Question

1) Considerando que a distância entre o ponto A(k,4) e a reta r, de equação 2x + 8y - 80 = 0, é igual a 6, qual é o valor de k?

2)Determine a altura relativa ao vértice A do triângulo A(1; 1), B(-1; -3) e C(2; -7).
Precisa dos cálculos, grato!

Answer 1

Resposta:

$\textsf{Leia abaixo}$

Explicação passo a passo:

$\sf{A(k,4)}$

$\sf{r:2x + 8y - 80 = 0}$

$\mathsf{d_{A,r} = 6}$

$\mathsf{d_{p,r} = |\:\dfrac{ax_0 + by_0 + c}{\sqrt{a^2 + b^2}}\:|}$

$\mathsf{6 = |\:\dfrac{2k + 8.4 - 80}{\sqrt{2^2 + 8^2}}\:|}$

$\mathsf{6 = |\:\dfrac{2k + 32 - 80}{\sqrt{4 + 64}}\:|}$

$\mathsf{6 = \dfrac{2k - 48}{2\sqrt{17}}}$

$\mathsf{12\sqrt{17} = 2k - 48}$

$\mathsf{2k = 48 + 12\sqrt{17}}$

$\boxed{\boxed{\mathsf{k = 24 + 6\sqrt{17}}}}$

$\textsf{A(1;1) B(-1;-3) C(2;-7)}$

$\textsf{reta suporte }\sf{\overline{BC}}:}$

$\sf{m = \dfrac{\Delta_Y}{\Delta_X} = \dfrac{y_C - y_B}{x_C - x_B} = \dfrac{-7-(-3)}{2 - (-1)} = \dfrac{-7 + 3}{2 + 1} = -\dfrac{4}{3}}$

$\mathsf{y - y_0 = m(x - x_0)}$

$\mathsf{y - (-7) = -\dfrac{4}{3}(x - 2)}$

$\mathsf{3y + 21 = -4x + 8}$

$\sf{r:4x + 3y + 13 = 0}$

$\textsf{reta perpendicular a }\sf{\overline{BC}}\textsf{ passando por A:}$

$\mathsf{m_1.m_2 = -1}$

$\mathsf{m_1.\left(-\dfrac{4}{3}\right) = -1}$

$\sf{m = \dfrac{3}{4}}$

$\mathsf{y - 1 = \dfrac{3}{4}(x - 1)}$

$\mathsf{4y - 4 = 3x - 3}$

$\mathsf{s:3x - 4y + 1 = 0}$

$\textsf{ponto de concorr{\^e}ncia entre as retas r e s:}$

$\begin{cases}\sf{4x + 3y = -13}\\\sf{3x - 4y = -1}\end{cases}$

$\begin{cases}\sf{12x + 9y = -39}\\\sf{-12x + 16y = 4}\end{cases}$

$\mathsf{25y = -35}$

$\mathsf{y = -\dfrac{7}{5}}$

$\mathsf{4x + 3\left(-\dfrac{7}{5}\right) = -13}$

$\mathsf{20x - 21 = -65}$

$\mathsf{20x = -44}$

$\mathsf{x = -\dfrac{11}{5}}$

$\mathsf{H\left(-\dfrac{11}{5};\:-\dfrac{7}{5}\right)}$

$\textsf{calcular a dist{\^a}ncia entre A e H:}$

$\sf{d_{AH} = \sqrt{(x_A - x_H)^2 + (y_A - y_H)^2}}}$

$\sf{d_{AH} = \sqrt{(1 - (-11/5))^2 + (1 - (-7/5))^2}}}$

$\sf{d_{AH} = \sqrt{(1 + (11/5))^2 + (1 + (7/5))^2}}}$

$\sf{d_{AH} = \sqrt{(16/5)^2 + (12/5)^2}}}$

$\sf{d_{AH} = \sqrt{(256/25) + (144/25)}}}$

$\sf{d_{AH} = \sqrt{(400/25)}$

$\sf{d_{AH} = \sqrt{16}$

$\boxed{\boxed{\sf{d_{AH} = 4}}}\leftarrow\textsf{altura relativa ao v{\'e}rtice A}$