Question

Qual é a altura do triângulo ABC, relativa ao lado BC,

sendo A(7, 9), B(–1, –3) e C(3, –2)?

Answer 1

Para achar a altura relativa ao lado BC Basta fazermos a distância do ponto A até a reta BC

Coeficiente angular da reta BC :

$\displaystyle \text m =\frac{-2-(-3)}{3-(-1)} \to \boxed{\text m = \frac{1}{4}}$

Reta BC passando pelo ponto (-1,-3)

$\displaystyle \text y-\text y_\text o=\text m(\text x-\text x_\text o)\\\\ \text y -(-3)=\frac{1}{4}(\text x -(-1)) \\\\ \text y +3=\frac{\text x+1}{4} \\\\ 4\text y+12=\text x+1 \\\\\\ \underline{\text{Reta BC}} :\\\\ \text x -4\text y-11=0$

Distância do ponto A até a reta BC :

$\displaystyle \text D = \frac{|\text{a.x}_\text o+\text{b.y}_\text o+\text c|}{\sqrt{\text a^2+\text b^2}} \\\\\\\ \text{a = 1 , b = -4 , c = -11 } \\\\ \text x_o = 7 \ , \ \text y_\text o = 9 \\\\\\ \text D = \frac{|1.7-4.9-11|}{\sqrt{1^2+(-4)^2}} \\\\\\ \text D = \frac{|7-36-11|}{\sqrt{1+16}} \\\\\\ \text D = \frac{|-40|}{\sqrt{17}} \\\\\\ \text D = \frac{40}{\sqrt{17}}.\frac{\sqrt{17}}{\sqrt{17}} \\\\\\ \huge\boxed{\text D = \frac{40\sqrt{17}}{17}\ }\checkmark$