Question

Determine a distância entre o ponto P e a reta r, sabendo que os pontos A(-6,0) e B(-1,7) pertençam a reta r. P(0,2)

Answer 1

Resposta:

$\textsf{Leia abaixo}$

Explicação passo a passo:

$\begin{bmatrix}\cancel x&\cancel y&\cancel1\\\cancel -6&\cancel0&\cancel1\\\cancel-1&\cancel7&\cancel1\end{bmatrix}$

$\mathsf{0 - y - 42 = 0 + 7x - 6y}$

$\boxed{\boxed{\mathsf{7x - 5y + 42 = 0}}}\leftarrow\textsf{reta r}$

$\boxed{\boxed{\mathsf{P(0;2)}}}$

$\mathsf{d_{p,r} = |\:\dfrac{a.x_0 + b.y_0 +c}{\sqrt{a^2 + b^2}}\:|}$

$\mathsf{d_{p,r} = |\:\dfrac{(7)(0) + (-5)(2) +42}{\sqrt{7^2 + (-5)^2}}\:|}$

$\mathsf{d_{p,r} = |\:\dfrac{0 - 10 +42}{\sqrt{49 + 25}}\:|}$

$\mathsf{d_{p,r} = |\:\dfrac{32}{\sqrt{74}}\:|}$

$\boxed{\boxed{\mathsf{d_{p,r} = \dfrac{16\sqrt{74}}{37}}}}$

Answer 2

$\large\boxed{\begin{array}{l}\rm\begin{vmatrix}\rm-6&\rm0&\rm1\\\rm-1&\rm7&\rm1\\\rm x&\rm y&\rm1\end{vmatrix}=0\\\rm -6(7-y)+1\cdot(-y-7x)=0\\\rm -42+6y-y-7x=0\\\rm r: 7x-5y+42=0\\\rm d_{P,r}=\dfrac{|ax_P+by_P+c|}{\sqrt{a^2+b^2}}\\\\\rm d_{P,r}=\dfrac{|7\cdot0-5\cdot2+42|}{\sqrt{7^2+(-5)^2}}\\\\\rm d_{P,r}=\dfrac{32}{\sqrt{49+25}}\\\\\rm d_{P,r}=\dfrac{32}{\sqrt{74}}\cdot\dfrac{\sqrt{74}}{\sqrt{74}}\end{array}}$

$\large\boxed{\begin{array}{l}\rm d_{P,r}=\dfrac{\diagdown\!\!\!\diagdown\!\!\!\!\!\!\!32^{16}\sqrt{74}}{\diagdown\!\!\!\diagdown\!\!\!\!\!\!74_{37}}\\\huge\boxed{\boxed{\boxed{\boxed{\rm\red{ d_{P,r}=\dfrac{16\sqrt{74}}{37}}}}}}\end{array}}$