Question

Determine a equação da bissetriz interna, por A, no triângulo de

vértices A(2, 2), B(−1, −2) e C(−10, −3).

Answer 1

Resposta:

$\textsf{Leia abaixo}$

Explicação passo a passo:

$\textsf{v{\'e}rtices do tri{\^a}ngulo}$

$\sf{A(2,2) \Leftrightarrow B(-1,-2) \Leftrightarrow C(-10,-3)}$

$\sf{r: reta\:suporte\:\overline{\rm AB} }$

$\sf{m = \dfrac{\Delta_Y}{\Delta_X} = \dfrac{y_B - y_A}{x_B - x_A} = \dfrac{-2-2}{-1 - 2} = \dfrac{-4}{-3} = \dfrac{4}{3}}$

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

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

$\sf{3y - 6 = 4x - 8}$

$\sf{r:4x - 3y - 2 = 0}$

$\sf{s: reta\:suporte\:\overline{\rm AC} }$

$\sf{m = \dfrac{\Delta_Y}{\Delta_X} = \dfrac{y_C - y_A}{x_C - x_A} = \dfrac{-3 - 2}{-10 - 2} = \dfrac{-5}{-12} = \dfrac{5}{12}}$

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

$\sf{y - 2 = \dfrac{5}{12}(x - 2)}$

$\sf{12y - 24 = 5x - 10}$

$\sf{s:5x - 12y + 14 = 0}$

$\sf{d_{P,R} = |\:\dfrac{ax_0 + by_0 + c}{\sqrt{a^2 + b^2}}\:|}$

$\sf{\dfrac{4x - 3y - 2}{\sqrt{4^2 + (-3)^2}} + \dfrac{5x - 12y + 14}{\sqrt{5^2 + (-12)^2}} = 0}$

$\sf{\dfrac{4x - 3y - 2}{\sqrt{16 + 9}} + \dfrac{5x - 12y + 14}{\sqrt{25 + 144}} = 0}$

$\sf{\dfrac{4x - 3y - 2}{\sqrt{25}} + \dfrac{5x - 12y + 14}{\sqrt{169}} = 0}$

$\sf{\dfrac{4x - 3y - 2}{5} + \dfrac{5x - 12y + 14}{13} = 0}$

$\sf{13(4x - 3y - 2) + 5(5x - 12y + 14) = 0}$

$\sf{52x - 39y - 26 + 25x - 60y + 70 = 0}$

$\sf{77x - 99y + 44= 0}$

$\boxed{\boxed{\sf{7x - 9y + 4= 0}}}$