Question

Determinar a altura do triângulo ABC,sendo A(3,2), B(1,-3), C(-4,-1).
Altura é relativa do vértice A.

Answer 1

Vamos começar calculando a reta suporte do lado BC utilizando o método do determinante:

$\left|\begin{array}{ccc}1&-3&1\\-4&-1&1\\x&y&1\end{array}\right|=0\\\\\\(~1.(-1).1+(-4).y.1+x.(-3).1~)~-~(~1.(-1).x+1.y.1+1.(-3).(-4)~)~=~0\\\\\\(~-1-4y-3x~)~-~(~-x+y+12~ )~=~0\\\\\\-1-4y-3x+x-y-12~=~0\\\\\\-2x-5y-13~=~0\\\\\\\boxed{2x+5y+13~=~0}$

Agora podemos utilizar a formulação da distancia de um ponto a uma reta:

$Distancia_{_{Ponto,Reta}}~=~\frac{\left|a.x_o+b.y_o+c\right|}{\sqrt{a^2+b^2}}\\\\\\onde~''a'',~ ''b''~e~"c" sao~os~coeficientes~da~reta~suporte~e~o~ponto\\(x_o,y_o)~\acute{e}~o~vertice~oposto~a~reta~suporte.\\\\Substituindo~os~dados:\\\\\\Distancia_{_{A,\overline{BC}}}~=~\frac{\left|2~.~(3)~+~5~.~(2)~+~13\right|}{\sqrt{2^2+5^2}}\\\\\\Distancia_{_{A,\overline{BC}}}~=~\frac{\left|6~+~10~+~13\right|}{\sqrt{4+25}}\\\\\\Distancia_{_{A,\overline{BC}}}~=~\frac{\left|29\right|}{\sqrt{29}}$

$Distancia_{_{A,\overline{BC}}}~=~\frac{29}{\sqrt{29}}\\\\\\Racionalizando\\\\Distancia_{_{A,\overline{BC}}}~=~\frac{29}{\sqrt{29}}.\frac{\sqrt{29}}{\sqrt{29}}\\\\\\Distancia_{_{A,\overline{BC}}}~=~\frac{29.\sqrt{29}}{\sqrt{29}^2}\\\\\\\boxed{Distancia_{_{A,\overline{BC}}}~=~\sqrt{29}}$