Question

As retas r: x + 3y – 11 = 0; s: 5x + 6y – 19 = 0 e t: 4x + 3y – 17 = 0 são as retas suportes dos lados de um triângulo. Determine os vértices dos triângulos.
COM URGÊNCIA!!!!

Answer 1

$\Large\boxed{\begin{array}{l}\begin{cases}\sf r:x+3y-11=0\\\sf s:5x+6y-19=0\\\sf t:4x+3y-17=0\end{cases}\end{array}}$

$\large\boxed{\begin{array}{l}\underline{\rm intersecc_{\!\!,}\tilde ao\,das\,retas\,r\,e\,s}\\\sf x+3y-11=0\implies x+3y=11\\\sf 5x+6y-19=0\\\sf 5x+6y=19\\\begin{cases}\sf x+3y=11\cdot(-2)\\\sf5x+6y=19\end{cases}\\\\+\underline{\begin{cases}\sf-2x-\diagdown\!\!\!\!\!6y=-22\\\sf5x+\diagdown\!\!\!\!\!\!6y=19\end{cases}}\\\sf 3x=-3\\\sf x=-\dfrac{3}{3}\\\\\sf x=-1\\\sf x+3y=11\\\sf -1+3y=11\\\sf 3y=11+1\\\sf 3y=12\\\sf y=\dfrac{12}{3}\\\\\sf y=4\\\sf A(-1,4)\end{array}}$

$\large\boxed{\begin{array}{l}\underline{\rm intersecc_{\!\!,}\tilde ao\,das\,retas\,r\,e\,t}\\\sf x+3y-11=0\implies x+3y=11\\\sf 4x+3y-17=0\implies 4x+3y=17\\\begin{cases}\sf x+3y=11\cdot(-1)\\\sf 4x+3y=17\end{cases}\\\\+\underline{\begin{cases}\sf -x-\diagdown\!\!\!\!\!3y=-11\\\sf4x+\diagdown\!\!\!\!\!3y=17\end{cases}}\\\sf 3x=6\\\sf x=\dfrac{6}{3}\\\\\sf x=2\\\sf x+3y=11\\\sf 2+3y=11\\\sf 3y=11-2\\\sf 3y=9\\\sf y=\dfrac{9}{3}\\\\\sf y=3\\\sf B(2,3)\end{array}}$

$\boxed{\begin{array}{l}\underline{\rm Intersecc_{\!\!,}\tilde ao\,das\,retas\,s\,e\,t}\\\sf 5x+6y-19=0\implies 5x+6y=19\\\sf 4x+3y-17=0\implies 4x+3y=17\\\begin{cases}\sf 5x+6y=19\\\sf 4x+3y=17\cdot(-2)\end{cases}\\\\+\underline{\begin{cases}\sf 5x+\diagdown\!\!\!\!\!6y=19\\\sf -8x-\diagdown\!\!\!\!\!6y=-34\end{cases}}\\\sf -3x=-15\cdot(-1)\\\sf 3x=15\\\sf x=\dfrac{15}{3}\\\\\sf x=5\\\sf 4x+3y=17\\\sf 4\cdot5+3y=17\\\sf 20+3y=17\\\sf 3y=17-20\\\sf 3y=-3\\\sf y=-\dfrac{3}{3}=-1\\\\\sf C(5,-1)\end{array}}$

$\large\boxed{\begin{array}{l}\sf Os\,v\acute ertices\,do\,tri\hat angulo\,s\tilde ao\\\sf A(-1,4),B(2,3)~e~C(5,-1)\end{array}}$