac{|-9|}{\sqrt{25}}$

$\sf d=\dfrac{9}{5}$

$\sf d=1,8$

Answer 2

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$\Large\boxed{\boxed{\begin{array}{c}\sf{\underline{Dist\hat{a}ncia~do~ponto~\grave{a}~reta}}\end{array}}}$

$\huge\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\sf{D_{p,r}=\dfrac{|ax_p+by_p+c|}{\sqrt{a^2+b^2}}}}}}}}}\\\sf{sendo~x_p~e~y_p~coordenadas~do~ponto~P}\\\sf{e~a,b,c~coeficientes~da~reta~r:ax+by+c=0}$

$\sf{D_{p,r}=\dfrac{|3\cdot2-4\cdot4+1|}{\sqrt{3^2+(-4)^2}}}\\\sf{D_{p,r}=\dfrac{9}{\sqrt{9+16}}}\\\sf{D_{p,r}=\dfrac{9}{\sqrt{25}}}\\\huge\boxed{\boxed{\boxed{\boxed{\sf{D_{p,r}=\dfrac{9}{5}}}}}}$