Question

1) determinar a equação de uma reta r que passa pelo ponto A (-1,4)e tem coeficiente angular 2.
2) determinar a equação da reta que passa pelo ponto p(2,5)e tem uma inclinação de 60°
3)os pontos A(1,2),B(3,1) e C (2,4)sao vértices de um triângulo. Determinar as equações das retas suportes dos lados desse triângulo. ​

Answer 1

$\Large\boxed{\begin{array}{l}\rm 1)~A(-1,4)\implies x_0=-1~~y_0=4\\\sf m=2\\\sf y=y_0+m\cdot(x-x_0)\\\sf y=4+2\cdot(x-[-1])\\\sf y=4+2\cdot(x+1)\\\sf y=4+2x+2\\\sf y=2x+6\\\rm 2)~P(2,5)\implies x_0=2~~y_0=5\\\sf \alpha=60^\circ\\\sf m=tg(\alpha)\\\sf m=tg(60^\circ)\\\sf y=\sqrt{3}\\\sf y=y_0+m\cdot(x-x_0)\\\sf y=5+\sqrt{3}\cdot(x-2)\\\sf y=5+\sqrt{3}x-2\sqrt{3}\\\sf y=\sqrt{3}x+5-2\sqrt{3}\end{array}}$

$\Large\boxed{\begin{array}{l}\rm3)\\\underline{\rm Equac_{\!\!,}\tilde ao~da~reta~suporte}\\\underline{\rm da~aresta~AB}\\\sf\begin{vmatrix}\sf 1&\sf2&\sf1\\\sf3&\sf1&\sf1\\\sf x&\sf y&\sf1\end{vmatrix}=0\\\sf 1\cdot(1-y)-2\cdot(3-x)+1\cdot(3y-x)=0\\\sf 1-y-6+2x+3y-x=0\\\sf x+2y-5=0\end{array}}$

$\Large\boxed{\begin{array}{l}\underline{\rm Equac_{\!\!,}\tilde ao~da~reta~suporte}\\\underline{\rm da~aresta~AC}\\\sf\begin{vmatrix}\sf1&\sf2&\sf1\\\sf2&\sf4&\sf1\\\sf x&\sf y &\sf1\end{vmatrix}=0\\\sf 1\cdot(4-y)-2\cdot(2-x)+1\cdot(2y-4x)=0\\\sf \backslash\!\!\!4-y-\backslash\!\!\!4+2x+2y-4x=0\\\sf -2x+y=0\end{array}}$

$\Large\boxed{\begin{array}{l}\underline{\rm Equac_{\!\!,}\tilde ao~da~reta~suporte}\\\underline{\rm da~aresta~BC}\\\sf\begin{vmatrix}\sf3&\sf1&\sf1\\\sf2&\sf4&\sf1\\\sf x&\sf y&\sf1\end{vmatrix}=0\\\sf 3\cdot(4-y)-1\cdot(2-x)+1\cdot(2y-4x)=0\\\sf 12-3y-2+x+2y-4x=0\\\sf -3x-y+10=0\cdot(-1)\\\sf 3x+y-10=0\end{array}}$