Question

Um triângulo tem vértices P = (2,1), Q= (2,5) e R= (x, 4). Sabendo-se que a área do triângulo é 20, calcule a abscissa do ponto R.

Answer 1

Utilizando o método do determinante, temos:

$Area~=~\frac{1}{2}~.~\left|\left|\begin{array}{ccc}x_P&y_P&1\\x_Q&y_Q&1\\x_R&y_R&1\end{array}\right|\right|\\\\\\\\20~=~\frac{1}{2}~.~\left|\left|\begin{array}{ccc}2&1&1\\2&5&1\\x&4&1\end{array}\right|\right|\\\\\\2~.~20~=~\left|~(~2.5.1+2.4.1+x.1.1~)~-~(~1.5.x+1.4.2+1.1.2~)~\right|\\\\\\40~=~\left|~(~10+8+x~)~-~(~5x+8+2~)~\right|\\\\\\40~=~\left|~10+8+x-5x-8-2~\right|\\\\\\40~=~\left|~8-4x~\right|$

$40~=~\left|~8-4x~\right|\\\\\\Resolvendo~a~equacao~modular:\\\\\\40~=~8-4x\\\\4x~=~8-40\\\\x~=~\frac{-32}{4}\\\\\boxed{x~=~-8}\\\\\\-40~=~8-4x\\\\4x~=~8+40\\\\x~=~\frac{48}{4}\\\\\boxed{x~=~12}$

Resposta: A abscissa de R pode ser x = -8  ou  x = 12.